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Solid States Devices => solid state devices => Topic started by: tinman on May 08, 2016, 10:42:41 AM

I have started this thread to solve MileHighs question below.
Please keep the insults down,and the language clean.
MH's question.
You have an ideal voltage source and an ideal coil of 5 Henrys. At time t=0 seconds the coil connects to the ideal voltage source. For three seconds the voltage is 4 volts. Then for the next two seconds the voltage is zero volts. Then for two seconds the voltage is negative three volts, and then for the next six seconds the voltage is 0.5 volts. Then after that the voltage is zero volts.
What happens from T=0 when the ideal voltage is connected to the ideal coil?.
My answer to this question isyou cannot place an ideal voltage across an ideal inductor.
The reason being,at T=0,when the ideal voltage is placed across the ideal inductor,the current would rise instantly to a value of infinity. The reason this cannot happen,is because an ideal inductor dose not dissipate any power in the form of heat,due to the fact that it has no resistance or hysteresis loss,as it is an ideal inductor. If an ideal voltage was placed across an ideal inductor(in theory),it would result in an explosion the likes the universe has not seen since the creation of itthe big bang all over again.
All are welcome to have a go at answering MHs question.
MH
For the record,could you please post your answer to your question above?
Brad

Brad:
Your answer is wrong and I already answered a more difficult version of the question on the other thread and I already told you I won't answer this question.
You are one strange egg Brad because you think you are "running the show" now but in fact the show is running you.
Re: This posting from the other thread: http://overunity.com/8341/joulethief101/msg483350/#msg483350 (http://overunity.com/8341/joulethief101/msg483350/#msg483350)
As I already stated, you make reference to an "R" resistance but there is none in the example being discussed. You also make reference to a dividebyzero for a time constant and state that it is "instantaneous" when it is infinity or undefined. You are told these two things are wrong in a later posting but like the Pope you pretend that you are infallible and can just "waive" past these two glaring errors because you are Brad.
Re: This posting from the other thread: http://overunity.com/8341/joulethief101/msg483376/#msg483376 (http://overunity.com/8341/joulethief101/msg483376/#msg483376)
You are all in a tizzy because I mention a "'trick' question." But unfortunately you have another case of crossed signals in your head. It's not my question in that posting that is the "trick," it's actually ION's followup question to my question that is the "trick." I say in jest to ION that his question is a "trick" question but that flies right over your head. Lo and behold, the signals get crossed in your head and you accuse me of asking a "trick" question when I did no such thing.
But the most mindblowing thing about that posting of yours is that you completely gloss over the extremely important and relevant technical information that is contained in that posting.
For me, there are only two outstanding issues and I will mention them again and I will put them in a better sequence this time:
1. Brad gets up the learning curve and understands the original question and then answers it correctly all by himself and clearly demonstrates that he understands what he is doing.
2. Brad admits that he is wrong when he stated that my response to the harder question is wrong.
You are stating that you want to try to answer the question. That's a good thing and your first try is wrong, good luck.
MileHigh

author=MileHigh link=topic=16589.msg483392#msg483392 date=1462705232]
MileHigh
Your answer is wrong and I already answered a more difficult version of the question on the other thread and I already told you I won't answer this question.
OK
If my answer is wrong,then it is now up to you to prove that by supplying the correct answer.
I have given my answer to your question,and you cannot just say it's wrong without being able to support that claim by supplying what you believe to be the correct answer.
We are only discussing the original question here ATM.
As I already stated, you make reference to an "R" resistance but there is none in the example being discussed.
MH
My reference to resistance is a value of 0 ohms.
Is not 0 ohms resistance no resistance? Your statement makes no sense ,as i do not state any resistance ,as 0 is none. I only say that,so as others reading the question understand that an ideal coil has 0 ohms of resistanceno resistance.
You are one strange egg Brad because you think you are "running the show" now but in fact the show is running you.
I am providing an answer to your question,and now you have to disprove my answer by way of supplying the one you think is correct.
You also make reference to a dividebyzero for a time constant and state that it is "instantaneous" when it is infinity or undefined. You are told these two things are wrong in a later posting but like the Pope you pretend that you are infallible and can just "waive" past these two glaring errors because you are Brad.
I am told i am wrong by who MH?
You are all in a tizzy because I mention a "'trick' question." But unfortunately you have another case of crossed signals in your head. It's not my question in that posting that is the "trick," it's actually ION's followup question to my question that is the "trick." I say in jest to ION that his question is a "trick" question but that flies right over your head. Lo and behold, the signals get crossed in your head and you accuse me of asking a "trick" question when I did no such thing.
OK,so lets say that your question above is not a !trick! question as i have stated.
So now i will ask you (as you think i have it wrong)to calculate the time from T=0 in your question,it takes the ideal inductor to reach it's maximum current flow value,once the ideal voltage of 4 volts is placed across that coil.
But the most mindblowing thing about that posting of yours is that you completely gloss over the extremely important and relevant technical information that is contained in that posting.
You mean IONs response to your question?,once again referring to an ideal voltage from an ideal power supply,being placed over an ideal inductor?.
Well let's leave ION out of this for a while,and let you answer the questions,as it is your question we are talking about here,and so you are the one that should be providing the required answers.
For me, there are only two outstanding issues and I will mention them again and I will put them in a better sequence this time:
1. Brad gets up the learning curve and understands the original question and then answers it correctly all by himself and clearly demonstrates that he understands what he is doing.
I have answered your question,and i stand by my answer.
2. Brad admits that he is wrong when he stated that my response to the harder question is wrong.
We are discussing your original question above only, in this thread at the moment.
You are stating that you want to try to answer the question. That's a good thing and your first try is wrong, good luck.
Regarding the question we are discussing in this thread(your original question to EMJ and Wattsup),i have answered the question correctly. If you believe i am wrong,then you must provide the evidence to show that.
So the questions i have for you MH are
1how is the current time constant calculated for your ideal inductor/
2What is the time taken for the current to rise to peak value from T=0,that moment when the ideal voltage of 4 volts is placed across the ideal inductor?.
Lets start with those two question's,and take it from there.
Brad

If there is no resistor in the schematic then you do not discuss an imaginary zeroohm resistor that you want to force into the schematic. It's a nonstarter.
I told you Brad to try to work it out with your peers or work it out by yourself. I am not going to supply you with the correct answer. If I did that you would balk anyway and fight over it for 100 postings. I am not doing that.
"You are wrong."
"Okay, now give me the right answer."
It's not going to work like that.
You should say, "Okay, I will discuss it with my peers and go do more research and learn more and improve my skills so that I can answer the question successfully by myself."
You clearly have no understanding at all about how an inductor actually works. That is the essence of the question. Why don't you start there?

If there is no resistor in the schematic then you do not discuss an imaginary zeroohm resistor that you want to force into the schematic. It's a nonstarter.
what do you call the opposing force within an conductor?
Would that opposition be from end to end and thus make that component value a resistance?
I told you Brad to try to work it out with your peers or work it out by yourself. I am not going to supply you with the correct answer. If I did that you would balk anyway and fight over it for 100 postings. I am not doing that.
"You are wrong."
"Okay, now give me the right answer."
It's not not going to work like that.
You should say, "Okay, I will discuss it with my peers and go do more research and learn more and improve my skills so that I can answer the question successfully by myself."
You clearly have no understanding at all about how an inductor actually works. That is the essence of the question. Why don't you start there?
MH,
You should say, "Okay, I will discuss it with my peers and go do more research and learn more and improve my skills so that I can answer the question successfully by myself."

Stop trolling Webby. Don't ruin a brand new thread.

If there is no resistor in the schematic then you do not discuss an imaginary zeroohm resistor that you want to force into the schematic. It's a nonstarter.
I told you Brad to try to work it out with your peers or work it out by yourself. I am not going to supply you with the correct answer. If I did that you would balk anyway and fight over it for 100 postings. I am not doing that.
"You are wrong."
"Okay, now give me the right answer."
It's not not going to work like that.
."
You clearly have no understanding at all about how an inductor actually works. That is the essence of the question. Why don't you start there?
So you are not here to debate or discuss your question,but only to leave it to me to do all the work.
So much for you insisting on debating things ::) ::)
You should say, "Okay, I will discuss it with my peers and go do more research and learn more and improve my skills so that I can answer the question successfully by myself
My skills are fine thank you MH.
My answer to your question above is correct,and as you have provided no answer other than to say mine is incorrect,we can safely say your answer is incorrect,as i believe mine is correct,and you disagree with my answer.
As you have not provided any proof that my answer is incorrect,nor have you attempted to answer a couple of simple questions,we can assume that you cannot answer them,nor can you provide evidence that my answer is incorrect.
So as you do not wish to be a part of this debate,or help others find the answer to your question,i will post some answers for you.
Q1can a voltage exist across an ideal inductor that has a DC current flowing through it.
Your answer MH (if you know your stuff)will be no,a voltage cannot exist across an ideal inductor that has a DC current flowing through it.
Have you ticked this box in the pole question?seems not.
So now that you agree that a voltage cannot exist across an ideal inductor that has a DC current flowing through it,how can you have a voltage of 4 volts existing across an ideal inductor for 3 second'san inductor that has no resistance?.
Brad

<<< no,a voltage cannot exist across an ideal inductor that has a DC current flowing through it. >>>
And that's probably the only thing that you have stated about this subject so far that is correct.
I will repeat to you: I already answered a variation on the question that is actually more difficult to answer. I gave a complete and full answer. It's up to you and your peers to try to answer the simpler question if you want to.
I am just waiting and hoping for a successful conclusion. I will repeat: You clearly have no understanding at all about how an inductor actually works. That is the essence of the question. Why don't you start there?
If you don't get a guru parachuting in to help you and you and your peers are unable to answer a question about one of the simplest circuits possible, so be it.
These are the only two things I am interested in hopefully seeing a successful resolution to:
1. Brad gets up the learning curve and understands the original question and then answers it correctly all by himself and clearly demonstrates that he understands what he is doing.
2. Brad admits that he is wrong when he stated that my response to the harder question is wrong.
MileHigh

MH,
does the question ask for when T=0
It does not start when T>0 the question is for when T=0

<<< no,a voltage cannot exist across an ideal inductor that has a DC current flowing through it. >>>
And that's probably the only thing that you have stated about this subject so far that is correct.
I will repeat to you: I already answered a variation on the question that is actually more difficult to answer. I gave a complete and full answer. It's up to you and your peers to try to answer the simpler question if you want to.
I am just waiting and hoping for a successful conclusion. I will repeat: You clearly have no understanding at all about how an inductor actually works. That is the essence of the question. Why don't you start there?
If you don't get a guru parachuting in to help you and you and your peers are unable to answer a question about one of the simplest circuits possible, so be it.
These are the only two things I am interested in hopefully seeing a successful resolution to:
1. Brad gets up the learning curve and understands the original question and then answers it correctly all by himself and clearly demonstrates that he understands what he is doing.
2. Brad admits that he is wrong when he stated that my response to the harder question is wrong.
MileHigh
MH
I would suggest at this point in time,you review your question,and not treat your inductor as a normal inductor that !dose! have a series/parallel resistance.
You clearly defined the inductor and voltage as being !ideal!.
You said your question was not a trick question,so i think you need to go and calculate the L/R time constant for your ideal inductorsome how ???
Once you have that L/R time constant value,we then only need multiply that by 5 to get a close approximation for the time it takes that 5 henry ideal inductor to reach it's steady state current value after T=0>the ideal voltage is applied across the ideal inductor.
All i ask is that you calculate the L/R time constant of your 5 henry ideal inductor.
Brad.
How are you to work out the L/R time constant when there is no R ???

MH,
does the question ask for when T=0
It does not start when T>0 the question is for when T=0
T=0 is the start of the cycle > the instant the ideal voltage is placed across the ideal inductor
Brad

You are lost Brad and you should take my advice and start from scratch. And like usual you are completely baffling because some big clues were presented to you and they passed right through you like you weren't even there. I am giving you a 10% chance to get it right.

A post from Magneticitist on the JT 101 thread
What I find curious about the whole 'Ideal' scenario no matter how useful it may be in real world applications.. An Ideal inductor is said to dissipate or radiate 0 energy, so technically it only passes current by putting faith in Ohms law's ability to handle the number 0.
I would think that in an ideal voltage source, connecting to an ideal inductor, nothing at all would happen because the voltage cannot waver, and the lack of resistance in the inductor would cause an infinite current were it not for an ideal inductor being unable to dissipate energy. If it cannot dissipate energy it perfectly contains on faith, we cannot possibly observe this energy and it might as well be at rest with no charge.
So even in the fantasy realm of imaginary voltage sources and coils that are ideal, an inductor can do no work unless it actually becomes something we cannot call 'ideal'. further evidence against this paradox of passing infinite current at 0 resistance.
MHs response to some one posting there thought'sand good ones at that.
I am just letting you know as a courtesy that you clearly have no idea what you are talking about.
So now myself,EMJ,Wattsup and Magneticitist are wrong,and MH is yet to post an answer to his own question,nor seems that he is willing to provide any information or answers to other simple questions asked of him.
Brad

You are lost Brad and you should take my advice and start from scratch. And like usual you are completely baffling because some big clues were presented to you and they passed right through you like you weren't even there. I am giving you a 10% chance to get it right.
Quote:
The time constant τ is an indicator of how long current takes to increase from zero to its steadystate value.
Here is a useful rule of thumb:
For most practical purposes, we may assume that all quantities in a DC RL circuit have reached their steadystate values after five time constants.
So if a circuit has a time constant of 1 millisecond, then it will take about 5 milliseconds for the circuit's currents and voltages to reach their steadystate values.
Since one time constant is equal to L÷R, we can write this rule of thumb as an equation:
Time to reach steady state ≈ 5×L÷R
So MH,if your !very large! 5 henry coil had a resistance value of say 5 ohms,then it would take 1 second to reach it's time constant,and 5 seconds to reach a steady state current flow.
If your 5 henry coil had just .0001 ohms resistance,then it would take 50,000 seconds to reach it's time constant,and 250,000 seconds to reach a steady state of current flow.
As your coil is ideal,it has no resistance.
So i ask againhow are you going to calculate the time constant of your ideal coil>?
Brad

The answer is that there is no time constant. A variation on the same question was already answered.
That's it, from this point on you can moan and groan and whine and complain all you want. A better scenario is you tasking yourself with this:
1. Brad gets up the learning curve and understands the original question and then answers it correctly all by himself and clearly demonstrates that he understands what he is doing.
If you try and make a good sincere effort of it but don't get there, then good show. If you get there then great show. But if you just whine for an answer and sit on your fanny and complain then you are going to have to hope that a guru comes along and spoon feeds you. But of course if you are spoon fed then chances are in two weeks you will be a blank slate.
It's up to you.

In an ideal voltage source the source Emf would be fixed and an ideal inductor would have virtually no losses. It seems to me no current could flow because the moment a charge tried to moved due to the ideal voltage source Emf the ideal inductor would produce an equal and opposite Cemf to oppose it. Ideally if the source Emf is always instantaneously opposed by the inductors Cemf then nothing can move, a stalemate.
AC

@AllCanadian seems a reasonable deduction that follows logic. welcome to the club of having no idea.

@Magneticists
seems a reasonable deduction that follows logic. welcome to the club of having no idea.
Logically, the fact that I expressed my idea here indicates I must have one and the presumption that just because my idea may not agree with yours that it's incorrect is false. However I have no indication that you have any idea which is why grown ups need to use "their words" and convey what they mean in an intelligent manner... such as technical debate on the issue at hand.
AC

Brad,
Here are two small but significant hints:
1) brush up on what it means mathematically when any number is divided by 0 (don't assume you are correct, verify it).
2) MH's question is regarding what happens at t=0, i.e. the instant the Vsource is connected (MH, please confirm).

@Magneticists
Logically, the fact that I expressed my idea here indicates I must have one and the presumption that just because my idea may not agree with yours that it's incorrect is false. However I have no indication that you have any idea which is why grown ups need to use "their words" and convey what they mean in an intelligent manner... such as technical debate on the issue at hand.
AC
Well said

Poynt:
Here is the question again:
<<<
You have an ideal voltage source and an ideal coil of 5 Henrys. At time t=0 seconds the coil connects to the ideal voltage source. For three seconds the voltage is 4 volts. Then for the next two seconds the voltage is zero volts. Then for two seconds the voltage is negative three volts, and then for the next six seconds the voltage is 0.5 volts. Then after that the voltage is zero volts.
What happens starting from t=0.
>>>
So the time frame is from 0 seconds to "beyond 13 seconds."
MileHigh

Brad,
Here are two small but significant hints:
1) brush up on what it means mathematically when any number is divided by 0 (don't assume you are correct, verify it).
2) MH's question is regarding what happens at t=0, i.e. the instant the Vsource is connected (MH, please confirm).
OK look I have to ask you out of genuine respect for this conversation.. How did you come to ask Brad those questions? In the beginning of this thread he stated:
"My answer to this question isyou cannot place an ideal voltage across an ideal inductor.
The reason being,at T=0,when the ideal voltage is placed across the ideal inductor,the current would rise instantly to a value of infinity."
with his own words he said he understands that MH's T=0 was in reference to the initial connection of the vsource.
the issue is an ideal source being placed across 0 resistance. whether it's a coil or straight wire doesn't really matter in this context.
to answer the question of what happens the very instant the ideal vsource is connected to the ideal coil?
T=0 THE VOLTAGE CONNECTS TO THE COIL, THAT'S WHAT HAPPENS.

Look at this posting: http://overunity.com/8341/joulethief101/msg483361/#msg483361

Thanks MH for correcting me.
Although the question involves t=0 to t=13s, knowing what happens right at t=0 (the moment Vsource connects the inductor) is important.
It's also important to know what results when a number is divided by 0.
Let's see if examining this causes Brad to come to a different answer.

Somebody has to go back to basics, it might as well be me.
The definition of an ideal inductor is a twoterminal device that obeys the current/voltage relationship:
V = L dI/dt where V is the applied voltage, L is the inductance in Henries and dI/dt is the rate of change of the current with time. The impedance of the voltage source and the resistance of the inductor are both assumed to be zero.
This definition can be rewritten as dI.dt = V/L. Given an initial current of zero, applying 4 V to a 5 H inductor leads to a current through the inductor that increases at 0.8 amps per second. After three seconds the current will be 2.4 amps. If the supply is not turned off the current will increase indefinitely at 0.8 A/second. There are no time constants involved.
For completeness, a capacitor is a twoterminal device the current through which is given by the equation:
I = C dV/dt where C is the capacitance in Farads. The current through the capacitor is proportion to the rate of change in the applied voltage times the capacitance. In this case applying a fixed current results in a voltage across the capacitor that increases indefinitely.

Thanks MH for correcting me.
Although the question involves t=0 to t=13s, knowing what happens right at t=0 (the moment Vsource connects the inductor) is important.
It's also important to know what results when a number is divided by 0.
Let's see if examining this causes Brad to come to a different answer.
I just googled a few math sites and they all said that you can not divide any number by 0. It is impossible and incorrect to do so.
So, now my question is...why is that? One site said that you would get infinity by doing this which is why you can't do this. Gee, and I thought I understood simple math, ha ha.
The intuitive answer, at least for me, would have been that division means breaking something into a certain number of equal parts and if there are no divisions into equal parts, then you are left with the original amount undivided. But, this fails the multiplication test when you multiply the answer to double check it....
10 divided by 0 = 10. Then 10 times 0 should equal 10, and of course it does not. So, 10 divided by 0 either equals infinity, 0 or, it is undefined. Some say it can't be done. All of this depends upon which math sites you want to believe.
Bill

In an ideal voltage source the source Emf would be fixed and an ideal inductor would have virtually no losses. It seems to me no current could flow because the moment a charge tried to moved due to the ideal voltage source Emf the ideal inductor would produce an equal and opposite Cemf to oppose it. Ideally if the source Emf is always instantaneously opposed by the inductors Cemf then nothing can move, a stalemate.
AC
Lol. I said that a few times on this site. Just did yesterday. ;) ;)
Like are we taking about an ideal inductor that has no magnetic fields? Like I said yesterday, it could be just the field charge before the ideal switch connects that could set up an ideal BEMF standoff and no current would ever flow, inductor or even a straight wire Im glad someone here is on the same level of thinking as I am on this. ;) ;)
Mags

I just googled a few math sites and they all said that you can not divide any number by 0. It is impossible and incorrect to do so.
So, now my question is...why is that? One site said that you would get infinity by doing this which is why you can't do this. Gee, and I thought I understood simple math, ha ha.
The intuitive answer, at least for me, would have been that division means breaking something into a certain number of equal parts and if there are no divisions into equal parts, then you are left with the original amount undivided. But, this fails the multiplication test when you multiply the answer to double check it....
10 divided by 0 = 10. Then 10 times 0 should equal 10, and of course it does not. So, 10 divided by 0 either equals infinity, 0 or, it is undefined. Some say it can't be done. All of this depends upon which math sites you want to believe.
Bill
Maybe we wouldnt have to worry with the 0 value for resistance, where it may be replaced by a time of electron travel over a distance. Can an electron move faster than the speed of light? Is the speed of light infinity or instantaineous? So there would still be a time thing to work with as I dont think T=0 can even exist. It means no time, nothing happened because that time didnt exist. ;)
Mags

Thanks MH for correcting me.
Although the question involves t=0 to t=13s, knowing what happens right at t=0 (the moment Vsource connects the inductor) is important.
It's also important to know what results when a number is divided by 0.
Let's see if examining this causes Brad to come to a different answer.
When you divide one number !x! by another number !y!,it is asking you to work out how many !y's! will fit into !x!,so how many 0's in 10?.
So to work out the L/R time constant,when the value of R is 0,then the answer is infinity,and so inductance in this case with the inductor being ideal, dose not play a part in current rise time. That being the case,you are placing an ideal voltage across a perfect dead short.
All L/R time constants are based around the ideal inductor where the resistance value is added in series with the inductor ,by way of a resistor that is external to the inductor. As the current rises,the voltage across the inductor drop's until it reaches a value of 0 volts. The problem here is that we have an ideal voltage that dose not drop,and there for the current will continue to rise to an infinite amount. If we remove the series resistor from our ideal inductor,we have a dead short due to the L/R time constant being infinite,and playing no part in current rise time.
Like MH saidthere is !no! time constant,as it is infinite.
And so we now have an ideal voltage>a voltage that will not change,being placed across a short that will not allow a voltage potential to exist across it.
So i stand by my answer>you cannot place an ideal voltage across an ideal inductor.
If you did(theoretically),the current would rise instantly to an infinite value.
Brad

When you divide one number !x! by another number !y!,it is asking you to work out how many !y's! will fit into !x!,so how many 0's in 10?.
So to work out the L/R time constant,when the value of R is 0,then the answer is infinity,and so inductance in this case with the inductor being ideal, dose not play a part in current rise time. That being the case,you are placing an ideal voltage across a perfect dead short.
All L/R time constants are based around the ideal inductor where the resistance value is added in series with the inductor ,by way of a resistor that is external to the inductor. As the current rises,the voltage across the inductor drop's until it reaches a value of 0 volts. The problem here is that we have an ideal voltage that dose not drop,and there for the current will continue to rise to an infinite amount. If we remove the series resistor from our ideal inductor,we have a dead short due to the L/R time constant being infinite,and playing no part in current rise time.
Like MH saidthere is !no! time constant,as it is infinite.
And so we now have an ideal voltage>a voltage that will not change,being placed across a short that will not allow a voltage potential to exist across it.
So i stand by my answer>you cannot place an ideal voltage across an ideal inductor.
If you did(theoretically),the current would rise instantly to an infinite value.
Brad
You may wish to exercise a few examples to see if your conclusion is correct. Your conclusion being that if the time constant (Tau) is infinite, the load immediately presents itself as a perfect short, meaning the current will be infinite and instantaneous.
Here are a few: (in all cases, L=6H)
1) R=1, Tau=6s
2) R=0.1, Tau=60s (1min)
3) R=1m Ohm, Tau=60ks (16.6 hours)
4) R=1u Ohm, Tau=6M(million)s (1667 hours)
5) etc.
What is happening to Tau as R decreases?
If R could be 0, Tau must be infinite.
What happens to the inductor current after t=0 when Tau=infinity?

then we have the extra confusing idea that whether a real coil or ideal coil, at T=0 the current is at absolute 0 because there is absolutely no resistance to yield a voltage drop, it's at a maximum across the coil. there is absolutely no current flow to yield a magnetic field, because the magnetic field that will not be created without current is opposing the current flow. hmm.
a superconductive coil will yield no external field because it perfectly contains it within the conductor resisting current change, perfectly. with absolute 0 resistance we are approaching an infinite magnetic field force and thus an infinite reactance. since infinity cannot overpower infinity, they simply cancel each other out in practice.
resistance is the factor that prevents a conductor from perfectly resisting an exterior magnetic field because its the factor that resists current.. with 0 resistance there is perfect opposition like a super cooled conductor opposing a magnet. in essence, can't an absolute 0 resistance in a conductor/inductor be considered both a dead short AND open circuit at the same time?

then we have the extra confusing idea that whether a real coil or ideal coil, at T=0 the current is at absolute 0 because there is absolutely no resistance to yield a voltage drop, it's at a maximum across the coil. there is absolutely no current flow to yield a magnetic field, because the magnetic field that will not be created without current is opposing the current flow. hmm.
a superconductive coil will yield no external field because it perfectly contains it within the conductor resisting current change, perfectly. with absolute 0 resistance we are approaching an infinite magnetic field force and thus an infinite reactance. since infinity cannot overpower infinity, they simply cancel each other out in practice.
resistance is the factor that prevents a conductor from perfectly resisting an exterior magnetic field because its the factor that resists current.. with 0 resistance there is perfect opposition like a super cooled conductor opposing a magnet. in essence, can't an absolute 0 resistance in a conductor/inductor be considered both a dead short AND open circuit at the same time?
There ya go. ;) Resistance is a necessity for current to flow. ;D Lol ;D
Mags

There ya go. ;) Resistance is a necessity for current to flow. ;D Lol ;D
Mags
I have actually always believed this and it's hard for me to think otherwise.
I can't help it no matter how foolish it may seem.

In all this, we havnt considered a load to be included. If we have the ideal supply and ideal wires and just a normal dc motor, then what? The ideal supply produces 12v and the normal motor is rated at 12v. So if current flows, we could run those wires a very long distance with no loss. So if we had the ideal supply and ideal say twisted pair run out to say 10miles, would there be 12v potential available at the end of those wires and would the motor run? Now I might think that the twisted pair would act like a cap, so there should be potential at the end, 10 miles down the road.
lol, soo if we twisted our wire first and then connected it to the ideal supply, would there be the end of the universe event? And even if we ran 1 wire N 10 miles and the other S 10 miles to reduce the capacitance as much as possible(and lets idealize that it was done in an ideally empty universe ::) will it ever end) there would still be a capacity. So Boom I suppose. ::)
There must be a need for ideal current limiters?? ??? ::) ;) Gees. Im going to think about this stuff every time I see or hear the word 'Ideal' ::)
Mags

I have actually always believed this and it's hard for me to think otherwise.
I can't help it no matter how foolish it may seem.
I have said it before also, considering... ;) Im just lol that someone else finally said it. ;) ;D
Mags

So the questions i have for you MH are
1how is the current time constant calculated for your ideal inductor/
Like any other inductor.
2What is the time taken for the current to rise to peak value from T=0,that moment when the ideal voltage of 4 volts is placed across the ideal inductor?.
Indeed this question is germane to the original question, What is your answer?

the question has 5H , time secs and +/ volts given , and zero volt crossing
1H = current changing @ 1 A per SECOND resulting in emf of 1V across an inductor

@Magneticitist
I have actually always believed this and it's hard for me to think otherwise. [/size]I can't help it no matter how foolish it may seem.
Think of space, charges move producing a magnetic field and there is no resistance. Supposedly empty space conducts just like a vacuum tube thus in 99% of the universe a complete lack of resistance is in fact the norm. The only problem with understanding nature seems to be human nature and their odd preoccupation with math, numbers and equations... the proverbial rabbit hole.
AC

Like any other inductor.
Indeed this question is germane to the original question, What is your answer?
How do you calculate the L/R time constant, when there is no R
My answer has already been postedthe current rise is instantor at the speed of lightclose enough,as there is no resistance to appose the current flow.
You posted a comment on th JT thread not long ago Poynt,and it was @T=0,one would think the universe would blow up.
Brad

T=0 is The Big Bang.
Think about it. There was no time prior to that event so it had to be 0.
Bill

You may wish to exercise a few examples to see if your conclusion is correct. Your conclusion being that if the time constant (Tau) is infinite, the load immediately presents itself as a perfect short, meaning the current will be infinite and instantaneous.
Here are a few: (in all cases, L=6H)
1) R=1, Tau=6s
2) R=0.1, Tau=60s (1min)
3) R=1m Ohm, Tau=60ks (16.6 hours)
4) R=1u Ohm, Tau=6M(million)s (1667 hours)
5) etc.
What is happening to Tau as R decreases?
If R could be 0, Tau must be infinite.
What happens to the inductor current after t=0 when Tau=infinity?
As i said before,if Tau is infinite,there is no inductance,and so there is nothing to create the BEMF that apposes the applied EMF,and so now the applied EMF is across a dead short,as an ideal inductor has no resistance.
This results in an instant current rise to an infinite value.
Brad

T=0 is The Big Bang.
Think about it. There was no time prior to that event so it had to be 0.
Bill
The big bang at T=0 was the creation of the universe. Poynt said that the universe would blow up at T=0,meaning that the universe was already in existance.
Brad

Come on tinman, the answer's in the question.
John.

the time constant for the inductor is based upon the concept that an inductor can never actually reach infinity, but never actually has a 'maximum' throttle. it just starts to become too difficult to distinguish the differences between the mathematical volumes of natural logarithm after the 5th time constant.
this does not help the hypothetical non resistant coil situation supplied with an infinite current/voltage ratio.
I personally do not quite understand the initial question as it seems vague, or I simply am not interpreting it properly being that as Brad said these types of exercises usually involve a series resistance. MH has made it clear a resistance is not required to run the test but even if our coil reaches a discernible infinity limit in 20 seconds that still doesn't change the basic principle of the question Brad has an issue with, the general idea of a proposed ideal voltage source meeting a proposed 0 resistance conductor, and substituting R=0 with R=1.

Come on tinman, the answer's in the question.
John.
The answer is not in the question,as the question is based around an inductor that dose not exist,and a voltage source that also dose not existthat being the !!ideal! part of the question.
There for,the answer to this question will only be a theoretical answer,as the circuit cannot be tested to obtain the correct answer.
I dont think MH gave much thought to his question,or the outcome of installing the !ideal! parts to this !so called! simple circuit. As soon as you add things that do not exist,the question becomes far from simple,as there are no results obtained from such a circuit in test before.
It's much the same as those that theorize that anything that travels at the speed of light,will have infinite massa theory that !as yet! cannot be backed up by a practical test. Light travels at the speed of light,and it dose not have infinite mass :o.
So when i say that you cannot hook an ideal voltage across an ideal coil,you actually cannot do it,as both the ideal coil and ideal voltage do not exist,so i am technically correct.
Then comes the theoretical conundrum of trying to work out the question.
MH states that there is no time constant for current rise within the inductor,and i agree on this. But here is the conundrum.
The time constant is either 0 or infinite for maximum current flow through the inductor at T=0>the moment the ideal voltage is placed across the ideal coil.
If the time constant is 0,that means that the current through the ideal inductor will rise instantly to an infinite valueas we have an ideal voltage source across the ideal inductor.
If the time constant is infinite for maximum current through the ideal inductor,then that means that no current flows through the inductorever,because if it takes an infinite amount of time to reach maximum current flow through the coil,then it also takes an infinite amount of time to reach a 10% value of maximum current flow through the ideal inductor,and it takes an infinite amount of time to reach 1% of the maximum amount of current flow through the ideal inductorand so on. So it will take an infinite amount of time before current even starts to flowso there will never be any current flow through the ideal inductor.
Now here comes the conundrum part. The ideal inductor has no resistance,and so now our ideal voltage is placed across a dead short,and that means an infinite amount of current will flow instantly.
At T=0,an ideal voltage of 4 volts is placed across the ideal inductor,for a period of 3 second's.
Either way you look at it,this is something that cannot happenas i have stated many times now in my answer to the question.
The next problem is this.
I can claim my answer to be correct,and no one can disprove it,as the ideal inductor and ideal voltage source do not exist.
MH could say his answer is correct,and i could not prove it to be wrong,as i have no way of testing his theory.
MH stated that the question is not a trick question,but based on the fact that he is using an inductor that dose not exist,being supplied a voltage from a source that also dose not exist,how is any body to answer the question correctly,when such a circuit cannot be tested?.
Brad

Just answer the question.

Brad,
Here are two small but significant hints:
2) MH's question is regarding what happens at t=0, i.e. the instant the Vsource is connected (MH, please confirm).
1) brush up on what it means mathematically when any number is divided by 0 (don't assume you are correct, verify it).
The only time 0 is used as a divisional number(that im aware of),is in computer arithmetic. But even then,it must be assigned with either a + or value,and even then,the resultant value of the division is either negative infinity,or positive infinity.
There are no other cases i know of,or could find,where anything can be divided by 0,and result in a defined number or amount. Every attempt at explaining that a number remains unchanged when divided by zero failed.
Brad

Just answer the question.
I have answered the question many times.
Where are you reading?.
Brad

Just answer the question.
Here is an idea John.
Instead of posting pictures of your farm animals,and non related commentspost your answer to the question here on this thread :D
Brad

I think the confusion lies in the general ridiculousness of the intended point.
The question is apparently geared to asses your knowledge of circuit operation by seeing how institutionalized you are regarding EE fundamentals as they are taught in a classroom setting.
no matter the logic you provide, until you work out some mathematical answer in detail based upon the numbers given your answer will not be considered. If you worked out the math to please them it would be pointless because as you have already stated you are in disagreement with the general nature of the question.
it's kind of like walking up to a mechanic who can take an engine apart and put it back together in better working condition than it started and saying "hey, work out how long this engine is going to last in minutes assuming I drove it 24hrs a day at at 50 mph, and there's no friction anywhere in the mechanical operation.. do it on paper and show me your entire work... oh? you can't? you clearly have no idea what you're doing then and should probably learn how to use a wrench before putting that engine back together"
MH and others want the question to be answered in the same form it would if you were a student in their EE class and it was assigned as a piece of homework. I'm not entirely sure the specifics of what they even want worked out, but I'd imagine a detailed explanation of the near characteristics of the coil over a scale of time given L is 5H. but is that even a common kind of EE question the way it was asked? Right now I'm in the position of just being curious as to how they want the question answered. I feel like I can actually learn something from this exercise but it's being made more difficult than it needs to be.

I don't want to spoil the fun, you'll get there in the end!
John.

Yes, all this does not detract from the ideal voltage source not being able to drop to anything else but its ideal voltage even in time. There would not be an eventual drop to 0 volts because the source would not care about an ideal inductor in any case, infinity or not. So............. even if the current rose to infinity and the inductor was totally destroyed, you would still have your source voltage unchanged. And there in lies the chicken and egg rub.
As well, you can analyze this until the cow jumps over the moon, but you will still not know what actually happens inside a coil and @MH, that includes you. Just because you can rationalize or try to rationalize these mind games, not one iota will be discovered on the coils true function, you are still stuck with only the mind effect.
Actually this is a good comparison for me because in SC current is the number of atoms active in energy conveyance and voltage is the depth at which conveyance occurs. Put 1 volt in a 5H coil that has a cvr tapped in the coil center and wait. The current having a finite number of atoms in the inductor, ideal or not, would be the limiting factor so you would not have any need to ponder ideal conditions to actually know what is happening inside your coil. Hence why I asked the question of wire gauge tables not defining their complete tests. Meanwhile you guys would still be scratching your heads wondering how a coil can handle infinite current.
So the question for you guys is simple. How can an ideal voltage provide infinite current. That is like saying a 46AWG wire should be able to render an infinite AWG wire red hot for infinity. hahaha And this is the same EE that calls OUers outofthebox crazy.
Maybe it's better to just chew on ideal gum with ideal jaws.
wattsup

I think the confusion lies in the general ridiculousness of the intended point.
The question is apparently geared to asses your knowledge of circuit operation by seeing how institutionalized you are regarding EE fundamentals as they are taught in a classroom setting.
no matter the logic you provide, until you work out some mathematical answer in detail based upon the numbers given your answer will not be considered. If you worked out the math to please them it would be pointless because as you have already stated you are in disagreement with the general nature of the question.
There is nothing ridiculous about the intended points. One point is to understand how an inductor works. Another point is to be able to look at a schematic diagram and have a better preliminary sense of what what might be going on in the schematic. Another point would be to be better able to design a circuit to do what you want it to do. Another point would be to be better able to analyze a circuit that you have built. Another point would be to be able to look at a scope trace as a passive observer vs. looking at a scope trace and relating it directly back to your circuit and understanding the waveform and why it is shaped the way it is.
Who says you have to give a mathematical answer for starters? Why not just try to answer the question on a conceptual level for starters and then see if you can move on to a more formal answer? The field is entirely wide open to you but I am not seeing much movement on that field.
On the other thread I clearly showed you how close an ideal inductor can be to a realworld inductor in an actual working circuit. I showed you that ideal voltage sources are available right now, and anybody can play with one.
What I an seeing from you so far is that you are just blindly believing what Brad is saying. The problem is that Brad is wrong. So why don't you try to brainstorm and come up with something new?
The simple truth is this: Brad has stuck with his wrong answer for a few days. Instead of sticking to the wrong answer he could have been researching this stuff over the past few days and who knows he could have already made a breakthrough and then he could have started moving towards the actual answer.
I will repeat, this question is valid and very important if you want to experiment with coils and make pulse motors, the whole nine yards. I admit it's a bit stark to come face to face with the fact that you play with coils all the time on your bench but you don't understand how they work. Do do you remain sour and suck on bitter lemons all day, or do you try to turn lemons into lemonade?
Certainly, just refusing to do anything except repeat the same thing is not productive.
MileHigh

I think the confusion lies in the general ridiculousness of the intended point.
The question is apparently geared to asses your knowledge of circuit operation by seeing how institutionalized you are regarding EE fundamentals as they are taught in a classroom setting.
no matter the logic you provide, until you work out some mathematical answer in detail based upon the numbers given your answer will not be considered. If you worked out the math to please them it would be pointless because as you have already stated you are in disagreement with the general nature of the question.
it's kind of like walking up to a mechanic who can take an engine apart and put it back together in better working condition than it started and saying "hey, work out how long this engine is going to last in minutes assuming I drove it 24hrs a day at at 50 mph, and there's no friction anywhere in the mechanical operation.. do it on paper and show me your entire work... oh? you can't? you clearly have no idea what you're doing then and should probably learn how to use a wrench before putting that engine back together"
MH and others want the question to be answered in the same form it would if you were a student in their EE class and it was assigned as a piece of homework. I'm not entirely sure the specifics of what they even want worked out, but I'd imagine a detailed explanation of the near characteristics of the coil over a scale of time given L is 5H. but is that even a common kind of EE question the way it was asked? Right now I'm in the position of just being curious as to how they want the question answered. I feel like I can actually learn something from this exercise but it's being made more difficult than it needs to be.
Like i said,i dont think MH thought about his question very well when he included ideals in it.
Do we even have an inductor in the question?.
Well from an EEs view ,they may think they do,but from an outsiders view,there may be no inductor at all.
What are the properties of an ideal inductor?
An ideal inductor has no resistance,no capacitance,but only inductance ???
How can it have inductance if it has no resistance?
Well before the EE guys gowhat the hell are you talking about,lets look at the meaning of resistance and inductance.
Resistance>Resistance is the friction in an electrical circuit that controls the flow of current
Inductance> Inductance is typified by the behavior of a coil of wire in resisting any change of electric current through the coil
So a resistor controls the flow of current,and inductance resist any change in current. So it would seem that the inductor also tries to control the flow of current by resisting any change to it.
So inductance is a resistance,and an ideal coil has no resistance.
So dose this mean that it has no inductance?
MH and others want the question to be answered in the same form it would if you were a student in their EE class and it was assigned as a piece of homework.
Im not interested in there !theoretical! answers,im looking for the correct answer to the question.
All circuits are modeled using a resistor in series with the ideal inductor,so as to represent real world inductors. We need to model this circuit without the resistor,and i know just the man for the job.
Poynt
This will be one of very few times you hear me say this ;D
Can you use your sim to model the circuit :o
I believe you have tried this before ;)
Now,you have said on many occasions that the sim will produce actual result's,so lets see it produce some results regarding MHs ideal inductor with an ideal voltage across it>lets see what happens.
No need to post the answerjust tell us if your sim was able to show real world results based around MHs parametersthat being a 5 Henry ideal coil,with a 4 volt ideal voltage across it for 3 seconds>the first part of the cycle in MHs question.
Brad

There is nothing ridiculous about the intended points. One point is to understand how an inductor works. Another point is to be able to look at a schematic diagram and have a better preliminary sense of what what might be going on in the schematic. Another point would be to be better able to design a circuit to do what you want it to do. Another point would be to be better able to analyze a circuit that you have built. Another point would be to be able to look at a scope trace as a passive observer vs. looking at a scope trace and relating it directly back to your circuit and understanding the waveform and why it is shaped the way it is.
Who says you have to give a mathematical answer for starters? Why not just try to answer the question on a conceptual level for starters and then see if you can move on to a more formal answer? The field is entirely wide open to you but I am not seeing much movement on that field.
What I an seeing from you so far is that you are just blindly believing what Brad is saying. The problem is that Brad is wrong. So why don't you try to brainstorm and come up with something new?
The simple truth is this: Brad has stuck with his wrong answer for a few days. Instead of sticking to the wrong answer he could have been researching this stuff over the past few days and who knows he could have already made a breakthrough and then he could have started moving towards the actual answer.
I will repeat, this question is valid and very important if you want to experiment with coils and make pulse motors, the whole nine yards. I admit it's a bit stark to come face to face with the fact that you play with coils all the time on your bench but you don't understand how they work. Do do you remain sour and suck on bitter lemons all day, or do you try to turn lemons into lemonade?
Certainly, just refusing to do anything except repeat the same thing is not productive.
MileHigh
On the other thread I clearly showed you how close an ideal inductor can be to a realworld inductor in an actual working circuit. I showed you that ideal voltage sources are available right now, and anybody can play with one.
No such voltage or inductor existnot even close.
The fact is MH,and this is an absolute fact,you cannot back up your answer with a real world test to show you are correct.
There is no such thing as an ideal inductor,nor is there any such thing as an ideal voltage.
An ideal inductor has no resistance,and inductance it self produces resistancea resistance to the change in current. So an ideal inductor dose not exist for that very reason,and there for your question cannot be answeredwell maybe in MH fairy land.
An ideal voltage also dose not exist. There is no device that can provide an infinite amount of current to maintain an ideal voltage across a dead short that is also idealbeing a short that will never collapse under any load.
"Can God create a stone so heavy that even God is not strong enough to lift it?" God, being almighty, should be able to create this stone but if He does, he cannot move it meaning he is not almighty. However, if God cannot create this stone, the same problem arises.
Brad

Yes, all this does not detract from the ideal voltage source not being able to drop to anything else but its ideal voltage even in time. There would not be an eventual drop to 0 volts because the source would not care about an ideal inductor in any case, infinity or not. So............. even if the current rose to infinity and the inductor was totally destroyed, you would still have your source voltage unchanged. And there in lies the chicken and egg rub.
As well, you can analyze this until the cow jumps over the moon, but you will still not know what actually happens inside a coil and @MH, that includes you. Just because you can rationalize or try to rationalize these mind games, not one iota will be discovered on the coils true function, you are still stuck with only the mind effect.
Hence why I asked the question of wire gauge tables not defining their complete tests. Meanwhile you guys would still be scratching your heads wondering how a coil can handle infinite current.
So the question for you guys is simple. How can an ideal voltage provide infinite current. That is like saying a 46AWG wire should be able to render an infinite AWG wire red hot for infinity. hahaha And this is the same EE that calls OUers outofthebox crazy.
Maybe it's better to just chew on ideal gum with ideal jaws.
wattsup
Lots of unproductive sour grapes. You have got to be kidding about the "mind games." The intention behind this question is to understand how a coil works in the real world. For many experimenters it's possible that mastering this will open up a whole new world of true understanding for them, and instead of just being passive builders and observers, they will become active designers and analyzers. What's the classic question among builders? "What wire gauge and how many turns?" They think that if they are mindless replicators then some magic will happen. For a given project, who says the wire gauge is that important? For a given project, who says the number of turns is that important? Well if you have no clue you may as well just replicate and do what you are told to do, right?
The issue of "infinite current" is irrelevant. The question of a certain wire gauge getting too hot at a certain current level is irrelevant.
There was a time when I explained to some beginners that the simple model for a battery that you use every day on your bench is an ideal voltage source in series with a resistor. That explains why the battery voltage drops under load. Look at that, an appearance of the dreaded ideal voltage source. They were aghast just by the notion of "There is a resistor inside the battery!!??" So for some people it's "radical" to actually try to understand how an inductor works. The wiser people will want to understand this and they will be able to apply it to the work they do on their bench.
MileHigh

No need to post the answerjust tell us if your sim was able to show real world results based around MHs parametersthat being a 5 Henry ideal coil,with a 4 volt ideal voltage across it for 3 seconds>the first part of the cycle in MHs question.
The sim will work perfectly and I am glad that you don't want to see it. The way you have to "see it" is to use your intellect to understand how an inductor works.
You are clearly still stuck like glue to your mistake. The only question is what do you do? Do you stick it out and refuse to explore other options until at the end you are spoon fed the answer? If you agree the answer is correct then you will have a rather embarrassing little apology to offer up. Or, do you do what I suggested to you: Assume for the sake of argument that I am right and for your own benefit start working and researching so that you come to a complete understanding on your own? Which of the two scenarios do you think will ultimately be better for you?

No such voltage or inductor existnot even close.
The fact is MH,and this is an absolute fact,you cannot back up your answer with a real world test to show you are correct.
There is no such thing as an ideal inductor,nor is there any such thing as an ideal voltage.
An ideal inductor has no resistance,and inductance it self produces resistancea resistance to the change in current. So an ideal inductor dose not exist for that very reason,and there for your question cannot be answeredwell maybe in MH fairy land.
An ideal voltage also dose not exist. There is no device that can provide an infinite amount of current to maintain an ideal voltage across a dead short that is also idealbeing a short that will never collapse under any load.
Brad
A real inductor can be 99.99% identical to an ideal inductor. Ideal voltage sources exist right now within certain limitations. A good bench power supply is an ideal voltage source. A car audio amplifier is an ideal voltage source that can output The Star Spangled Banner as an ideal voltage. I explain that all on the other thread.
The realworld test can easily be done as explained on the other thread. The only limitation is that the realworld inductor will behave a tiny smidgen differently from the ideal inductor.
You are just making phony dismissive arguments that take you off track. The point of the exercise is perfectly clear: To understand how an inductor works.

The way you have to "see it" is to use your intellect to understand how an inductor works.
The only question is what do you do? Do you stick it out and refuse to explore other options until at the end you are spoon fed the answer? If you agree the answer is correct then you will have a rather embarrassing little apology to offer up. Or, do you do what I suggested to you: Assume for the sake of argument that I am right and for your own benefit start working and researching so that you come to a complete understanding on your own? Which of the two scenarios do you think will ultimately be better for you?
The sim will work perfectly and I am glad that you don't want to see it.
I think you will find that it will not,well at least it did not on the thread i read on OUR.
Poynt had to add a resistance in series with the ideal inductor,and as your question dose not include a resistance,then the sim must represent your ideal inductor that has no resistance.
You are clearly still stuck like glue to your mistake.
As you ,nor anyone else has proven that i have made a mistake,then my answer standsyou cannot connect an ideal voltage across an ideal inductor.
Brad

Considering a sim to provide an answer here is a good thing because all simulators I've used will not allow an inductor with zero resistance. Some low value will be assigned if the operator fails to fill in the blank. So, what we have to consider is what happens as the resistance approaches zero.
For example, what will the current be at T=0 when we apply the assumed ideal voltage source of 4 volts to the 5 henry inductance with a dc resistance of 1 ohm? Will it be 4 amps? How about .01 ohm, will it be 400 amps or? How about 1e10 ohms, will it be 4e10 amps?
Looking at the problem from another angle, if we assume we had somehow achieved a current level of 2.4 amps in our ideal inductor with zero resistance and we then shorted it with a perfect conductor having zero resistance, what happens to the inductor current?
partzman

Somebody has already posted on this thread very recently how an ideal inductor behaves. I am sure that you can sense that Poynt is in agreement with me.
The best course of action for you would be to understand what inductance really is an how it behaves. You are arguing that there is a radical discontinuity between how an ideal inductor behaves and a realword inductor with a 0.0000000001 ohm wire resistance behaves. Nature does not like discontinuities at all. Just that fact should get you thinking again and hopefully find you going in the right direction.
"Nature does not like discontinuities at all."

I am really surprised that a sim program would balk at an ideal inductor but if that's the case there must be a circuit condition where there is a problem. However, in my my example I can't see the sim encountering a problem like a dividebyzero error or whatever.

author=MileHigh link=topic=16589.msg483533#msg483533 date=1462796493]
You are just making phony dismissive arguments that take you off track. The point of the exercise is perfectly clear: To understand how an inductor works.
No MH. You are making claims you cannot back up,as you do not have access to an ideal inductor.
A real inductor can be 99.99% identical to an ideal inductor.
99.99% of infinity is an infinite amount away from ideal MH. This may shock you ,but the difference between 99.99% and absolute can be extremely large.
When i asked what the L/R time constant was,you said there isnt one.
With your non ideal coil(being the one that is 99.99% close to ideal)there is an L/R time constant,and so that coil has a finite L/R time constant,where as the ideal coil has either a time constant of 0,or it is infinite. These two values are far from your 99.99% close enough is near enough coil,as it is not even close.
Like i said,you should have thought about your question a little better.
Ideal voltage sources exist right now within certain limitations. A good bench power supply is an ideal voltage source. A car audio amplifier is an ideal voltage source that can output The Star Spangled Banner as an ideal voltage. I explain that all on the other thread.
They are not ideal at all.
An ideal voltage dose not changeat all,when a load is placed across itnot even by .0000001 of a volt. The ideal power supply would need an ideal transformer,and ideal FWBR,and all other components would have to be idealincluding the power station,the grid supply,and the meter on your house.
All these factors you fail to take into account when you stated an ideal voltage and an ideal inductor.
The realworld test can easily be done as explained on the other thread.
You cannot do a real world test on fictional components and values.
The only limitation is that the realworld inductor will behave a tiny smidgen differently from the ideal inductor.
Nothe difference is !infinite!you just dont get this,do you ?.
Brad

If the time constant is infinite for maximum current through the ideal inductor,then that means that no current flows through the inductorever,because if it takes an infinite amount of time to reach maximum current flow through the coil,then it also takes an infinite amount of time to reach a 10% value of maximum current flow through the ideal inductor,and it takes an infinite amount of time to reach 1% of the maximum amount of current flow through the ideal inductorand so on. So it will take an infinite amount of time before current even starts to flowso there will never be any current flow through the ideal inductor.
;)
My blown universe post was a bit of a ruse. In fact it implied the opposite of what would theoretically happen, that is, "nothing".

How do you calculate the L/R time constant, when there is no R
L/0=infinite

A sim of an ideal V source across an ideal inductor will crash the sim.
A tiny bit of resistance must be added as the computational resolution is finite.

Brad:
<<< Nothe difference is !infinite!you just dont get this,do you ?. >>>
You are just getting belligerent for no reason and you are twisting your logic and not really making sense.
I said that a realworld coil can be 99.99% identical in behaviour to an ideal coil. That means the difference in their behaviour on the bench will be one part in 10,000. That's pretty damn similar.
Any good bench power supply with a big capacitor on the output, a huge transformer or huge switching power supply, and a welldesigned negative feedback control system to maintain a constant output voltage will be 99.99% identical to an ideal voltage source so the same argument applies.
The point being again that discussing ideal coils or ideal voltage sources is not far fetched at all.
This is the real topic of discussion: a) What is inductance? b) Demonstrate your understanding of inductance by solving for the current for an ideal inductor in a simple circuit.
That is what you need to focus on.
MileHigh

You may wish to exercise a few examples to see if your conclusion is correct. Your conclusion being that if the time constant (Tau) is infinite, the load immediately presents itself as a perfect short, meaning the current will be infinite and instantaneous.
Here are a few: (in all cases, L=6H)
1) R=1, Tau=6s
2) R=0.1, Tau=60s (1min)
3) R=1m Ohm, Tau=60ks (16.6 hours)
4) R=1u Ohm, Tau=6M(million)s (1667 hours)
5) etc.
What is happening to Tau as R decreases?
If R could be 0, Tau must be infinite.
What happens to the inductor current after t=0 when Tau=infinity?
This is the very reason that MHs question cannot be answered,as i have stated before.
The conundrum kicks in at T=0.
If R = 0,which id dose,as the inductor is ideal,then no current flows through the ideal inductor.
If Tau is infinite,then 10% of maximum current flow is also an infinite amount of time,and so is 1% of maximum current flow,and so is .1% of maximum current flowETC. This means that it will also take an infinite amount of time before current start to flow,as there is no division of infinite that it self is not infinite. This means that there is no inductance,as there is no current flow to create a magnetic field,and there for no CEMF to resist the change in current that isnt flowing anyway :o
But there is also no resistance in an ideal coil,and so the ideal voltage is now across a dead short.
So now we have an infinite amount of current flow :o
MH quoted him selfQuote: there is no time constantpost 14
So,the current either rises instantly,or the current rise time is infinite,which means there is no current flowing through the ideal coil.
Brad

L/0=infinite
As i thought
Brad

A sim of an ideal V source across an ideal inductor will crash the sim.
A tiny bit of resistance must be added as the computational resolution is finite.
So this means we cannot sim MHs question as it is defined ?
Brad

<<< So,the current either rises instantly,or the current rise time is infinite,which means there is no current flowing through the ideal coil. >>>
Sorry but I have to lol, when the "answer" is two completely contradictory answers.

author=MileHigh link=topic=16589.msg483543#msg483543 date=1462799110]
That is what you need to focus on.
MileHigh
[/quote]
You are just getting belligerent for no reason and you are twisting your logic and not really making sense.
No i am not MH. I am working on your question using the values and components you described.
I said that a realworld coil can be 99.99% identical in behaviour to an ideal coil. That means the difference in their behaviour on the bench will be one part in 10,000. That's pretty damn similar.
I think you will find that there is a huge difference between an ideal inductor,and a real world inductor MH. And once this is resolved,then you will find that your near enough to ideal power supply is not even close to being ideal.
The point being again that discussing ideal coils or ideal voltage sources is not far fetched at all.
If we are going to be accurate and true to our selves in this discussion,then i think you are going to find that there is an infinite gap between real and ideal.
This is the real topic of discussion: a) What is inductance? b) Demonstrate your understanding of inductance by solving for the current for an ideal inductor in a simple circuit.
Your simple circuit requires resistance,and is normally a resistor added in series with an ideal inductor. But your question and comments that follow say there is no resistance,and so, it cannot be resolved,as poynt has just stated in regards to the sim test.
Brad

<<< So,the current either rises instantly,or the current rise time is infinite,which means there is no current flowing through the ideal coil. >>>
Sorry but I have to lol, when the "answer" is two completely contradictory answers.
Exactly MH.
As i said,you did not think about your question to well when you added ideals to it.
It is also the reason i stated that an ideal voltage cannot be placed across an ideal inductor.
You did see the word conundrum attached to that postalong with several other postsdidnt you?.
Your question cannot be answered,as it is a contradiction to it self.
You stated as well that there is no time constant. What dose that mean?dose the current rise instantly,or is the L/R time constant infinite?.
Brad

@ MH
for many years Poynt has told me that a sim can simulate any real world circuit. Poynt also knows how much i hate the use of sim's when dealing with real world circuit's. But time after time ,Poynt has shown me that the sim has been successful 100% of the time in replicating anything i (and others) have built,and has come to more exacting results than i could obtain on the bench with real world devices.
Poynt has just stated that your circuit as defined cannot be simulated,and so your circuit dose not represent a real world device.
But i would like to continue with this discussion ,as it has really made me (and others here)think of what the outcome of your defined circuit would be.
At this point in time,i am sticking to my answers givenboth the real world answer>you cannot place an ideal voltage across an ideal inductor,and also my theoretical answer,being the current would rise instantly,to an infinite value.
The reason for my theoretical answer is because of the resistance value of zero when calculating the L/R time constant.
I have found no math that allows for any value to be divided by zero without the answer being infinite. If the current rise time is infinite,then there is no rise in current as far as the inductor side of it go's. This means that there is no inductor. But as there is an ideal resulting in a zero value of resistance,the ideal voltage is now placed across a dead shortthe big conundrum of the whole question using your ideal values.
And so my answer of an instant current rise of an infinite value.
Brad

;)
My blown universe post was a bit of a ruse. In fact it implied the opposite of what would theoretically happen, that is, "nothing".
So you are saying that at T=0the instant the ideal voltage is placed across the ideal inductor,nothing would happen?no voltage would appear across the ideal inductor?.
Brad

<<< Poynt has just stated that your circuit as defined cannot be simulated,and so your circuit dose not represent a real world device. >>>
Of course it can be simulated. To my surprise in this case he has to add a resistor. We have seen him add little phantom resistors many times in the past to get the simulator to run so ultimately there is no real surprise.
The place to simulate this setup is in your head, like I explained before. Then there are no issues. I am attaching the "short answer that solves everything" again to this posting.

Brad:
How about this: Take the identical question and add a 0.00000000001 ohm resistor in series with the 5 Henry ideal coil. Now you have a realworld coil.
Can you answer the question now? If you answer that properly then as far as I am concerned your answer will be perfectly legitimate.
MileHigh

Brad:
How about this: Take the identical question and add a 0.00000000001 ohm resistor in series with the 5 Henry ideal coil. Now you have a realworld coil.
Can you answer the question now? If you answer that properly then as far as I am concerned your answer will be perfectly legitimate.
MileHigh
Yes MH,now that question can be answered,as we have the needed value to calculate the L/R time constant.
But first i want to resolve your original questionthe one this thread is about,as i have done nothing but think about the situation that could exist in this ideal case,and the outcome of such an event.
I am hoping that Poynt will keep working on this with the rest of us here,as i have a feeling that even he may not be sure of the outcome,due to the result of the attempted simulation.
At this point in time,i will stick with my theories and answers given,but it is not yet resolved,and so will continue research toward the answer.
Brad

<<< Poynt has just stated that your circuit as defined cannot be simulated,and so your circuit dose not represent a real world device. >>>
Of course it can be simulated. To my surprise in this case he has to add a resistor. We have seen him add little phantom resistors many times in the past to get the simulator to run so ultimately there is no real surprise.
The place to simulate this setup is in your head, like I explained before. Then there are no issues. I am attaching the "short answer that solves everything" again to this posting.
Your short answer dose not solve your question.
Below you can see what i mean that a small difference in resistance value can have an infinite difference in the end resulting value. One has a finite value,and the other has an infinite value,even though the two resistance values are very close. This is the difference between a real world inductor and an ideal inductor.
Brad

Brad:
How about this: Take the identical question and add a 0.00000000001 ohm resistor in series with the 5 Henry ideal coil. Now you have a realworld coil.
Can you answer the question now? If you answer that properly then as far as I am concerned your answer will be perfectly legitimate.
MileHigh
So using your resistance value MH,i have calculated the L/R time constant to be 500,000,000,000.00 seconds or 1388888888.88 hours.
So we can safely say(as poynt did),that next to nothing will happen when the ideal voltage is placed across the inductor and series resistor>!OR!,the inductor having a very low resistance value,will causes the current to rise to 63% of 400000000 kiloamps,keeping in mind that we still have our ideal voltage that will drop to a value of 0 across the inductor,but remain at 4 volts acros the resistor /inductor ,now that we have the series resistor,as it dose when using real world circuits.
Brad

I've attached an LtSpice sim of a 5 henry inductor with a dc resistance of 1e11 connected to a 4 volt ideal voltage source for 3 secs.
At the end of three secs we see the inductor current has reached 2.4 amps. This is as close to a perfect inductor as one could want.
Using the formula that MH supplied that is, delta I = Et/L, this results in delta I = 4*3/5 = 2.4 amps.
What happens if we increase the dc resistance to a higher resistance? The peak current reached will simply be lower than the above depending on the value of the resistance under otherwise the same conditions.
partzman

Brad,
Yes, nothing will happen with the ideal inductor, i.e. it will have 4V (or whatever the voltage is at any point in time) across it and zero current through it for ever and ever. (You can't have both answers.)
Now, replace the ideal inductor with an ideal discharged capacitor, what is the outcome at t=0?
Do you still think with the inductor scenario current instantly going to infinity is a possibility?

I've attached an LtSpice sim of a 5 henry inductor with a dc resistance of 1e11 connected to a 4 volt ideal voltage source for 3 secs.
At the end of three secs we see the inductor current has reached 2.4 amps. This is as close to a perfect inductor as one could want.
Using the formula that MH supplied that is, delta I = Et/L, this results in delta I = 4*3/5 = 2.4 amps.
What happens if we increase the dc resistance to a higher resistance? The peak current reached will simply be lower than the above depending on the value of the resistance under otherwise the same conditions.
partzman
Unfortunately partzman,it is no where near an ideal inductors outcome,as an ideal inductor never has any current passing through it.
Brad

Brad,
Now, replace the ideal inductor with an ideal discharged capacitor, what is the outcome at t=0?
Do you still think with the inductor scenario current instantly going to infinity is a possibility?
Yes, nothing will happen with the ideal inductor, i.e. it will have 4V across it and zero current through it for ever and ever. (You can't have both answers.)
Well this as you know,was one of my answers. My other answer is because there is no resistance with an ideal inductor,and there for it is a dead short.
Your self,verpies,MH and ION have all stated that there is no voltage across an ideal inductor that has a steady DC current flowing through it. So if it is as you saynothing will happen when an ideal voltage is placed across an ideal inductorno current will flow,how can you then say that no voltage will be across an ideal inductor when there is a DC current flowing through it. How did the current flow take place if no current flow will take place when the ideal voltage was placed across it?.
In regards to the ideal capacitor,would we not have the same conundrum?
Brad

Brad:
You have actually been given part of the answer, and you see that you were dead wrong. Turn that into a learning experience.
These are the two challenges for you:
1. Brad gets up the learning curve and understands the original question and then answers it correctly all by himself and clearly demonstrates that he understands what he is doing.
2. Brad admits that he is wrong when he stated that my response to the harder question is wrong.
I will repost my answer to the harder version of the question on this thread.
MileHigh

Here is the harder version of the question and the answer:
You have an ideal voltage source and an ideal coil of 5 Henrys. At time t=0 seconds the coil connects to the ideal voltage source. The voltage source waveform is 20*t^2. So as the time t increases, the voltage increases proportional to the square of the time.
The question is what happens starting at t = 0
The answer:
The current through the ideal coil starts from zero at time t = 0 and then increases with this formula: i = 1.33*t^3.
Time..........Voltage.........Current
0...............0.................0
1...............20...............1.33
5...............500.............166.67
10.............2000............1333.33
20.............8000............10666.67
50.............50000..........166666.7
Brad, you need to try to get up the learning curve such that you get to the point where you come back and acknowledge the answer given above is correct.

I've attached an LtSpice sim of a 5 henry inductor with a dc resistance of 1e11 connected to a 4 volt ideal voltage source for 3 secs.
At the end of three secs we see the inductor current has reached 2.4 amps. This is as close to a perfect inductor as one could want.
Using the formula that MH supplied that is, delta I = Et/L, this results in delta I = 4*3/5 = 2.4 amps.
What happens if we increase the dc resistance to a higher resistance? The peak current reached will simply be lower than the above depending on the value of the resistance under otherwise the same conditions.
partzman
Thank you partzman for running the simulation. I am hoping that this gives Brad and others enough information to answer the full question and demonstrate that they understand the concepts and show competence in the subject matter. In that sense please refrain from providing further answers to the complete question that defines a multipart voltage waveform. The hope is that the interested parties undertake to teach themselves as apposed to being spoon fed the answers.
MileHigh

I just want to point out this argument started over an actual inductor. like, a non imaginary, real world, coil that is part of a real circuit.
The argument began when the group was discussing just how the circuit operated. Now I'm no expert, but I'm fairly sure
this inductor has an inductance and a resistance, and I'm sure some capacitance in the windings.. I'm also pretty sure we can track down
the rest of the factors in the circuit such as source voltage.. frequency.. etc.. I have seen no one do that. Instead days have been spent
arguing over a completely fictitious scenario that basically boils down to a basic philosophical argument about duality or the chicken and the egg.
At this point I don't see how we are any closer to agreeing just how the JT circuit originally in question fires below 300mV.
Instead we have moved to the fantasy realm of an ideal hypothetical scenario because MH want's to prove that
Brad just blindly threw the circuit together and has no idea how it works.
At least I learned something.. I learned that even though one can understand the characteristics of an inductor, how to use it
in a circuit, how it behaves in comparison to other components, and how to work out its role in a circuit over time.. none of that
has absolutely anything to do with how an inductor really works unless you can answer some ambiguous question about a
hypothetical coil.
what happens after T=0? well, there's a dog somewhere that probably took a dump, someone's favorite tv show probably came on,
somewhere along the lines the inductor started charging.. oh and it also continued to display characteristics of any other inductor even though
we have to assume that entirely of faith.
I'm assuming since MH has finished his little test exercise, can we go ahead and start figuring out
what's going on in the actual circuit now?

Here is the harder version of the question and the answer:
You have an ideal voltage source and an ideal coil of 5 Henrys. At time t=0 seconds the coil connects to the ideal voltage source. The voltage source waveform is 20*t^2. So as the time t increases, the voltage increases proportional to the square of the time.
The question is what happens starting at t = 0
The answer:
The current through the ideal coil starts from zero at time t = 0 and then increases with this formula: i = 1.33*t^3.
Time..........Voltage.........Current
0...............0.................0
1...............20...............1.33
5...............500.............166.67
10.............2000............1333.33
20.............8000............10666.67
50.............50000..........166666.7
Brad, you need to try to get up the learning curve such that you get to the point where you come back and acknowledge the answer given above is correct.
look.. I'm just some dude that loves electronics tinkering and never claimed to be any kind of electrical engineer, but this version actually seems a simpler question. Though I'm extremely confused as to how there is 0 voltage at 0 seconds when at 0 seconds the coil is connected to the source.
For those of us who are trying to interpret your original question properly out of simple respect, any more details as to 'how' you wanted the question answered would be of great help.

Magneticitist:
You are way off. For starters, you threw a lot of mumbojumbo talk at this subject earlier when discussing this test. You are talking more mumbojumbo talk now.
Like I already clearly stated to you in an earlier posting, this example has many applications in the real world of your bench. What we are talking about is real, and there are no philosophical arguments to be made.
We are not interested in the resistance or the capacitance between the windings, at all. That just makes everything more complicated. We are not talking about frequency at all. Nor are we talking about a Joule Thief. Forget about the usual buzz words and the usual box of tricks, time to get real.
At least I learned something.. I learned that even though one can understand the characteristics of an inductor, how to use it in a circuit, how it behaves in comparison to other components, and how to work out its role in a circuit over time.. none of that has absolutely anything to do with how an inductor really works unless you can answer some ambiguous question about a hypothetical coil.
It has every damn thing to do with how an inductor really works. Your problem is that you have been taken out of your comfort zone. You like to fire off philosophical musings when it comes to coils and circuits, it gives the impression that you have a "higher order" discussion going on about electronics. The problem here is that we were trying to figure out how a very simple circuit works and you had no idea yourself. Don't be dismissive of real solid knowledge and make silly cracks about a dog taking a dump.
The message to you is to come down to Earth and if you want to do better work on your bench, then you want to understand this stuff and have complete mastery over it. Pay attention and try to learn something important. Drop the mumbojumbo talk and be real.
The question still has not been answered, do you want to take a shot at it?
MileHigh

well at least we can agree, that both sides of this argument feel as if they are speaking to a brick wall that's reading a magazine.
(btw that was sarcasm about the inductor..)

look.. I'm just some dude that loves electronics tinkering and never claimed to be any kind of electrical engineer, but this version actually seems a simpler question. Though I'm extremely confused as to how there is 0 voltage at 0 seconds when at 0 seconds the coil is connected to the source.
For those of us who are trying to interpret your original question properly out of simple respect, any more details as to 'how' you wanted the question answered would be of great help.
The formula for the voltage source that connects to the coil is v = 20*t^2. Therefore when t=0 the voltage is zero. This is like setting the voltage on your bench power supply to zero and connecting up a coil. Then you turn the knob to increase the voltage, and you turn it faster as time progresses. So the question is what is the current through the coil when you do this?
Do you perhaps think that is a dumb thing to do? Millions would disagree with you. What if you play the Star Spangled Banner across the coil? Well, there will be a current waveform in that case also, and it would not be that hard to determine. Notice that I say "determine" and NOT "measure." You don't need to measure the current when the Star Spangled Banner is playing across the coil, you can determine what it is using your brain and a few tools. That's the whole point in this discussion. Note that sometimes you simply can't measure the current, you need to use your wits instead to determine the current.
At this point how to answer the original question should be pretty obvious to most. Please check with Brad first and get his take on how to answer the question. He should know how to respond to that. If it is still not clear after that then I will answer you.
The way the question was posed was intentional. You are supposed to have the electronics aptitude and experience to know what to say for the answer. I am not saying this to give you a hard time. Rather, I am just stating this to illustrate how far many bench experimenters have to go to get to a better and more productive place. There is no point in playing with coils if you don't really know what you are doing. It might be tough medicine to say that but it is good medicine. And that brings us full circle back to the question that I asked EMJunkie being repeated here. He had been playing with coils on his bench for ten years and he couldn't answer the question either.
MileHigh

Brad,
Now, replace the ideal inductor with an ideal discharged capacitor, what is the outcome at t=0?
Do you still think with the inductor scenario current instantly going to infinity is a possibility?
Yes, nothing will happen with the ideal inductor, i.e. it will have 4V (or whatever the voltage is at any point in time) across it and zero current through it for ever and ever. (You can't have both answers.)
Ok,in this point in time,i will provide my real world answer,along with my !two! theoretical answers.
My real world answer is(and has been throughout this thread)that you cannot place an ideal voltage across an ideal inductor,as an ideal inductor dose not and never will exist.
Theoretical answer 1> As R = 0,then the L/R time constant is infiniteas you have just stated is correct Poynt. The first thing this means is that the question asked by MH with the values given in that question,results in nothing at all happeningas you also just stated would be the situation.
What we now need to do to understand my second theoretical answer, is define the properties of an ideal inductor. An ideal inductor has no resistance or capacitance,but only inductance. This is now when we find out that the ideal inductor has no inductance at allregardless of the inductance value of that ideal inductor. For inductance to occur,current must flow into that inductor,and as we have both just established,no current will into that inductor ,regardless of the voltage across that ideal inductor,or the time the voltage is across that inductor,as the current rise time is infinite,and there for there is an infinite time before current starts to flow>there is no current flowing into that inductorever,as we know know.
As we have no capacitance value,then the voltage across the inductor is not stored across the inductor,so this is omitted. But what we do have,due to there being no resistance,is a dead short across the inductors terminals,which brings rise to my theoretical answer 2.
Theoretical answer 2>. As it has now been concluded that there is no current flowing into the ideal inductor,then there is no inductance taking place. We are now left with an ideal voltage across a dead short,due to the zero value of resistance. This leads to an instant current rise to an infinite value due to the ideal voltage now being across a dead short. I will define instant being the ideal speed at which current can flow,that being the speed of light.
So you see now why i called this a conundrum theory.
But what these two theories prove,is my real world answer,and that is,as you know>that you cannot place an ideal voltage across an ideal inductor,as an ideal inductor dose not and never will exist.
There is also a second conundrum,and that being that we all agree that a voltage cannot exist across an ideal inductor that has a steady DC current flowing through it. If an ideal inductor,with an ideal voltage placed across it,never has current flowing into or through it,how did the steady state current get induced into the ideal inductor,that dose not allow a voltage to be seen across it?.
We have worked out that current will never flow into an ideal inductor,and now all we have to do is work out the rest of what i postedor should i say Poynt,have your opinion toward what i have posted above.
Brad

Brad:
You have actually been given part of the answer, and you see that you were dead wrong. Turn that into a learning experience.
These are the two challenges for you:
1. Brad gets up the learning curve and understands the original question and then answers it correctly all by himself and clearly demonstrates that he understands what he is doing.
2. Brad admits that he is wrong when he stated that my response to the harder question is wrong.
I will repost my answer to the harder version of the question on this thread.
MileHigh
MH
Did you not see Poynts post (80)
Quote: Yes, nothing will happen with the ideal inductor, i.e. it will have 4V (or whatever the voltage is at any point in time) across it and zero current through it for ever and ever.
This in itself says that the results to your question will benothing happens. It also shows that one of my theoretical answers was correct,and my real world answer was also correctwhen you read my last pots,you may start to understand the complications with your !original! question.
To quote your post 69 Sorry but I have to lol, when the "answer" is two completely contradictory answers.,i think once again,you have laughed at me to soon.
Brad

Here is the harder version of the question and the answer:
You have an ideal voltage source and an ideal coil of 5 Henrys. At time t=0 seconds the coil connects to the ideal voltage source. The voltage source waveform is 20*t^2. So as the time t increases, the voltage increases proportional to the square of the time.
The question is what happens starting at t = 0
The answer:
The current through the ideal coil starts from zero at time t = 0 and then increases with this formula: i = 1.33*t^3.
Time..........Voltage.........Current
0...............0.................0
1...............20...............1.33
5...............500.............166.67
10.............2000............1333.33
20.............8000............10666.67
50.............50000..........166666.7
Brad, you need to try to get up the learning curve such that you get to the point where you come back and acknowledge the answer given above is correct.
As i clearly stated MH,we are not resolving your modified version until your original question is answered.
Please stop trying to confuse people in this thread,that is dedicated to resolving your original question.
Brad

Here is the harder version of the question and the answer:
The question is what happens starting at t = 0
The answer:
The current through the ideal coil starts from zero at time t = 0 and then increases with this formula: i = 1.33*t^3.
Time..........Voltage.........Current
0...............0.................0
1...............20...............1.33
5...............500.............166.67
10.............2000............1333.33
20.............8000............10666.67
50.............50000..........166666.7
Brad, you need to try to get up the learning curve such that you get to the point where you come back and acknowledge the answer given above is correct.
You have an ideal voltage source and an ideal coil of 5 Henrys. At time t=0 seconds the coil connects to the ideal voltage source. The voltage source waveform is 20*t^2. So as the time t increases, the voltage increases proportional to the square of the time.
If the voltage increases,then it is not an ideal voltage,as an ideal voltage dose not change in time.
Your original question clearly states an ideal voltage of 4 volts for 3 seconds.
The rest of your answer is not related to the original question,as the voltage is not 4 volts for 3 seconds.
https://en.wikipedia.org/wiki/Voltage_source
An ideal voltage source is a twoterminal device that maintains a fixed voltage drop across its terminals. It is often used as a mathematical abstraction that simplifies the analysis of real electric circuits.
Brad

Brad:
Re: Poynt's comment:
Yes, nothing will happen with the ideal inductor, i.e. it will have 4V (or whatever the voltage is at any point in time) across it and zero current through it for ever and ever.
I am not sure of the context for that or he may simply have been sarcastic.
What I can say to you is this:
The question was posed to you with an ideal inductor and you objected to that. So I then reposed the question to you and added a resistor that gives you a time constant of 158,440 years and you agreed in principle that that variant of the question can be answered. On top of that, you have already been given a partial correct answer.
So what you (or anyone else) need to do is try to answer either question, the original or the variant. Both answers will essentially be the same.
So there is only one real answer, even if you use the variant question. And the answer is still not forthcoming from you or from anyone else. That means you have an opportunity to shine and demonstrate that you understand what is going on, you understand inductance, and you understand how an inductor works.
So forget about your long and complicated answer in post #91. A simple question can be answered with a simple answer, and then also demonstrate your competence on the subject matter. Give it your best shot.
MileHigh

author=MileHigh link=topic=16589.msg483572#msg483572 date=1462810403]
Thank you partzman for running the simulation.
MileHigh
Partzmans answer dose not represent your original question.
If partzman runs the same simulation with the resistance value removed,so as the circuit represents your ideal inductor,he will derive the same answer as myself and Poynt.
I am hoping that this gives Brad and others enough information to answer the full question and demonstrate that they understand the concepts and show competence in the subject matter.
The rest of us are hoping that MH learns that when you add ideals into questions,it changes everything drastically,and the situation in no way represents real worl outcomes.
In that sense please refrain from providing further answers to the complete question that defines a multipart voltage waveform. The hope is that the interested parties undertake to teach themselves as apposed to being spoon fed the answers.
All the information i have provided has been under my own understandings and research. There has been no spoon feeding thank you very much MH.
Brad

Brad:
<<< If the voltage increases,then it is not an ideal voltage,as an ideal voltage dose not change in time. >>>
Actually, that's a "rule" that you seem to have made up for yourself. An ideal voltage source can be a function of time. i.e.; v(t) = 7*sin(omega*t).
That's a +/7volt sine wave at the angular frequency omega that has a zero output impedance.
I can't fathom how you "lock" yourself sometimes. An ideal voltage source is a voltage source that has a zero output impedance. It has nothing to do with it possibly varying in time.
Like I said before, you can have an ideal voltage source that outputs an Ozzy Osbourne song or The Star Spangled Banner  it's still an ideal voltage source.
MileHigh

I just want to point out this argument started over an actual inductor. like, a non imaginary, real world, coil that is part of a real circuit.
The argument began when the group was discussing just how the circuit operated. Now I'm no expert, but I'm fairly sure
this inductor has an inductance and a resistance, and I'm sure some capacitance in the windings.. I'm also pretty sure we can track down
the rest of the factors in the circuit such as source voltage.. frequency.. etc.. I have seen no one do that. Instead days have been spent
arguing over a completely fictitious scenario that basically boils down to a basic philosophical argument about duality or the chicken and the egg.
At this point I don't see how we are any closer to agreeing just how the JT circuit originally in question fires below 300mV.
Instead we have moved to the fantasy realm of an ideal hypothetical scenario because MH want's to prove that
Brad just blindly threw the circuit together and has no idea how it works.
At least I learned something.. I learned that even though one can understand the characteristics of an inductor, how to use it
in a circuit, how it behaves in comparison to other components, and how to work out its role in a circuit over time.. none of that
has absolutely anything to do with how an inductor really works unless you can answer some ambiguous question about a
hypothetical coil.
what happens after T=0? well, there's a dog somewhere that probably took a dump, someone's favorite tv show probably came on,
somewhere along the lines the inductor started charging.. oh and it also continued to display characteristics of any other inductor even though
we have to assume that entirely of faith.
I'm assuming since MH has finished his little test exercise, can we go ahead and start figuring out
what's going on in the actual circuit now?
MH ask this question to a couple of members of this forum,and he has been ranting on ever since how they could not answer his question correctly.
I decided to have a go at answering the question,which is posted in the first post of this thread.
I opened this thread to discuss this original question,and not MHs modified version that now has a non ideal inductor.
I told MH that his question is more complicated than he believes it is.
It has already been established that from T=0 to T=13 seconds,nothing will happen,as current will not flow through an ideal inductor. This opens cans of worms all over the place,as you can see in my previous 4 to 5 posts.
Brad

MH you have to understand, the problems I have with your 'ideal' argument and your test question are entirely based upon the context in which you are using them..
As I said I totally understand the entire use of this circuit theory. Why would I argue that? Of course, by knowing the nature of what you are playing with, you can use certain tools to approximate possible outcomes.. essentially design a functioning circuit on paper and know how it will work before even having to wire it.. I get it.. The argument I have against the 'ideal' 0 resistance situation is simple.. we don't know for sure how a circuit would act with 0 resistance. we just don't. we lie to ourselves and say once we get to a point where from our perspective things become far too 'small' to say they matter anymore, we can just go ahead and say they don't exist.
Anyway like I said it seemed as if you had presented this question like it somehow related to the circuit that was originally being discussed. I'm trying to explain how your pop quiz had entirely nothing to do with the real world circuit in question and you take that as me trying to crap on the basic circuit theory you had to learn and have grown to respect as a useful tool in EE, all talks of 0 resistance aside.
If it would make you feel any better we can dispense with the trivial motions we are sure to play out.. I'm dumb, you're smart. I have absolutely no idea what an inductor is and you're the master of the electronic universe. I resign to the fact that I am utterly incapable of designing even the simplest functional circuit so long as it contains an inductor, and you know them inside out, top to bottom, with no mysteries to uncover.
Now that we have all that out of the way, I'd really like to get back to the joule thief discussion myself.. Didn't you and Brad have a disagreement about the circuit operation?
Is there some further explanation that proves your reasoning somewhere buried underneath this pop quiz question we have spent pages and pages discussing?

The formula for the voltage source that connects to the coil is v = 20*t^2. Therefore when t=0 the voltage is zero. This is like setting the voltage on your bench power supply to zero and connecting up a coil. Then you turn the knob to increase the voltage, and you turn it faster as time progresses. So the question is what is the current through the coil when you do this?
Do you perhaps think that is a dumb thing to do? Millions would disagree with you. What if you play the Star Spangled Banner across the coil? Well, there will be a current waveform in that case also, and it would not be that hard to determine. Notice that I say "determine" and NOT "measure." You don't need to measure the current when the Star Spangled Banner is playing across the coil, you can determine what it is using your brain and a few tools. That's the whole point in this discussion. Note that sometimes you simply can't measure the current, you need to use your wits instead to determine the current.
Please check with Brad first and get his take on how to answer the question. He should know how to respond to that. If it is still not clear after that then I will answer you.
The way the question was posed was intentional. You are supposed to have the electronics aptitude and experience to know what to say for the answer. I am not saying this to give you a hard time. Rather, I am just stating this to illustrate how far many bench experimenters have to go to get to a better and more productive place. There is no point in playing with coils if you don't really know what you are doing. It might be tough medicine to say that but it is good medicine. And that brings us full circle back to the question that I asked EMJunkie being repeated here. He had been playing with coils on his bench for ten years and he couldn't answer the question either.
MileHigh
At this point how to answer the original question should be pretty obvious to most.
Except to you MH.
The first answer we have so far isnothing will happen.
I am awaiting Poynts review on the reasons for my second theoretical answerand my theories on the rest of it.
Brad
Brad

Brad:
Partzmans answer dose not represent your original question.
If partzman runs the same simulation with the resistance value removed,so as the circuit represents your ideal inductor,he will derive the same answer as myself and Poynt.
I will repeat this a second and last time. The addition of the resistance to the simulation is to solve a constraint in the way the software works. It has nothing whatsoever to do with what you are referring to.
You are in for an unpleasant surprise when Poynt comes back. You think that both of you are on the same page but you are not. Like I said, I can't explain his quote but in due time we will find out.
You literally have been given a partial correct answer from partzman in two forms. Someone else earlier in the thread described an ideal inductor behaviour perfectly. It looks like it's going to take a few more days before you come round and it may have to end up being a forced spoon feeding session from a guru.
MileHigh

MH ask this question to a couple of members of this forum,and he has been ranting on ever since how they could not answer his question correctly.
I decided to have a go at answering the question,which is posted in the first post of this thread.
I opened this thread to discuss this original question,and not MHs modified version that now has a non ideal inductor.
I told MH that his question is more complicated than he believes it is.
It has already been established that from T=0 to T=13 seconds,nothing will happen,as current will not flow through an ideal inductor. This opens cans of worms all over the place,as you can see in my previous 4 to 5 posts.
Brad
one thing that majorly confused me, and apparently you as well, was the 'ideal voltage' source also being a variable supply. so not only does this magic power supply provide infinite current with 0 vdrop it can change voltage at any time. I was under the impression an ideal vsource was a fixed source for all intents and purposes, though apparently not. this is why I was saying that kind of question is more what you would expect from your EE professor, in class. without being primed with the basics of that class, it's quite hard to know where one is supposed to start answering, or how.
IMO that does not mean a person is completely void of electronics knowledge of circuit building capability.
people still make fun of Ed Leedskalnin but could he not manipulate magnetic current to do his bidding? sure, maybe he could have done so much more had he earned a degree, but his lack of degree by no means made him a clueless idiot.

Magneticitist:
The question defines a voltage source connected to an inductor. It defines the voltage waveform.
The only unknown in the setup is the current. You have to solve for the current.
MileHigh

Brad:
Re: Poynt's comment:
I am not sure of the context for that or he may simply have been sarcastic.
What I can say to you is this:
The question was posed to you with an ideal inductor and you objected to that. So I then reposed the question to you and added a resistor that gives you a time constant of 158,440 years and you agreed in principle that that variant of the question can be answered. On top of that, you have already been given a partial correct answer.
So what you (or anyone else) need to do is try to answer either question, the original or the variant. Both answers will essentially be the same.
So there is only one real answer, even if you use the variant question. And the answer is still not forthcoming from you or from anyone else. That means you have an opportunity to shine and demonstrate that you understand what is going on, you understand inductance, and you understand how an inductor works.
So forget about your long and complicated answer in post #91. A simple question can be answered with a simple answer, and then also demonstrate your competence on the subject matter. Give it your best shot.
MileHigh
As i have now stated 3 times MH,we will first answer your original question,and then we will look at your modified question,which now includes a resistance,and so is no longer related to the original question,as the inductor is no non ideal. The two answers and outcomes will be infinitely differentthat you can be assured.
I do not believe Poynt is being sarcastic,as this is a very important subject as far as im concerned.
I am no longer interested in proving you wrong,as your original question has me fired up like i have not been in a long time. My brain has gone into overdrive on this one,and it has been a long time since i have had so much interest in a subject. It is now 2.15 am,and i cant sleep. I just lye in been thinking about this subjectgoing over and over my thoughts on it. I have to start a 9 hour shift in 4 hours,but i cant sleep.
The ramifications of my theories being correct,change everything as far as what is believed to be an ideal inductor. If Poynt was being sincere with his answer about there being no current flow through an ideal inductor,which as you know,was also !one! of my answers,then we have started to travel down a road no one has traveled yet.
I have to wait for Poynts view on my answers i provided.
Maybe even PW,Vortex1,verpies would like to join in on the discussion?.
Brad

Brad:
Re: Poynt's comment:
I am not sure of the context for that or he may simply have been sarcastic.
I'm not being sarcastic at all.
Unless I'm overlooking something, or don't correctly understand the question, there will be no current, and the voltage across the inductor will be whatever the voltage source is at any one time, from t=0 to the end of time.
Given that tau=infinity, there can be no change in either I or V. Indeed if R is nonzero, there will be a current and the voltage change over time.

one thing that majorly confused me, and apparently you as well, was the 'ideal voltage' source also being a variable supply. so not only does this magic power supply provide infinite current with 0 vdrop it can change voltage at any time. I was under the impression an ideal vsource was a fixed source for all intents and purposes, though apparently not.
That's an ironic twist for you. The supposedly "openminded researchers that go where conventional scientists don't dare go" can sometimes be so closedminded that they can't conceive of something as simple as an ideal voltage source that varies in time.

That's an ironic twist for you. The supposedly "openminded researchers that go where conventional scientists don't dare go" can sometimes be so closedminded that they can't conceive of something as simple as an ideal voltage source that varies in time.
it's quite simple.. having not invested any real time into learning about 'ideal' components, I had to look up what constituted an ideal voltage source.. The information I gathered stated an ideal voltage source was a fixed source that can supply infinite current without dropping.. Obviously that was an incorrect or incomplete description.
(i will say that yes, the possibility of a variable supply did indeed fly over my head. I won't cry about it!)
Now you're just being spiteful.

I'm not being sarcastic at all.
Unless I'm overlooking something, or don't correctly understand the question, there will be no current, and the voltage across the inductor will be whatever the voltage source is at any one time, from t=0 to the end of time.
Given that tau=infinity, there can be no change in either I or V. Indeed if R is nonzero, there will be a current and the voltage change over time.
I am going to have to assume that you don't understand the question. Before t=0 the voltage source and the coil are disconnected. The only reason for putting it that way is to implicitly establish that the current through the coil is zero at the start of the test. At t=0 you then have a 4volt source connected to a 5Henry inductor, and then the voltage varies in time as described.
When it is reworded as above, is it clear now?

Brad,
I don't support your answer regarding the inductor acting like an instantaneous short. Just because there is no current flow, doesn't mean the load presented to the voltage source is no longer "inductive".
There is only one answer, that being the inductor will essentially look like a perpetual "open" to the voltage source. At least that is my thinking on this until MH corrects me, and I welcome being corrected if I am in fact wrong on this.

it's quite simple.. having not invested any real time into learning about 'ideal' components, I had to look up what constituted an ideal voltage source.. The information I gathered stated an ideal voltage source was a fixed source that can supply infinite current without dropping.. Obviously that was an incorrect or incomplete description.
Now you're just being spiteful.
They say fixed source meaning that the voltage will not change under load. They don't say anything else.

Brad:
I will repeat this a second and last time. The addition of the resistance to the simulation is to solve a constraint in the way the software works. It has nothing whatsoever to do with what you are referring to.
You literally have been given a partial correct answer from partzman in two forms. Someone else earlier in the thread described an ideal inductor behaviour perfectly. It looks like it's going to take a few more days before you come round and it may have to end up being a forced spoon feeding session from a guru.
MileHigh
MH
Maybe you can work out the L/R time constant of an ideal coil.
As Poynt stated,and as i stated,it will then be 5/0=infinity.
You have already stated that there is no time constant,and the value could be infinity or undefinedcorrect?,or do i have that wrong?.
If it is infinity,then there is going to be no current flow into the ideal inductorcorrect?
So unless you know some sort of math that allows a division of 5/0,and provides a value we can work with,then i will stick with my claim.
You are in for an unpleasant surprise when Poynt comes back. You think that both of you are on the same page but you are not. Like I said, I can't explain his quote but in due time we will find out.
Well i dont believe that Poynt would lead me down the garden path,or post sarcastic results,just because that is what i believe. Sure,we have had our moments,but we have also had our times where we get along just fine,like in the Lewin topic not long ago. I listened to what he was saying,and carried out the test as he stated,and it resulted in being exactly what he said it would be. There is always going to be times when we dont agree on something,but when it comes time to work on thing's like this,then we should both be striving toward a correct answer and outcome.
Brad.

Brad:
<<< If it is infinity,then there is going to be no current flow into the ideal inductorcorrect? >>>
Here is something for you to ponder: A straight line is a circle with a radius of infinity.

I also want to point out in a non sarcastic manner, that this is what it's all about. despite our little beefs we are trying to put our minds together to debate and brainstorm what's going on here... From the EE professionals to the casual tinkerers it's not so cut and clear... this is how complex electricity can be..
I'm not saying a simple agreeable answer is impossible... but clearly if one 'side' has yet to clearly come to an agreeable conclusion then it must be, at the very least, a little confusing.

They say fixed source meaning that the voltage will not change under load. They don't say anything else.
well in all honestly my mistake then.. but I took 'fixed source' as being utterly fixed.

Brad,
I don't support your answer regarding the inductor acting like an instantaneous short. Just because there is no current flow, doesn't mean the load presented to the voltage source is no longer "inductive".
There is only one answer, that being the inductor will essentially look like a perpetual "open" to the voltage source. At least that is my thinking on this until MH corrects me, and I welcome being corrected if I am in fact wrong on this.
OK,so here is the part in that reply i dont understand.
If the inductor looks like a perpetual open (circuit i guess?)to the voltage source,then what is the load?
Brad

Brad:
<<< If it is infinity,then there is going to be no current flow into the ideal inductorcorrect? >>>
Here is something for you to ponder: A straight line is a circle with a radius of infinity.
A straight line is not a circle,as it has a beginig and endunless it is infinitely long. A circle is an infinite loop.
Brad

OK,so here is the part in that reply i dont understand.
If the inductor looks like a perpetual open (circuit i guess?)to the voltage source,then what is the load?
Brad
Zero current means no load. So we have an infinite load, i.e. an "open" circuit.

I'm not being sarcastic at all.
Unless I'm overlooking something, or don't correctly understand the question, there will be no current, and the voltage across the inductor will be whatever the voltage source is at any one time, from t=0 to the end of time.
Given that tau=infinity, there can be no change in either I or V. Indeed if R is nonzero, there will be a current and the voltage change over time.
Thanks Poynt
I did not believe that you would answer questions this important in a sarcastic manner as i pointed out to MH.
So this would mean that MHs original question would result in the applied voltage being seen across the ideal inductor,and nothing more than that.
So rather that slam all my theories in one post,maybe a step at a time would be better.
So,if there is no current flowing through the inductor,that would mean no inductance/no resulting rising magnetic field?correct?
Brad

<<< You have already stated that there is no time constant,and the value could be infinity or undefinedcorrect?,or do i have that wrong?. >>>
Here is the reason I am saying there is no time constant: It's because in the ideal coil case you will never have a situation that meets the standard definition of a time constant. So in that sense there is no time constant.
Take the example of a car that is front wheel drive. A car with frontwheel drive will not handle the same way as a car with rearwheel drive. So there is no point in talking about handling problems with rearwheel drive cars to someone driving a frontwheel drive car.
And yes, the time constant is infinity. It's another way of saying the same thing. The real question is what does that mean for this test?

Poynt, I hope you did not miss my reply #108.

Zero current means no load. So we have an infinite load, i.e. an "open" circuit.
Ok,im going to need help understanding this Poynt
The zero current meaning no load is no problem,but having an infinite load(a load of infinite proportions),would this not require a current flow of infinite magnitude?,and we have no current flow.
Brad

MH,
For your test (13s of a few different voltages), the different voltages have no affect on the outcome.
So the question could have been asked with one voltage and still have the same answer, correct?
I did the simulation with 0.0000000001f (femto) Ohms, which is very very small, and ran it for 100s, far exceeding the limit of your test. The current stayed flat at 0A for the full 100s. Needless to say the voltage across the coil also remained at 4V for the entire 100s.
I think we can conclude from these results that my answer is correct.

I am going to have to assume that you don't understand the question. Before t=0 the voltage source and the coil are disconnected. The only reason for putting it that way is to implicitly establish that the current through the coil is zero at the start of the test. At t=0 you then have a 4volt source connected to a 5Henry inductor, and then the voltage varies in time as described.
When it is reworded as above, is it clear now?
Im going to have to correct that MH.
At T=0,an ideal voltage of 4 volts is placed across an ideal inductor of 5 Henrys for 3 seconds. This is how your question reads.
To quote again,An ideal voltage source is a twoterminal device that maintains a fixed voltage drop across its terminals. It is often used as a mathematical abstraction that simplifies the analysis of real electric circuits.
Brad

MH,
For your test (13s of a few different voltages), the different voltages have no affect on the outcome.
So the question could have been asked with one voltage and still have the same answer, correct?
I did the simulation with 0.0000000001f (femto) Ohms, which is very very small, and ran it for 100s, far exceeding the limit of your test. The current stayed flat at 0A for the full 100s. Needless to say the voltage across the coil also remained at 4V for the entire 100s.
I think we can conclude from these results that my answer is correct.
Poynt
Thank you for your time on this,i am really enjoying this topic.
But i have to get some sleepwork in 3 hour time ::)
I hope we can look into this further together,along with anyone else that is interested.
Cheers
Brad

Ok,im going to need help understanding this Poynt
The zero current meaning no load is no problem,but having an infinite load(a load of infinite proportions),would this not require a current flow of infinite magnitude?,and we have no current flow.
Brad
Don't get too hung up on my use of the term "infinite load". In electronics when we say that we mean an "open circuit", not a "short circuit". I know it seems counter intuitive. If we mean to say a "short", we say "short, "short circuit" or "zero Ohm" load.

MH,
For your test (13s of a few different voltages), the different voltages have no affect on the outcome.
So the question could have been asked with one voltage and still have the same answer, correct?
I did the simulation with 0.0000000001f (femto) Ohms, which is very very small, and ran it for 100s, far exceeding the limit of your test. The current stayed flat at 0A for the full 100s. Needless to say the voltage across the coil also remained at 4V for the entire 100s.
I think we can conclude from these results that my answer is correct.
Poynt, I am really surprised and I suspect that you are a victim of GIGO from your simulator. You put in a resistance value that was too low for the sim and it got scrambled brains. (a bug!) That affected you also I am afraid! lol I almost suspect you may have worked 36 hours nonstop or something.
Perhaps forget about the sim and do it in your head?
Hold on, I will find the posting number of another sim also. ===>>> Please look at post #79.

wait, are we now saying that the answer MH was looking for, was to clarify that no current would flow and it would act like an open circuit?
this is confusing because I recall at least a couple of us saying that and it was some ol whackadoo or crazy stuff, or whatever.
edit nevermind I see MH is in disagreement, now idk what is going on

wait, are we now saying that the answer MH was looking for, was to clarify that no current would flow and it would act like an open circuit?
this is confusing because I recall at least a couple of us saying that and it was some ol whackadoo or crazy stuff, or whatever.
edit nevermind I see MH is in disagreement, now idk what is going on
It could be sleep deprivation, you never know. It's like the old cliche that when you factor in all of the rewrites and recompiles and bug fixes, a software engineer can only write 10 lines of code a day.
However, the other issue is that we are getting way off the beaten track. We still haven't really scratched the surface when it comes to actually understanding the how and why of what is going on in this trivial circuit. I suppose we need to actually get it answered also. The how and why is the real point in this whole exercise.

Poynt, I am really surprised and I suspect that you are a victim of GIGO from your simulator. You put in a resistance value that was too low for the sim and it got scrambled brains. (a bug!) That affected you also I am afraid! lol I almost suspect you may have worked 36 hours nonstop or something.
Perhaps forget about the sim and do it in your head?
Hold on, I will find the posting number of another sim also. ===>>> Please look at post #79.
I saw that post.
His R value is too big. I can obtain similar results as I increase the resistor value. 1p Ohm is nowhere near 0 Ohms. Have Partzman try 0.00000000000000000000000000000000000001f Ohms. He will see a flatline of current (0A) for at least 100 seconds.
I did work this out in my head first, and the sim supports my conclusion.

Well now I feel like I am in the Twilight Zone.
Poynt: The current is one over "L" integral v dt.
That's 1/5 * integral (4) dt.
That's 1/5 * 4t.
That's 4/5*t.
When t = 3 seconds that's 12/5 = 2.4 amps.
Can we get out of the Zone now?

What does old Lewin think?
YT Kirchoffs rule is for the birds. Stripey shirt 16.01.

Perhaps the sim skids out at these lower R settings, and defaults to 0 rather than crashing as it does with no resistor at all.
It does seem to settle at the 2.4A point when R is relatively small though. So, yeah it must be right.
It doesn't makes sense to me from the infinite tau perspective though.

The Tau business is fairly simple to explain.
When you transition from a finite Tau to an infinite Tau the current waveform goes from an inverse exponential curve to a straight line. Note that it is a straight line with a constant slope of V/L.
So Tau at infinity just means the current trace is a perfectly straight line. Since it is a straight line the concept of "reaching 63% of the maximum value" does not apply anymore because that concept does not exist when the current waveform is a perfectly straight line. i.e.; "There is no time constant."
So Tau being infinity does not mean stopping current flow, it means linearly increasing current flow.
Since we are discussing limits, the only possible way for the current to flatline at zero "forever" would be for the inductance to be infinity. Then you have a "more real" Tau = infinity because this time L/R becomes infinity/R.
So when Tau = infinity/R that gives you the horizontal current trace stuck at zero with a slope of zero (V/infinity), whereas when Tau = L/0, you get a current trace that is a straight line with with a slope of V/L.

Thanks for the explanation MH. ;)

My idea was to stick in a series resistor of 166666666 Ohms.
John.

For Brad:
Don't get put off with what may look like a bunch of Engineering mumbojumbo talk. I never intended to go in this direction. All of the mumbojumbo talk can be expressed in ordinary simple English that anybody that wants to learn can truly learn. I am glad that you are all fired up about this subject.
So I am going to ask you again to rethink your stance. Accept the fact that at 3 seconds the current has climbed to 2.4 amps in a perfectly straight line. Assuming that you do accept this fact, then the next step for you and your peers is to figure out how and why, and how you can apply this to real life.
I am still hoping that you and your peers will be able to answer the question and explain why. At this point the thread has been littered with clues. The bestcase scenario would be for you to leverage off of this information and brainstorm with your friends and answer the simple question about the current waveform and demonstrate that you understand the how and why of what's going on.
MileHigh

Oh, when I do 4/1.66666 I get something like 2.4A. Amazing!
John.

The Tau business is fairly simple to explain.
When you transition from a finite Tau to an infinite Tau the current waveform goes from an inverse exponential curve to a straight line. Note that it is a straight line with a constant slope of V/L.
So Tau at infinity just means the current trace is a perfectly straight line. Since it is a straight line the concept of "reaching 63% of the maximum value" does not apply anymore because that concept does not exist when the current waveform is a perfectly straight line. i.e.; "There is no time constant."
So Tau being infinity does not mean stopping current flow, it means linearly increasing current flow.
Since we are discussing limits, the only possible way for the current to flatline at zero "forever" would be for the inductance to be infinity. Then you have a "more real" Tau = infinity because this time L/R becomes infinity/R.
So when Tau = infinity/R that gives you the horizontal current trace stuck at zero with a slope of zero (V/infinity), whereas when Tau = L/0, you get a current trace that is a straight line with with a slope of V/L.
is it really that simple to you? how can this not be construed as utterly confusing and even contradictory? I suppose I just have it completely backwards, but isn't our ability to calculate the current at a given time based upon the constant? but yet we remove the constant and untether it to infinity like what would happen to any normal inductor without a time constraint with which to calculate it? how do we know the rate of charge would follow a straight line incline, isn't the magnetic field supposed to be able to follow an equally opposing one making it just a flatline forever?
if you were to 'zoom in' at roughly the 99.5% point of the inductors charge you may see that you have only just begun charging it. it would almost be like looking at this examples straight incline with an infinite or 'no time constraint'. the idea of 'infinity' or 'zero' seem to introduce some natural 'stalemate' as another user put it. we are expected to look at 0 as a balance of opposing forces, or a neutral. an atrest. yet in this case the 0 is just supposed to represent a lack thereof as if there is a difference. it's not something I'm tying to argue to death it's just very counter intuitive in my opinion sorry of that offends you.

Ok,im going to need help understanding this Poynt
The zero current meaning no load is no problem,but having an infinite load(a load of infinite proportions),would this not require a current flow of infinite magnitude?,and we have no current flow.
Brad
Personally I think the term: "infinite load" is just silly.
I would prefer the term: "infinite resistance" as Current (I) is directly proportional to the Resistance (R) of the load and the Voltage (V) applied accross the Load. Ohms Law Exquation: V/R = Current (I)
With: "infinite resistance", no current can flow.
I know this is what Poynt meant, but this mix in words is confusing for others that dont know.
Chris Sykes
hyiq.org

is it really that simple to you? how can this not be construed as utterly confusing and even contradictory? I suppose I just have it completely backwards, but isn't our ability to calculate the current at a given time based upon the constant? but yet we remove the constant and untether it to infinity like what would happen to any normal inductor without a time constraint with which to calculate it? how do we know the rate of charge would follow a straight line incline, isn't the magnetic field supposed to be able to follow an equally opposing one making it just a flatline forever?
if you were to 'zoom in' at roughly the 99.5% point of the inductors charge you may see that you have only just begun charging it. it would almost be like looking at this examples straight incline with an infinite or 'no time constraint'. the idea of 'infinity' or 'zero' seem to introduce some natural 'stalemate' as another user put it. we are expected to look at 0 as a balance of opposing forces, or a neutral. an atrest. yet in this case the 0 is just supposed to represent a lack thereof as if there is a difference. it's not something I'm tying to argue to death it's just very counter intuitive in my opinion sorry of that offends you.
For starters, just forget about all of the talk about an inductor being an infinite value. The take away from this is as follows: As the resistance gets lower and lower, the time constant gets longer and longer, and the inverse exponential curve for the current waveform gets flatter and flatter. When the resistance drops to zero it becomes a straight line and then the concept of a time constant becomes invalid.
So, how much of a mind bender is it really to go from a sloped current waveform that is nearly perfectly flat and you only start to detect a slight curve in the waveform after five hours and a sloped current waveform that is perfectly straight?
What you really want to do if you are interested is work with your peers and focus on trying to understand what the circuit is doing and and answering the question. Focus on doing the research and trying to answer the question. If you know what you are talking about, this is a very simple problem with a very simple explanation.
Also as a side note, if you are serious, you have to work on your language and start using commonly accepted terms and phrases to describe electronic circuits. You need to reign in the meaningless expressions and the mixing up of variables and concepts and use the proper term for the proper concept.

Well this is what I wanted to do from the start, but not with this imaginary test circuit.
How about we do that with the original circuit in question, with your input included?
no offense but I didn't come here to simulate how your test circuit may possibly function.
I don't really want to argue what 0 means in the real world either.
if we were to let's say, graph out the joule thiefs inductor voltage to current ratio over time until
we can see how many seconds it takes to become what we can consider fully charged, is that going to show us what's firing the transistor? according to what it seems like you were originally alluding to, then no. in order to really make use of this lesson on inductors and how we can understand what may be happening in let's say, our joule thief circuits, we need to add more information into the scenario than inductance and voltage in ideal terms. there are a lot of different ways inductors can act under varying circumstances and to assume some simple ideal DC over an ideal inductor of 5H scenario really paints some kind of picture as to 'how we should proceed' fine tuning our joule thief seems like a lesson tailored more toward spiteful cerebralmeasuring than genuine intentions of shedding light on the topic (that was) at hand.

Brad:
I am going to do a preemptive strike to get you off of your fixation so that you start to move forward and both answer the question but more importantly understand the answer and the underlying concepts.
For the first part of the question the cat is out of the bag and the formula for the current is i = 0.8*t.
It seems crazy doesn't it? All that arguing and anguish and wild imaginary theories and the answer for the first three seconds is i = 0.8*t.
Okay, so after one second the current is 0.8 amperes.
I attached the formula from Hyperphysics for a voltage source energizing an RL circuit. You can see the red trace for the current and the formula. You can see the "V/R" in the formula and that might tempt you to say that that's proof that the current goes to infinity but that would be dead wrong. The red formula does not work when the resistance becomes zero because the term "V/0" is invalid and undefined.
So now I am going to prove to you that the first part of the answer given to you is correct. We are going to solve the formula for t=1 second, L=5 Henrys and R=0.001 ohms. Sound fair enough? If I am correct the two answers for the current value at one second should be very close to each other.
From the red formula:
i = 4/0.001 * (1  2.7182818^(0.001/5))
i = 4000 * (1  2.7182818^0.0002)
i = 4000 * (1  0.99980001999866673333)
i = 4000 * 0.000199980013332666693332
i = 0.799920005333 amperes.
So as you can see, when we introduced a resistor of 0.001 ohms, after one second the current was just a tiny tiny smidgen less than the 0.8 amperes you get when there is no resistor. You note that the calculation for when there is no resistor is 10 times easier as compared to when there is a resistor.
So there is your proof in the pudding. It's time for you to move on and try to answer the question and more importantly understand the how and the why.
MileHigh

The Tau business is fairly simple to explain.
When you transition from a finite Tau to an infinite Tau the current waveform goes from an inverse exponential curve to a straight line. Note that it is a straight line with a constant slope of V/L.
So Tau at infinity just means the current trace is a perfectly straight line. Since it is a straight line the concept of "reaching 63% of the maximum value" does not apply anymore because that concept does not exist when the current waveform is a perfectly straight line. i.e.; "There is no time constant."
So Tau being infinity does not mean stopping current flow, it means linearly increasing current flow.
Since we are discussing limits, the only possible way for the current to flatline at zero "forever" would be for the inductance to be infinity. Then you have a "more real" Tau = infinity because this time L/R becomes infinity/R.
So when Tau = infinity/R that gives you the horizontal current trace stuck at zero with a slope of zero (V/infinity), whereas when Tau = L/0, you get a current trace that is a straight line with with a slope of V/L.
Im sorry MH,but that dose not work either,as the V has no slope ,due to that fact that the voltage is ideal.
I will once again post the definition of an ideal voltage.
Quote: An ideal voltage source is a twoterminal device that maintains a fixed voltage drop across its terminals. It is often used as a mathematical abstraction that simplifies the analysis of real electric circuits.
This is the definition that is used to define behaviours in circuits such as yours.
The voltage of 4 volts from your ideal voltage supply,will not alter during the 3 seconds after T=0.
Regardless of whether it is L/0 or R/0,Tau is always infinite,meaning that the current will not rise in the case of an ideal inductor.
You are trying to use real world calculations to define outcomes in an ideal situation that dose not exist,and then you are surprised by the outcome.
Poynts simulation shows the results expected,and if he ran that simulation for !lets say! 100 years,we would see the current rise expected for such a low value of resistancethere is nothing wrong with his sim.
Brad

Well this is what I wanted to do from the start, but not with this imaginary test circuit.
How about we do that with the original circuit in question, with your input included?
no offense but I didn't come here to simulate how your test circuit may possibly function.
I don't really want to argue what 0 means in the real world either.
if we were to let's say, graph out the joule thiefs inductor voltage to current ratio over time until
we can see how many seconds it takes to become what we can consider fully charged, is that going to show us what's firing the transistor? according to what it seems like you were originally alluding to, then no. in order to really make use of this lesson on inductors and how we can understand what may be happening in let's say, our joule thief circuits, we need to add more information into the scenario than inductance and voltage in ideal terms. there are a lot of different ways inductors can act under varying circumstances and to assume some simple ideal DC over an ideal inductor of 5H scenario really paints some kind of picture as to 'how we should proceed' fine tuning our joule thief seems like a lesson tailored more toward spiteful cerebralmeasuring than genuine intentions of shedding light on the topic (that was) at hand.
I will repeat to you, the circuit is only "imaginary" if you have a bad attitude and refuse to open up your mind and learn something new and you refuse to apply your knowledge. I will repeat to you, that there is almost no difference between an inductor with zero ohms resistance and an inductor with 0.001 ohms resistance and I just proved it to you and Brad in my previous posting where I did the "real" calculation.
If you can't understand how closely an ideal inductor matches a real inductor and how much can be learned from trying to understand how an ideal inductor works then read over the thread again. I am not here to argue that anymore.
<<< if we were to let's say, graph out the joule thiefs inductor voltage to current ratio over time until
we can see how many seconds it takes to become what we can consider fully charged >>>
The quoted sentence above does not really make any sense and that is the whole point of this exercise. Understand how an inductor works so you can talk about using one sensibly and also understand how a Joule Thief works and then build better Joule Thieves.
Yes there are different ways of using inductors in circuits. A good start is right here. This example applies directly to the energizing phase of a Joule Thief. If you understand the basics then moving to regular inductors is easier. And like I told you in many circuits real inductors for all practical intents and purposes act exactly like ideal inductors.
"Spiteful cerebralmeasuring" is you just being defensive because you are in the realm of real electronics here, not the usual throwing around of meaningless catch phrases.
The best thing you could do for yourself is have the same spark as Brad and follow it through and learn something real and substantial about electronics. I will remind you, you read the question and you did not have the slightest idea what to do. I am hoping that you and Brad arrive at a successful conclusion to this debate.
I am not going to touch any Joule Thief thread, sorry.

Brad:
I am going to do a preemptive strike to get you off of your fixation so that you start to move forward and both answer the question but more importantly understand the answer and the underlying concepts.
For the first part of the question the cat is out of the bag and the formula for the current is i = 0.8*t.
It seems crazy doesn't it? All that arguing and anguish and wild imaginary theories and the answer for the first three seconds is i = 0.8*t.
Okay, so after one second the current is 0.8 amperes.
I attached the formula from Hyperphysics for a voltage source energizing an RL circuit. You can see the red trace for the current and the formula. You can see the "V/R" in the formula and that might tempt you to say that that's proof that the current goes to infinity but that would be dead wrong. The red formula does not work when the resistance becomes zero because the term "V/0" is invalid and undefined.
So now I am going to prove to you that the first part of the answer given to you is correct. We are going to solve the formula for t=1 second, L=5 Henrys and R=0.001 ohms. Sound fair enough? If I am correct the two answers for the current value at one second should be very close to each other.
From the red formula:
i = 4/0.001 * (1  2.7182818^(0.001/5))
i = 4000 * (1  2.7182818^0.0002)
i = 4000 * (1  0.99980001999866673333)
i = 4000 * 0.000199980013332666693332
i = 0.799920005333 amperes.
So as you can see, when we introduced a resistor of 0.001 ohms, after one second the current was just a tiny tiny smidgen less than the 0.8 amperes you get when there is no resistor. You note that the calculation for when there is no resistor is 10 times easier as compared to when there is a resistor.
So there is your proof in the pudding. It's time for you to move on and try to answer the question and more importantly understand the how and the why.
MileHigh
I have filled in the blanks in your diagram,and also noted that there will be no voltage curve when the voltage is ideal,and set at 4 volts for 3 secondsas per your original question.
I am pretty sure Poynt has not yet finished looking into this,and i think you will find that i am correct when i say that your ideal voltage of 4 volts across the ideal inductor for the first 3 seconds from T=0 ,remains 4 volts,as an ideal voltage dose not change in value over timeregardless of the load,and so there is no such voltage curve as you have depicted in your diagram .
Brad

<<< Im sorry MH,but that dose not work either,as the V has no slope ,due to that fact that the voltage is ideal. >>>
There is no slope associated with the voltage so I don't know what you are talking about. Forget the Tau business for now, it's not relevant.
Read again, Poynt stated that his sim did not run right the first time and when he tweaked the R value it ran as expected and he is in agreement with me now.
The best thing you could do for yourself is followthough and answer the complete question and demonstrate competence in the subject matter. You have been given a very generous start, now it's up to you and your peers to follow through. You still have a long way to go.
The question still has not been answered....

I will repeat to you, the circuit is only "imaginary" if you have a bad attitude and refuse to open up your mind and learn something new and you refuse to apply your knowledge. I will repeat to you, that there is almost no difference between an inductor with zero ohms resistance and an inductor with 0.001 ohms resistance and I just proved it to you and Brad in my previous posting where I did the "real" calculation.
If you can't understand how closely an ideal inductor matches a real inductor and how much can be learned from trying to understand how an ideal inductor works then read over the thread again. I am not here to argue that anymore.
<<< if we were to let's say, graph out the joule thiefs inductor voltage to current ratio over time until
we can see how many seconds it takes to become what we can consider fully charged >>>
The quoted sentence above does not really make any sense and that is the whole point of this exercise. Understand how an inductor works so you can talk about using one sensibly and also understand how a Joule Thief works and then build better Joule Thieves.
Yes there are different ways of using inductors in circuits. A good start is right here. This example applies directly to the energizing phase of a Joule Thief. If you understand the basics then moving to regular inductors is easier. And like I told you in many circuits real inductors for all practical intents and purposes act exactly like ideal inductors.
"Spiteful cerebralmeasuring" is you just being defensive because you are in the realm of real electronics here, not the usual throwing around of meaningless catch phrases.
The best thing you could do for yourself is have the same spark as Brad and follow it through and learn something real and substantial about electronics. I will remind you, you read the question and you did not have the slightest idea what to do. I am hoping that you and Brad arrive at a successful conclusion to this debate.
I am not going to touch any Joule Thief thread, sorry.
why do you always do that? is this sentence REALLY that hard for you to understand?
"<<< if we were to let's say, graph out the joule thiefs inductor voltage to current ratio over time until we can see how many seconds it takes to become what we can consider fully charged >>>"
is that really such a laymans way of stating it that you simply refuse to attempt comprehending it?
What are we doing in this exercise? Are we not basically graphing an inductor charging according to a time constant? (which in this case doesn't exist) you just happen to use a 5H inductor, nowhere near what the JT inductor is in Brads circuit, and a 0 resistance which has caused a conundrum you are still not understanding the nature of. And yet and still you are claiming to know how an inductor with 0 resistance will act.
I'm literally stating what we just did in practice by analyzing this exercise, and am comparing it to how we can use this same circuit theory analysis to gain an understanding about what is going on
with the joule thief circuit you no longer wish to shed light on.
I'm not going to keep telling you that I'm not attacking your reasoning for using such theoretical models.. It's common practice for even the greenest of EE students. I keep telling you it's not helping us understand the confusion with Brads circuit. You could have easily generated a different question that was more tailored to that specific circuit if you had some revelation you wished to portray in lesson form.

Below i have added the correct voltage and current traces as would be seen using an ideal voltage,and an ideal inductor.
You will note that it is also the same as what Poynt's sim showed during his 100 second run time.
Brad
P.S
I have incorrectly worded current trace !across! inductor,when it should be current through inductoras im sure you are well aware of.

I have filled in the blanks in your diagram,and also noted that there will be no voltage curve when the voltage is ideal,and set at 4 volts for 3 secondsas per your original question.
I am pretty sure Poynt has not yet finished looking into this,and i think you will find that i am correct when i say that your ideal voltage of 4 volts across the ideal inductor for the first 3 seconds from T=0 ,remains 4 volts,as an ideal voltage dose not change in value over timeregardless of the load,and so there is no such voltage curve as you have depicted in your diagram .
Brad
Read my posting again and understand it, I solved for R=0.001 ohm so I was not discussing an ideal inductor there.

Magneticitist:
Sorry for being picky but I am not going to discuss Joule Thieves on this thread. For the main coil in a Joule Thief, there is no such thing as a "inductor voltage to current ratio" and I am making a point to you about communicating effectively. How can you expect to talk shop about Joule Thieves if the person you are talking to is a beginner that would not have a clue what you are saying when you say "inductor voltage to current ratio" which itself doesn't really make sense. I won't go after you anymore about this but now you are fully aware.
Note, this thread is all about the energizing of an inductor. My advice to you is to forget about Joule Thieves on this thread and focus on trying to answer the question and understand what is going on.
MileHigh

Read my posting again and understand it, I solved for R=0.001 ohm so I was not discussing an ideal inductor there.
Good Lord man... did you not just ever so thoroughly stress there is no real difference?
What is the point of this? Why would you even continue to prod him into solving for R=.001 if you are
satisfied with his answer for R=0? We are now practicing an exercise in futility.
I see maybe you are trying to prove a point? show that there is no real difference?
but once again, it wouldn't. you cannot compare two things by only having one of them.

<<< Im sorry MH,but that dose not work either,as the V has no slope ,due to that fact that the voltage is ideal. >>>
The best thing you could do for yourself is followthough and answer the complete question and demonstrate competence in the subject matter. You have been given a very generous start, now it's up to you and your peers to follow through. You still have a long way to go.
The question still has not been answered....
Read again, Poynt stated that his sim did not run right the first time and when he tweaked the R value it ran as expected and he is in agreement with me now.
That is correct,his sim ran correct the second time when he placed a very small resistance in with the ideal inductor,and the result was no flow of current.
I have not seen Poynt agree with you. I read This dose not agree with the Tau being infinity,andthank you for the explanation.
There is no slope associated with the voltage so I don't know what you are talking about. Forget the Tau business for now, it's not relevant.
Your diagram shows a clear slope in the voltage across the inductor.
How can Tau not be relevant?.
Brad

Read my posting again and understand it, I solved for R=0.001 ohm so I was not discussing an ideal inductor there.
Please solve for R=0,so as we are defined to the parameters of your original question,as i have shown you already the infinite difference this makes even at a resistance far less than .001 ohms
Brad

Good Lord man... did you not just ever so thoroughly stress there is no real difference?
What is the point of this? Why would you even continue to prod him into solving for R=.001 if you are
satisfied with his answer for R=0? We are now practicing an exercise in futility.
I see maybe you are trying to prove a point? show that there is no real difference?
but once again, it wouldn't. you cannot compare two things by only having one of them.
If you read the posting again you will see that I compared the 5 Henry coil with no resistor (ideal) and the 5 Henry coil with a 0.001 ohm resistor (real) and I clearly show that the behaviour between the two coils is nearly identical at t=1 second.

Magneticitist:
Sorry for being picky but I am not going to discuss Joule Thieves on this thread. For the main coil in a Joule Thief, there is no such thing as a "inductor voltage to current ratio" and I am making a point to you about communicating effectively. How can you expect to talk shop about Joule Thieves if the person you are talking to is a beginner that would not have a clue what you are saying when you say "inductor voltage to current ratio" which itself doesn't really make sense. I won't go after you anymore about this but now you are fully aware.
Note, this thread is all about the energizing of an inductor. My advice to you is to forget about Joule Thieves on this thread and focus on trying to answer the question and understand what is going on.
MileHigh
well this is the golden point we have both made now. this thread was made out of respect to your question.. a question Brad never had to make any attempt at answering.. A question that was conceived by you to somehow prove your view on his circuit was correct and superior. That was a joule Thief circuit. at least, a circuit using a transformer. Why you decided to then go into an inductors 101 is obviously beyond me as we have clearly discussed I am out of my element just here to throw around catch phrases and impress total strangers.
I have been trying to figure your reason for bringing up this inductor lesson since I first read about
your question and I still don't get how it could possibly relate to the original conversation.
I understand you want to knit pick such statements as 'voltage to current ratio' but I don't see how that is confusing.. at the time it seemed a fairly straightforward way to word it. I'm pretty sure I could come up with a number of ways to literate an inductor charging, and how it has a voltage unit and current unit we could measure.. I don't think you realize saying such things to me doesn't offend me. I can only gather that conclusion since you so frequently want to point out that all of my thoughts here are just random things I pulled out of my buttcrack on the spot for jollies that not even the most remedial of minds can understand.

Please solve for R=0,so as we are defined to the parameters of your original question,as i have shown you already the infinite difference this makes even at a resistance far less than .001 ohms
Brad
You are shooting blanks, we are not discussing resistors. What I did was compare the behaviour of the ideal 5 Henry coil and a real 5 Henry coil (0.001 series resistance) and showed you that their behaviour is nearly identical at t=1 second. Do you get that?
If you get that then it proves to you that the formula for the current through the ideal coil of i = 0.8*t is correct.
In other words the explained behaviour for the ideal coil is not a short circuit or an open circuit, it's i = 0.8t. Do you get that?
If you get all of this then you should move forward and try to answer the full question and demonstrate that you know what you are doing.

Magneticitist:
Stop the psychobabble and the drama, ok? The thread is here and Brad started it and that's all there is to it.
Do you want to contribute to it and learn? Then work with Brad and see what you can come up with. Try to answer the question and use that as a vehicle to learn.
MileHigh

You are shooting blanks, we are not discussing resistors. What I did was compare the behaviour of the ideal 5 Henry coil and a real 5 Henry coil (0.001 series resistance) and showed you that their behaviour is nearly identical at t=1 second. Do you get that?
If you get that then it proves to you that the formula for the current through the ideal coil of i = 0.8*t is correct.
In other words the explained behaviour for the ideal coil is not a short circuit or an open circuit, it's i = 0.8t. Do you get that?
If you get all of this then you should move forward and try to answer the full question and demonstrate that you know what you are doing.
Tau=L/R is how the time constant for the rise of current flow is determined,and that time constant has a value of infinity,and only changes to a finite value when a small amount of resistance is added to the inductor.
The second thing to ask your self,is how is the inductor defined as being a value of 5 Henry's when you have no resistive value to make that definition?. As you can see,the inductance in all three example in the link below,rely on measuring a voltage drop across a resistance to obtain the current flow to obtain inductive value. So how is this done when there is no resistance? When you work this out,you may begin to understand as to why an ideal inductor dose not dissipate any power,which also lines up with your very own statement that a voltage cannot be measured across an ideal inductor when a DC current if flowing through it,as current alone is not power,and hence the reason that an ideal inductor dose not dissipate power. But now your question also posses the problem that we !do! have an ideal voltage across the ideal inductor that you say has a DC current flowing through it. It can only have a voltage across it if there is no current flowing through itand visa versa .
http://www.wikihow.com/MeasureInductance
It is like my answer says it isyou cannot place an ideal voltage across an ideal inductor,as an ideal inductor dose not exist.
I have also shown that regardless of how little the resistance value may be,it will lead to a value that is infinitely different to that of an ideal inductor that has no R value.
Brad

What does DC current have to do with anything when you put voltage across an ideal coil? You will have to contemplate that.
You have been given more than enough information to do some hunting and researching on your own so that you answer the question correctly. You still haven't taken a stab at the full question.
After all this, it still feels to me like there is faint hope that you are going to achieve those two tasks I listed for you with your current attitude. You never know though like I already said. I think that you are going to have to hope a guru comes in to help you because I am jumping off the train and switching to passive observer mode.
I hope one day in the not too distant future to see you followup on both tasks successfully and demonstrate full competence in this very important subject matter.
MileHigh

I'm just an oaf but I get what you're saying.. an inductor requires at least some form of 'resistance' to have a measurable inductance otherwise we are just saying "hey this is a 5H inductor" and base that entirely off of speculation.

What does DC current have to do with anything when you put voltage across an ideal coil? You will have to contemplate that.
You have been given more than enough information to do some hunting and researching on your own so that you answer the question correctly. You still haven't taken a stab at the full question.
After all this, it still feels to me like there is faint hope that you are going to achieve those two tasks I listed for you with your current attitude. You never know though like I already said. I think that you are going to have to hope a guru comes in to help you because I am jumping off the train and switching to passive observer mode.
I hope one day in the not too distant future to see you followup on both tasks successfully and demonstrate full competence in this very important subject matter.
MileHigh
The fact that you have dismissed the L/R time constant to answer your original question is troubling.You have dismissed using this methodwhich is used in any other situation,based on the fact that there is no R value. You know that if you did use this conventional method,that the result would be exactly as myself and Poynt said it would bethere would be no flow of current,due to the infinite time constant value. You have dismissed the method that gives us the time constant required to answer your question correctly. This method (Tau=L/R) is the correct method to use in regards to your question. The fact that it results in no current flowing through the ideal inductor ,is inline with the fact that an ideal inductor dose not dissipate energy. Also to back it up is the fact that we have an ideal voltage of 4 volts across that inductor for a period of 3 seconds. This also states that during this time,no current can be flowing through that inductor,as there is a voltage across it,and as you your self have stated,if a DC current was flowing through that inductor,then there would be no voltage across itbut we have 4 volts across it,and so no current is flowing through it. This also falls in line with Poynts sim test.
All circuits used to make this definition are based around an ideal inductor,but with an added series resistor to simulate the resistance that would exist in a real world inductor. If we are to define the outcome of placing an ideal voltage across an ideal inductor,then there is no reason at all that the series resistor just be removed to obtain the correct evaluation for an ideal inductor.
The only reason you do not wish to use this method of Tau=L/R,is because that then puts you in a position of being incorrect. this being the case,i find that a sad time indeed in the history of this forum,where people come to learn and solve such problems using what we know and undderstand to be correct.
Quote: http://www.learnaboutelectronics.org/ac_theory/dc_ccts45.php
When a current is applied to an inductor it takes some time for the current to reach its maximum value, after which it will remain in a "steady state" until some other event causes the input to change. The time taken for the current to rise to its steady state value in an LR circuit depends on:
•The resistance (R)
This is the total circuit resistance, which includes the DC resistance of the inductor (RL) itself, plus any external circuit resistance.
• The inductance of L
Which is proportional to the square of the number of turns, the cross sectional area of coil and the permeability of the core.
I am standing firm on my answers,and i hope Poynt(and others) takes the time to have another look at this,and not just accept your example as a reality.
Brad

guys let's just make a deal and be honest. MH wants to act like you are a complete loony for believing what you do, but I'm fairly sure this is not some extremely rare argument and has been argued before, probably to this very day by many others.. there are others in this thread that share Brads view as well. does that mean MH is wrong? of course not. but until some kind of all ending smoking gun evidence can be presented (which it cannot) what is the point? ok let's say we're using a real R value then. still, what's the point? even plugging in the numbers and calculating for current or the rate of current change, or the voltage, or whatever.. at this point it's just going through the motions.
How about we agree to make a real conscious effort learning about inductors the same way MH did when he was being indoctrinated so long as he agrees to stop using circuit examples with values of 0 and expect a real provable formulaic outcome or omitting variables from formulas altogether.
Then hopefully we can go back to the original topic loool. I'm genuinely interested I mean when do you see me here, that's what I came for and for like 10 pages there it looked like you guys were creeping ever closer to a real provable answer which is what I thought MH had up his sleeve with this whole question thing but oh well can we move on now.

guys let's just make a deal and be honest. MH wants to act like you are a complete loony for believing what you do, but I'm fairly sure this is not some extremely rare argument and has been argued before, probably to this very day by many others.. there are others in this thread that share Brads view as well. does that mean MH is wrong? of course not. but until some kind of all ending smoking gun evidence can be presented (which it cannot) what is the point? ok let's say we're using a real R value then. still, what's the point? even plugging in the numbers and calculating for current or the rate of current change, or the voltage, or whatever.. at this point it's just going through the motions.
How about we agree to make a real conscious effort learning about inductors the same way MH did when he was being indoctrinated so long as he agrees to stop using circuit examples with values of 0 and expect a real provable formulaic outcome or omitting variables from formulas altogether.
Then hopefully we can go back to the original topic loool. I'm genuinely interested I mean when do you see me here, that's what I came for and for like 10 pages there it looked like you guys were creeping ever closer to a real provable answer which is what I thought MH had up his sleeve with this whole question thing but oh well can we move on now.
I started this thread to remove the debate on MHs ideal coil and voltage question from the JT 101 thread,so as discussion on the topic of JTs could continue without unrelated topics being discussed there,and splitting up the thread. But it seems that now capacitors has also taken hold in that threadperhaps a dedicated thread on capacitors should also be opened.
Brad

I will drop in and say this out of frustration:
1. Talking about DC current through an ideal coil and no resultant voltage drop is meaningless and has nothing to do with this exercise.
2. When you apply voltage across an ideal coil you get changing and increasing current. This simple fact is escaping you and as a result you are leading yourself down a garden path. Just like when you apply voltage across a real coil you get changing and increasing current until the current levels off.
It's pull yourself up by your own bootstraps time, or stick in mud and be wrong and go nowhere time, or wait for a guru to save you time.

This is true I don't mean to disrespect this thread or MH's wishes to avoid further JT discussion... but I have to ask since you two are honestly the best people to ask at this point.. I was eagerly reading
your discussion in the JT thread before this topic and didn't quite realize how far off topic it was actually leading. I guess I may have to try to steal the JT thread back on topic but then I don't want to interrupt the capacitor talk after it has gained some charge. this topic actually has merit too though I suppose, I just don't see how one side could really disprove the other.

Ideal inductors do exist in our society today. I am certainly no expert in the subject but examine the superconducting electromagnets used in MRI. Once below their critical superconducting temperature, certain materials exhibit zero resistance and yet maintain inductance. The inductive fields exist outside the confines of the wire but they do exist. A 5 henry superconducting coil is still 5 henries even with zero resistance and they will "store" their current with zero voltage drop for extremely long periods of time. There are qualifications for these ideal conductors to work as they do but they are in use everyday.
Resistance of a coil does not determine it's inductance, it simply hinders pure inductance.
I have attached another sim using a coil resistance of 1e110. This parameter may have passed a preset limit in LtSpice however.
partzman

The Tau business is fairly simple to explain.
When you transition from a finite Tau to an infinite Tau the current waveform goes from an inverse exponential curve to a straight line. Note that it is a straight line with a constant slope of V/L.
So Tau being infinity does not mean stopping current flow, it means linearly increasing current flow.
Since we are discussing limits, the only possible way for the current to flatline at zero "forever" would be for the inductance to be infinity. Then you have a "more real" Tau = infinity because this time L/R becomes infinity/R.
So when Tau = infinity/R that gives you the horizontal current trace stuck at zero with a slope of zero (V/infinity), whereas when Tau = L/0, you get a current trace that is a straight line with with a slope of V/L.
MH
Your explanation above only explains the trace or wave form seen on the scope. It dose not explain away the actual resultant math value. The math is precise,and defines the actual time taken for the current to rise,and mathematically that time is infiniteyou cannot redefine math at your will.
So Tau at infinity just means the current trace is a perfectly straight line. Since it is a straight line the concept of "reaching 63% of the maximum value" does not apply anymore because that concept does not exist when the current waveform is a perfectly straight line. i.e.; "There is no time constant."
Tau at infinity dose not just mean the current trace is just a flat line. Tau at infinity is a mathematical calculated value using the the equation that is always used to calculate the Tau time constant> Tau=L/R.
There is a time constant,and that time constant is infinity. A flat line on a scope dose not just dismiss this mathematically calculated time constant. In fact,that flat line seen on the scope ,represents the infinite flat line that would be seen as an infinite time value. If the scope had an infinitely long screen,then you would see that trace remain flat at a zero value throughout the entire time a voltage is placed across the ideal inductor.
Brad

So I guess the question is how much time does it take for an ideal inductor to reach a particular current over time if resistance was not an obstacle. This would mean that the ideal inductor is functional as an inductor.....
Well the ideal inductors current rise when the ideal input voltage is applied will be a straight line increase and not a curve because the absence of voltage division because of no resistance. So the current could rise indefinitely over time, directly related to time and the resistance value does not need to be in the equation L/R. Correct?
That is if the ideal inductors bemf ends up not being equal to the input and the ideal inductor actually works.
Mags

the superconductors are a good example but they are like comparing 0 resistance with .0000001 or any of the other ridiculously low resistances mentioned to sim with.
and how do we know if R was 0 that line wouldn't be level flat or plumb straight up.
when we super cool the conductors we are probably just reallocating that resistive variable elsewhere. maybe much less measurable resistance but maybe a much less hindered ability to create
a field. idk I'm no expert either lol. at any rate it makes me think of how the curie point is sort of like
the opposite effect and maybe we can take magnetic inductor with very little resistance and heat it to a point where it has very high resistance and has a harder time creating a field if that doesn't sound stupid.

So I guess the question is how much time does it take for an ideal inductor to reach a particular current over time if resistance was not an obstacle. This would mean that the ideal inductor is functional as an inductor.....
Well the ideal inductors current rise when the ideal input voltage is applied will be a straight line increase and not a curve because the absence of voltage division because of no resistance. So the current could rise indefinitely over time, directly related to time and the resistance value does not need to be in the equation L/R. Correct?
That is if the ideal inductors bemf ends up not being equal to the input and the ideal inductor actually works.
Mags
yea my beef is if the ideal coil aka absolute 0 R coil cannot dissipate or radiate anything whatsoever, then to me that's like another way of saying it has immeasurable current. infinite, 0, immeasurable either way and might as well be non existent. you would never have the field or current to begin with and the lack of a time constant would just basically make all the equations equal 0 since progress over time is out of the window and no longer even discernible from 0 time progression at all. it's like we are talking about trying to apply current to something that can perfectly and absolutely resist current change aren't we?

Ideal inductors do exist in our society today. I am certainly no expert in the subject but examine the superconducting electromagnets used in MRI. Once below their critical superconducting temperature, certain materials exhibit zero resistance and yet maintain inductance. The inductive fields exist outside the confines of the wire but they do exist. There are qualifications for these ideal conductors to work as they do but they are in use everyday.
Resistance of a coil does not determine it's inductance, it simply hinders pure inductance.
I have attached another sim using a coil resistance of 1e110. This parameter may have passed a preset limit in LtSpice however.
partzman
A 5 henry superconducting coil is still 5 henries even with zero resistance and they will "store" their current with zero voltage drop for extremely long periods of time.
Partzman
A super conductor that has a steady DC current flowing through it,has no voltage across it. If it did,then it would be dissipating power,and super conductors do not dissipate power.
Brad

So I guess the question is how much time does it take for an ideal inductor to reach a particular current over time if resistance was not an obstacle. This would mean that the ideal inductor is functional as an inductor.....
Well the ideal inductors current rise when the ideal input voltage is applied will be a straight line increase and not a curve because the absence of voltage division because of no resistance. So the current could rise indefinitely over time, directly related to time and the resistance value does not need to be in the equation L/R. Correct?
That is if the ideal inductors bemf ends up not being equal to the input and the ideal inductor actually works.
Mags
You have nailed it on the heador near to Mag's.
Unfortunately MH is just not getting it,and he is trying to use a math function that dose not account for the voltage and inductor on being ideal.
Well now I feel like I am in the Twilight Zone.
Poynt: The current is one over "L" integral v dt.
That's 1/5 * integral (4) dt.
That's 1/5 * 4t.
That's 4/5*t.
When t = 3 seconds that's 12/5 = 2.4 amps.
The above is not applicable to an ideal inductor with an ideal voltage across it.
The math above is based on the premise that the inductor will eventually reach a maximum current value in Tau x 5s. We already know that by using the L/R time constant,that Tau is infinity.
We can also solve this a second way. That is to place the ideal voltage across the ideal inductor,and time how long it takes for the maximum current value to be reached. We then divide this time by 5 to obtain our Tau time constant. This also results in an infinite time,as the voltage is ideal,and the inductor has no resistance. It also means once again that there will be no current flowing through the ideal inductor :o.
As i said,and have all alongyou cannot place an ideal voltage across an ideal inductor,because as you see,you are left with a paradox.
If an ideal voltage is placed across an ideal inductor(that has no resistance to control the flow of current),then the current would take an infinite amount of time to reach it's peak level.
This then means that the current would also take an infinite amount of time to start to flow into that inductor at T=0,as when you divide an infinite amount of time by any other amount of time,you end up with an answer that is also infinite.
So that is the paradox,but it is also correct,and once again backs up all my answers i have given in regards to the original question.
Even MH cannot deny that it would take an infinite amount of time for the current to reach maximum value,when an ideal voltage(a voltage that dose not change over time) is placed across an ideal inductor that has no resistance(dose not limit or control the flow of current)
If he understands this,then he may also understand the conundrum/paradox associated with his!so called! simple circuit,being that,from the above,we also know it would also take an infinite amount of time for the current to start to flow through the ideal inductor.
This will lead to the point where he now understands my answeryou cannot place an ideal voltage across an ideal inductor
Can i explain both of my theoretical answers?yes i can,and i am surprised Poynt did not pick up on itthe paradox. MH laughed at my two theoretical answers being totally different.
1being that there is no current flowing through the inductorthat Poynt did agree withnot sure what he thinks now after MH threw in his wobbly mathematics.The reason being>as above. If it takes an infinite time for the current to reach it's maximum,then it also takes an infinite amount of time before the current starts to flow.
2@ T=0,when the ideal voltage is placed across the ideal inductor,the current will rise instantly to an infinite value. Answer one makes answer two also correct,in that,as the current is going to rise for an infinite time to an infinite value,then at the very start of that infinite time ,T=0,the current will also be at an infinite amount. As i said,no matter how many parts you divide an infinite value by,each part in itself will also have an infinite value.
This all sounds crazy i know,and hence the reason i included the word conundrum and/or paradox with my answers.
This also shows that MHs question cannot be answered,as it cannot exist.
Changing values around,and changing from an ideal to a non ideal,and using math that is based around non ideal situations,is not going to make the original question answerable.
Brad

electric field moves at light speed doesn't it? how would we know it lagged over distance without resistance.
yet a graph can somehow show some kind of change over time when time has already been omitted due to there
being no relative way to measure it.
one day they'll get a super conductor 100% ideal with absolute 0 resistance, and it will be when the entire universe
is one giant solid piece of silver, and there's absolutely nothing else in the universe to compare it to.
to agree with Albert Einstein, or to disagree with Albert Einstein?..
That is the question.

OK, it's 2:30 AM and I am sleepy so here goes nothing.
The ideal business is just a show piece. You have this ideal voltage meaning it provides anywhere from none to infinite current. As long as your wire is fat enough and let's say 200 turns, that fixed voltage will be applied and you will have enough current in the source to heat up those fat wires. Since those wires of the coil as so fat, the resistance is almost nothing even though nothing can be at an absolute zero resistance.
Now put that same ideal voltage into a coil of thinner wire and only 20 turns. Now that wire has some resistance because it is much thinner then that fat wire coil. Now the voltage is applied and the current, even though it could go to infinity if your wire was of infinite diameter, the current will stabilize at a given level and stay there. The wire determines the current because in any case, the wire has x number of copper atoms and cannot invent or materialize any more so there are only an x amount of atoms conveying and hence x current in the coil. With only 20 turns that 4 volts has a chance to make it to the end of the coil so that whole coil with have current conveyed. I am saying conveyed because I do not believe in flow but that should not detract from this.
Now take that same thin wire coil that now has 2000 turns hence higher henries. Apply your ideal voltage of 4 volts. Even if the current can go to infinity, the coil is only hit with 4 volts and 4 volts in a 2000 turn coil is nothing, probably won't even be conveyed to the end of the coil so the current will be greater at the start of the coil and very weak by the time it gets to the end of the coil hence the current at the start will be greater then the current at the end of the coil. Yes this is counter andwould need to be tested with a multi tapped coil.
Yes this is not an ideal coil. Unfortunately yes, @MH did not need to include an ideal coil in this question and that is a mistake he needs to man up to but that's his business not mine. It would have saved 100 pages of nullisms.
Just for the record I actually do know that DC does not work like that in our coils but that's another topic and should again not detract from this subject as an EE discourse goes.
So again, what @MH just wants to explain is actually very rudimentary but by using the ideal voltage construct just makes the none to infinite current available to match the coils wire AWG and length and topology. Just apply the voltage and the current will find its own level. Yes a DC power supply will do that for you as well. Even if the coil had wire the diameter of the sun and the length of our solar system it will still have resistance so it would be better to remove the ideal from the coil and leave it with the voltage source.
But I still think @MH needs to concede that the original question should not have employed two ideal subjects in the same question. One is enough to explain the process which is again so rudimentary that is is sort of insulting but again, maybe guys take is for granted that you really do not need to always measure currents to know what's going on going your coils.
I rarely use anything to measure watts. My LEDs or bulbs say it all already. When one of them blows, I'll know something is really good. hahahahahaha You don't need to be so obsessed with measurements as I have found it takes time, it already shows what you should already know and it especially eats away at your bench time, patience and distracts you from the most important part of testing a build and that is.................... working the variables. You make a build and test it. You learned one thing. You then test variables on the build as comparisons and now you are learning multiples more on effects. That's where the gold is.
GGGGGGHHHHHHHHHHHHHHHH.
wattsup

We can also solve this a second way. That is to place the ideal voltage across the ideal inductor,and time how long it takes for the maximum current value to be reached. We then divide this time by 5 to obtain our Tau time constant. This also results in an infinite time,as the voltage is ideal,and the inductor has no resistance. It also means once again that there will be no current flowing through the ideal inductor :o .
Since an ideal inductor must have a zero resistance, this means that it must be shorted (if it ain't shorted, it ain't ideal) and it becomes physically impossible to connect any real voltage sources in series with it.
Otherwise, I agree with the above statement. Not only an ideal inductor is devoid of an asymptotic V/R current limit but also the current through an inductor of infinite inductance, that is somehow connected to an ideal voltage source, could never change because of the implied zero di/dt at any voltage.
Of course, it is debatable whether an ideal inductor must have an infinite inductance. Some would say that it is enough for it to have zero resistance and zero parasitic capacitance.
However it is possible to externally change the magnetic flux penetrating a shorted ideal inductor. Doing so will instantaneously cause a current to circulate through it ^{*}, in order to maintain the previous flux level penetrating its windings. This is a voltageless current!  it cannot be measured by a voltmeter and it was not caused by a voltage source.
Last but not least  inductors are current devices and voltage creates no effects in them. Voltage cannot even be measured in shorted ideal inductors (neither practically nor theoretically!). Measurement of voltage (emf) is meaningful only for nonideal inductors (e.g. open inductors or inductors with series resistances). Open inductors or inductors without current flowing though them are dummy inductors  they create no effects on the environment. Voltmeter deflection notwithstanding.
P.S.
I'm just replying to Tinman's post and I have not read what others wrote in this thread.
^{*} (without delay and regardless of its inductance)

@All  Dont forget, Impedance has two different types of Resistance, Real and Imaginary.
Imaginary consists of Inductive Reactance and Capacitive Reactance  This is considered to be a Resistance also.
For DC, it does not apply unless youre looking at the rise and fall times as a Frequency Component. So youre pretty safe to say DC there is none.
However, at any frequency there will be a value of Impedance even if the Real Resistance is 0.
Chris Sykes
hyiq.org
see: https://www.researchgate.net/figure/237776087_fig9_Figure9ImpedanceandESRvsFrequencyforT520vsT528equalparts
P.S: Impedance (Z) and Equivalent Series Resistance (ESR) run the same race, they are parts of the samething. So, really, at any Frequency, there can be no Ideal Inductor with Zero Resistance. At least according to theory. Only at DC. Which some have already explained.

@All  Dont forget, Impedance has two different types of Resistance, Real and Imaginary.
To be technically correct it should've been written:
"Impedance has two different types of Ohms  Real and Imaginary"
OR
"Impedance has two different components  Real and Imaginary"
...because the word "Resistance" is reserved for the real component of Impedance.
This is just a terminological correction  not a conceptual one.

To be technically correct it should've been written:
"Impedance has two different types of Ohms  Real and Imaginary"
OR
"Impedance has two different components  Real and Imaginary"
...because the word "Resistance" is reserved for the real component of Impedance.
This is just a terminological correction  not a conceptual one.
Yes, of course, thanks Verpies, both are measured in Ohms: ΩjΩ or Ω+jΩ
If you see something like:
Z = 10j10 (j = Inductive Reactance)
or
Z = 10+j10 (+j = Capacitive Reactance)
This is where 10 Ohms of Real Resistance and 10 Ohms of Reactance: (X_{L} Reactive Iductance or X_{C} Reactive Capacitive).
See: http://www.saylor.org/site/wpcontent/uploads/2011/07/ME301vol2.pdf
http://www.allaboutcircuits.com/textbook/alternatingcurrent/chpt4/seriesresistorcapacitorcircuits/
Chris Sykes
hyiq.org

Otherwise, I agree with the above statement. Not only an ideal inductor is devoid of an asymptotic V/R current limit but also the current through an inductor of infinite inductance, that is somehow connected to an ideal voltage source, could never change because of the implied zero di/dt at any voltage.
Of course, it is debatable whether an ideal inductor must have an infinite inductance. Some would say that it is enough for it to have zero resistance and zero parasitic capacitance.
However it is possible to externally change the magnetic flux penetrating a shorted ideal inductor. Doing so will instantaneously cause a current to circulate through it ^{*}, in order to maintain the previous flux level penetrating its windings. This is a voltageless current!  it cannot be measured by a voltmeter and it was not caused by a voltage source.
Last but not least  inductors are current devices and voltage creates no effects in them. Voltage cannot even be measured in shorted ideal inductors (neither practically nor theoretically!). Measurement of voltage (emf) is meaningful only for nonideal inductors (e.g. open inductors or inductors with series resistances). Open inductors or inductors without current flowing though them are dummy inductors  they create no effects on the environment. Voltmeter deflection notwithstanding.
P.S.
I'm just replying to Tinman's post and I have not read what others wrote in this thread.
^{*} (without delay and regardless of its inductance)
Since an ideal inductor must have a zero resistance, this means that it must be shorted (if it ain't shorted, it ain't ideal) and it becomes physically impossible to connect any real voltage sources in series with it.
Thank you verpies for joining in on this discussion.
You have confirmed my real world answeran ideal voltage cannot be applied to/placed across an ideal inductor.
Last but not least  inductors are current devices and voltage creates no effects in them. Voltage cannot even be measured in shorted ideal inductors (neither practically nor theoretically!). Measurement of voltage (emf) is meaningful only for nonideal inductors (e.g. open inductors or inductors with series resistances). Open inductors or inductors without current flowing though them are dummy inductors  they create no effects on the environment. Voltmeter deflection notwithstanding.
I only hope Poynt reads what both you and i have stated,and revisits his thoughts on the question presented by MH,and understands that the math MH is using to make his calculations do not apply when dealing with ideal inductor's.
Just another proof that placing a voltage across an ideal inductor dose not create a current flow through that ideal inductor.
Being an ideal inductor,means that it dose not dissipate power,and that also means the CEMF is also ideal,> equal to that which creates it,and thus no current flows when a voltage is placed across that ideal inductor.
A non ideal inductor dose have an R value,and this means it dose dissipate power. This also means that the CEMF value is not as high as the EMF that created it,and so current will flow through a non ideal inductoras we know.
Brad

To be technically correct it should've been written:
"Impedance has two different types of Ohms  Real and Imaginary"
OR
"Impedance has two different components  Real and Imaginary"
...because the word "Resistance" is reserved for the real component of Impedance.
This is just a terminological correction  not a conceptual one.
Verpies
Would you care to answer the question below,and give the reason for your answer.
You have an ideal voltage source and an ideal coil of 5 Henrys. At time t=0 seconds the coil connects to the ideal voltage source. For three seconds the voltage is 4 volts. Then for the next two seconds the voltage is zero volts. Then for two seconds the voltage is negative three volts, and then for the next six seconds the voltage is 0.5 volts. Then after that the voltage is zero volts.
What happens from T=0 when the ideal voltage is connected to the ideal coil?.
Brad

However it is possible to externally change the magnetic flux penetrating a shorted ideal inductor. Doing so will instantaneously cause a current to circulate through it ^{*}, in order to maintain the previous flux level penetrating its windings. This is a voltageless current!  it cannot be measured by a voltmeter and it was not caused by a voltage source.
Better yet, the current induced in the above scenario is independent of the rate of change of flux (dΦ/dt) penetrating that inductor.
On the other hand, the emf induced across an open inductor is not independent of the rate of change of flux (dΦ/dt) penetrating that inductor.

Brad,
To be clear, I am in agreement with MH. It makes sense to me now, and apparently when I answered the question years back on OUR, I also got the answer correct.
Regarding my simulation, when using such small resistance values without changing some settings in SPICE (LT Spice must already be set to handle this), the simulation engine runs out of computational precision, which is why it "flatlines" with very low values. When the value is too low, the sim runs out of gas and starts making gross approximations, which is evident below with R=1f Ohm. But you do see that it is honing in on the 2.4A value? Any smaller in value and the trace just flatlines.
For an ideal inductor, yes Tau is infinite, but this has does not preclude current flow through the inductor. If however the inductance was an unrealistically large value like 1 million Henries, then yes the current would essentially be zero for a relatively long duration of time. That was my confusion. Tau simply determines the rise time, and since it is infinitely long, the trace becomes a nice straight line rather than the curve we normally see.
I was wrong in my analysis, but it is clear to me now.

Better yet, the current induced in the above scenario is independent of the rate of change of flux (dΦ/dt) penetrating that inductor.
On the other hand, the emf induced across an open inductor is not independent of the rate of change of flux (dΦ/dt) penetrating that inductor.
That is because one is closed,and the other open.
Ifas in the question i posted in my last post,the inductor is a closed loop,in that it has an ideal voltage source that completes the loop. Being an !ideal! voltage source,it too must be void of resistance and resistive losses,and so becomes part of the ideal inductor.
So now that the inductor and voltage source are a complete ideal loop,void of resistance,across no two points in that loop can a voltage exist.
And hence,once again,you cannot place a voltage across an ideal inductor,when current is flowing through that closed inductor loop.
Brad
So the question asked is referring to a closed loop scenario.

Brad,
To be clear, I am in agreement with MH. It makes sense to me now, and apparently when I answered the question years back on OUR, I also got the answer correct.
Regarding my simulation, when using such small resistance values without changing some settings in SPICE (LT Spice must already be set to handle this), the simulation engine runs out of computational precision, which is why it "flatlines" with very low values. When the value is too low, the sim runs out of gas and starts making gross approximations, which is evident below with R=1f Ohm. But you do see that it is honing in on the 2.4A value? Any smaller in value and the trace just flatlines.
For an ideal inductor, yes Tau is infinite, but this has does not preclude current flow through the inductor. If however the inductance was an unrealistically large value like 1 million Henries, then yes the current would essentially be zero for a relatively long duration of time. That was my confusion. Tau simply determines the rise time, and since it is infinitely long, the trace becomes a nice straight line rather than the curve we normally see.
I was wrong in my analysis, but it is clear to me now.
Rise time is an actual value,not a trace on a scope.
Is verpies also wrong? Quote: Since an ideal inductor must have a zero resistance, this means that it must be shorted (if it ain't shorted, it ain't ideal) and it becomes physically impossible to connect any real voltage sources in series with it.
Otherwise, I agree with the above statement. Not only an ideal inductor is devoid of an asymptotic V/R current limit but also the current through an inductor of infinite inductance, that is somehow connected to an ideal voltage source, could never change because of the implied zero di/dt at any voltage.
How long will it take for the current to reach it's peak in an ideal inductor(regardless of inductance value),when that inductor has no resistance,and is supplied with an ideal voltage?
Once you have this value,you can divide it by 5 to obtain your time constant for the rise of current in that ideal inductor. Mh is using math that applies to an inductor on the understanding that that inductor will reach a maximum current value in a finite time. This cannot be applied to an ideal inductor,where the current will never reach a maximum value when an ideal voltage is placed across it.
Brad

. . . “Mad Hatter: “Why is a raven like a writingdesk?”
“Have you guessed the riddle yet?” the Hatter said, turning to Alice again.
“No, I give up,” Alice replied: “What’s the answer?”
“I haven’t the slightest idea,” said the Hatter”

Brad,
I don't understand what verpies is trying to say, so I can't say if he is in agreement with MH or not.
You could ask him.

hmm, I thought the EMF followed current. It's the electric field that's the precursor. No EMF if the current is not allowed to change. Electric field see's whats going on the entire time tho.
 question:
is a rise in current a change in current, or not?
the inverse exponential curve to straight line makes perfect sense though..
that is, if current were allowed to flow or change. we have to assume that
is the case first.
*unlike a capacitor, imaginarily we have started at source voltage in our scenario across the inductor. well, again, unlike the capacitor, mission accomplished. no need for current flow because no lenz effect.

is a rise in current a change in current, or not?
It is

Magneticitist
hmm, I thought the EMF followed current. It's the electric field that's the precursor. No EMF if the current is not allowed to change. Electric field see's whats going on the entire time tho.
You may want to rethink your statement. The ElectroMotive Force is the force which causes the charges to move and this motion is called an electric current. The EMF is a Force on the charges and without a force acting on something it has no reason to move.
I cannot even begin to explain how completely messed up your post is... review Coulombs Law, Faradays Law, Electron flow notation, an ElectroMotive Force (Emf).
AC

I don't understand what you mean. I'm saying EMF is a result of current. Well, 'result of' is not quite what I mean but it's hard for me to word it otherwise.
How can there be an electromagnetic field without current? There is however an electric field, the voltage.
maybe I have this wrong but I think you may be missing that we are essentially saying the same thing. damn maybe MH is right and I need to start honing my word choices but can see how I was confusing EMF magnetic field, but I thought I was being clear by also mentioning the electric field as being the established circuit voltage. I believe someone else put it, a currentless voltage.

I don't understand what you mean. I'm saying EMF is a result of current. Well, 'result of' is not quite what I mean but it's hard for me to word it otherwise.
How can there be an electromagnetic field without current? There is however an electric field, the voltage.
maybe I have this wrong but I think you may be missing that we are essentially saying the same thing. damn maybe MH is right and I need to start honing my word choices.
EMF=ElectroMotive Force not electromagnetic field...maybe this will help your confusion?
Bill

EMF=ElectroMotive Force not electromagnetic field...maybe this will help your confusion?
Bill
you are right thank you I was just realizing that
(i will admit I really need to brush up on what to call certain things MH! but in theory I feel
I understand what is going on.. for example I want to call the voltage the electric field in nature when that's not even correct after checking)

you are right thank you I was just realizing that
(i will admit I really need to brush up on what to call certain things MH! but in theory I feel
I understand what is going on.. for example I want to call the voltage the electric field in nature when that's not even correct after checking)
No problem. This can be very confusing...and...just when I think I am starting to understand something...I find out that I really am not, ha ha.
Bill

No problem. This can be very confusing...and...just when I think I am starting to understand something...I find out that I really am not, ha ha.
Bill
it is indeed very confusing and hard to convey thoughts without looking like an idiot if you're like
me and have not dedicated enough time to properly learning the right terminology for everything.
but in an even harder way to explain it seems like this argument involves a basic paradox that
was at least evident enough to warrant many pages of debate. at the end of it all while I personally
struggle to convey my thoughts, I just think the damn thing will do absolutely nothing. Sit at source voltage and act as if it wasn't even there to begin with.

With things changing who knows where you are unless you are on the bleeding edge :)
Is the charge carrier still considered to be the electron??
I ask because if it is then would there not be a limit to the current that can pass through the inductor?
I think that would look like the total electrons in the inductor moving at C,, it may be a huge number but it is a finite number.
I was also thinking,, if that is the case, then any more current flow would charge the inductor negative and then it would see the source as positive in a capacitive way and thus limiting current flow as well.

I thought the EMF followed current.
An EMF measured across an open inductor is proportional to the rate of change of flux penetrating that inductor (dΦ/dt).
EMF measured across an inductor is not proportional to the current flowing through it and in an open inductor the current cannot flow at all.
It's the electric field that's the precursor.
Precursor of what?

Brad,
To be clear, I am in agreement with MH. It makes sense to me now, and apparently when I answered the question years back on OUR, I also got the answer correct.
Regarding my simulation, when using such small resistance values without changing some settings in SPICE (LT Spice must already be set to handle this), the simulation engine runs out of computational precision, which is why it "flatlines" with very low values. When the value is too low, the sim runs out of gas and starts making gross approximations, which is evident below with R=1f Ohm. But you do see that it is honing in on the 2.4A value? Any smaller in value and the trace just flatlines.
For an ideal inductor, yes Tau is infinite, but this has does not preclude current flow through the inductor. If however the inductance was an unrealistically large value like 1 million Henries, then yes the current would essentially be zero for a relatively long duration of time. That was my confusion. Tau simply determines the rise time, and since it is infinitely long, the trace becomes a nice straight line rather than the curve we normally see.
I was wrong in my analysis, but it is clear to me now.
I dont think it is clear Poynt,and your original thought (current will not flow)is correct.
If your sim can replicate any real electrical event in a circuit,why dose it crash when you use an ideal inductor that no resistance?. Why is your sim unable to carry out a simple analysis such as you think MH has successfully done?.
The answer is simple,and as i stated. MH is using math that is based on the premise that the inductor will reach a maximum current level. !!Can you calculate what the maximum current level will be of an ideal inductor using MHs calculations? Will this answer (When t = 3 seconds that's 12/5 = 2.4 amps) then be correct?.
When modeling a circuit ,an ideal inductor is usedalong with a series resistor that mimics the resistance that would exist in a real world inductor. This is why you have to place some resistance in series with your ideal inductor in your sim to stop it crashing.
Why dose your sim crash without it?
Because your using an ideal inductor. An ideal inductor has ideal inductance,no resistance,no capacitance,and dose not dissipate power. This means that the CEMF is also ideal,and so is equal to the EMF ,and so an equal current will flow in the opposite direction to that of the current produced by the EMF.
Here is verpies statement.
Since an ideal inductor must have a zero resistance, this means that it must be shorted (if it ain't shorted, it ain't ideal) and it becomes physically impossible to connect any real voltage sources in series with it.Not only an ideal inductor is devoid of an asymptotic V/R current limit but also the current through an inductor of infinite inductance, that is somehow connected to an ideal voltage source, could never change because of the implied zero di/dt at any voltage.
Rememberit is only the resistance and parasitic capacitance that allows the EMF to be greater than the CEMF,and allow the flow of current,something that an ideal inductor is void of.
Brad

An EMF measured across an open inductor is proportional to the rate of change of flux penetrating that inductor (dΦ/dt).
EMF measured across an inductor is not proportional to the current flowing through it and in an open inductor the current cannot flow at all.
Precursor of what?
well, look.. I hadn't thought about this kind of stuff in a while and in this debate tried to quickly brush up on my terminology online, and read how the EMF is actually the potential. I was thinking the electric field was the voltage when apparently that's not a metaphor. Now I don't really know what to call what at the moment because I am getting more confused at the terminology other people are using that seems to contradict what I just read. I need to spend some more time reading this in detail before I know exactly what do call what here but I think of it like electric voltage, magnetic  current. maybe that is a complete falsehood that has emerged from too much misinterpreted tinkering and not enough reading. at any rate the point I was trying to make is that any magnetic force that is present is present when there is current flowing and if not we just have voltage from the source.
there's no di/dt to even consider, no magnetic field to consider, no lenz to consider, because it's all perfectly working against itself in theory and pretty much just not existing at all. like any inductor the voltage is the precursor unlike the opposite in a capacitor.

@Tinman
Rememberit is only the resistance and parasitic capacitance that allows the EMF to be greater than the CEMF,and allow the flow of current,something that an ideal inductor is void of.
I would agree and if the resistance is defined as zero in the question then Ohms law has no application... let's move on. No resistance and no capacitance which leaves an ElectroMotive Force from the source which are Coulomb forces due to the Ideal voltage source electric field. The source Emf acts forward however the moment something tries to move a magnetic field evolves producing an equal and opposite CounterForce, our Cemf, which opposes the charges motion.
Logic suggest that if nothing can dissipate and energy is perfectly conserved then the Cemf must balance perfectly with the source Emf... remember these are perfectly conserved forces. If an ideal superconductor produces a perpetual lossless current closed loop then an ideal superconducting coil must also produce a perpetual lossless Cemf countering our source Emf. One cannot say the rules always apply then change the rules simply because they do not like the answer. Either the forces balance perfectly and energy is conserved or energy is not conserved in a lossless system and we have problems. Ideally it must be ideal because we have already defined it as such.
AC

@Tinman
I would agree and if the resistance is defined as zero in the question then Ohms law has no application... let's move on. No resistance and no capacitance which leaves an ElectroMotive Force from the source which are Coulomb forces due to the Ideal voltage source electric field. The source Emf acts forward however the moment something tries to move a magnetic field evolves producing an equal and opposite CounterForce, our Cemf, which opposes the charges motion.
Logic suggest that if nothing can dissipate and energy is perfectly conserved then the Cemf must balance perfectly with the source Emf... remember these are perfectly conserved forces. If an ideal superconductor produces a perpetual lossless current closed loop then an ideal superconducting coil must also produce a perpetual lossless Cemf countering our source Emf. One cannot say the rules always apply then change the rules simply because they do not like the answer. Either the forces balance perfectly and energy is conserved or energy is not conserved in a lossless system and we have problems. Ideally it must be ideal because we have already defined it as such.
AC
Absolutely AC
Ideal means perfecta losless conversion between EMFforward currentcounterEMFreverse current.of the same amount.
So that would mean a dead short when an ideal voltage from an ideal source is placed across the ideal inductor,as as much current would be trying to flow back into the ideal voltage source,as the ideal voltage source is trying to deliver.
My answer standsyou cannot place an ideal voltage across an ideal inductor.
Brad

@Tinman
I would agree and if the resistance is defined as zero in the question then Ohms law has no application... let's move on. No resistance and no capacitance which leaves an ElectroMotive Force from the source which are Coulomb forces due to the Ideal voltage source electric field. The source Emf acts forward however the moment something tries to move a magnetic field evolves producing an equal and opposite CounterForce, our Cemf, which opposes the charges motion.
Logic suggest that if nothing can dissipate and energy is perfectly conserved then the Cemf must balance perfectly with the source Emf... remember these are perfectly conserved forces. If an ideal superconductor produces a perpetual lossless current closed loop then an ideal superconducting coil must also produce a perpetual lossless Cemf countering our source Emf. One cannot say the rules always apply then change the rules simply because they do not like the answer. Either the forces balance perfectly and energy is conserved or energy is not conserved in a lossless system and we have problems. Ideally it must be ideal because we have already defined it as such.
AC
I don't see why we should have to take it further than "perfectly and absolutely resists current change". This automatically means no EMF or counter EMF doesn't it?

Absolutely AC
Ideal means perfecta losless conversion between EMFforward currentcounterEMFreverse current.of the same amount.
This would more so describe an ideal inductor that also has an infinite amount of inductance. As such, when connected across a voltage source, no current would ever flow as the time constant would also be infinite. An ideal inductor with infinite inductance would appear to be a continuous open circuit when connected in parallel with a voltage source.
Consider the superconducting magnet used in an MRI machine. Typically, they have a total inductance of around 6000Hy. They are "charged" very slowly to a couple hundred amps or so with an adjustable current limited source to prevent an excessive overvoltage condition.
So that would mean a dead short when an ideal voltage from an ideal source is placed across the ideal inductor,as as much current would be trying to flow back into the ideal voltage source,as the ideal voltage source is trying to deliver.
My answer standsyou cannot place an ideal voltage across an ideal inductor.
You seem to be describing an ideal capacitor with an infinite amount of capacitance. An ideal capacitor with an infinite amount of capacitance would have an infinite time constant and appear to be a continuous short circuit when connected in parallel with a voltage source.
That's my 2 cents for now. I've been away and will try to catch up on this thread as time allows. Very busy...
PW

This would more so describe an ideal inductor that also has an infinite amount of inductance. As such, when connected across a voltage source, no current would ever flow as the time constant would also be infinite. An ideal inductor with infinite inductance would appear to be a continuous open circuit when connected to a voltage source.
Consider the superconducting magnet used in an MRI machine. Typically, they have a total inductance of around 6000Hy. They are "charged" very slowly to a couple hundred amps or so with an adjustable current limited source to prevent an excessive overvoltage condition.
You seem to be describing an ideal capacitor with an infinite amount of capacitance. An ideal capacitor with an infinite amount of capacitance would have an infinite time constant and appear to be a continuous short circuit when a voltage is applied across it.
That's my 2 cents for now. I've been away and will try to catch up on this thread as time allows. Very busy...
PW
let's say this is true, and it honestly makes a degree of sense even though I disagree in principle..
Wouldn't this mean the inductor is 'perfectly' accomplishing the exact opposite of what it's supposed to be perfectly doing in nature? Wouldn't the EMF and *CEMF be increasing against each other the entire time to infinity? And wouldn't that have to mean a continuously rising current?
How does that not constitute current change?

This photo is for the Tinman.
Basically we get too much overunity and have to handrear the triplets.
These are a few of this years pets.
How about this for a business idea Tinman? Start the manufacture of
braking systems for overunity machines.
The one thing Wayne Travis got right on his early plans for his church
was the provision of a cooling system!!!

I don't see why we should have to take it further than "perfectly and absolutely resists current change".
This automatically means no EMF or counter EMF doesn't it?
What it means ,is that there can be no voltage across the ideal inductorso yes.
It also means,as i said before,the current would be instant,and infinitebut no current flow. It would be much like supplying each end of a water pipe with water at 40psi. You would have pressure,but no flow.
Brad

Consider the superconducting magnet used in an MRI machine. Typically, they have a total inductance of around 6000Hy. They are "charged" very slowly to a couple hundred amps or so with an adjustable current limited source to prevent an excessive overvoltage condition.
You seem to be describing an ideal capacitor with an infinite amount of capacitance. An ideal capacitor with an infinite amount of capacitance would have an infinite time constant and appear to be a continuous short circuit when connected in parallel with a voltage source.
That's my 2 cents for now. I've been away and will try to catch up on this thread as time allows. Very busy...
PW
This would more so describe an ideal inductor that also has an infinite amount of inductance. As such, when connected across a voltage source, no current would ever flow as the time constant would also be infinite. An ideal inductor with infinite inductance would appear to be a continuous open circuit when connected in parallel with a voltage source.
Thanks for joining PW.
I have to say that i dont agree with the ideal inductor needing to have an infinite inductance value.
Any iductor that is ideal is loss less dose not dissipate power,due to having no resistance or capacitance. It converts or stores 100% of the energy it receivesideal. Being the case,the current created when the v ideal voltage is placed across the ideal inductor,would create an CEMF and current of the same value100% conversionideal.
Due to the fact that the voltage is also ideal,the source of that voltage must also be ideal.
Lets say it's an ideal battery,meaning that that battery could deliver 4 volts across any load without a voltage drop. The battery would also have to be ideal,and there for would have no internal resistance,and would not dissipate power by way of waste heat,or radiated energy of any type.
Once this ideal voltage source is hooked across the ideal inductor,as soon as any magnetic field started to form,it would produce an exact opposite,in the way of CEMF and currentideal conversion. We know this to be true in an ideal situation,as everything has an equal and opposite reaction. In a non ideal situation,some of that equal and opposite is lost to heat due to resistance and parasitic capacitancebut not in an ideal situation,which is what the original question states the inductor and voltage source isideal.
The result would be an instant and infinite current build up between the ideal voltage source,and the ideal inductor,but no current would flow.And as there is no resistance throughout the circuit,no voltage would appear anywhere across that loop.
It is hard for some to understand what !ideal! mean's,but think about it long enough,and you begin to put all the pieces together.
Every action has an equal and opposite reactionthis dose not change.
And so,in this ideal situation,the CEMF must be equal and opposite to the EMF,and the current produced by the CEMF that apposes the current produced by the EMF must also be equal and opposite,as an ideal inductor dose not dissipate power.
Brad

This photo is for the Tinman.
Basically we get too much overunity and have to handrear the triplets.
These are a few of this years pets.
How about this for a business idea Tinman? Start the manufacture of
braking systems for overunity machines.
The one thing Wayne Travis got right on his early plans for his church
was the provision of a cooling system!!!
Seems you have hit the bottle again John.
Any chance you could provide useful input into this thread?.
Brad

let's say this is true, and it honestly makes a degree of sense even though I disagree in principle..
Wouldn't this mean the inductor is 'perfectly' accomplishing the exact opposite of what it's supposed to be perfectly doing in nature? Wouldn't the EMF and CMF be increasing against each other the entire time to infinity? And wouldn't that have to mean a continuously rising current?
How does that not constitute current change?
Because the current produced by the inductor is equal and opposite to that being provided by the ideal voltage source,and so no current flows,but it dose rise to an infinite amount.
I have asked this question beforewhat happens when an unstoppable force meets an unmovable object?.
Brad

Well now I feel like I am in the Twilight Zone.
Poynt: The current is one over "L" integral v dt.
That's 1/5 * integral (4) dt.
That's 1/5 * 4t.
That's 4/5*t.
When t = 3 seconds that's 12/5 = 2.4 amps.
Can we get out of the Zone now?
And when t= 20 seconds that's 9714.05 amps.
Poynt ran his sim for 100 second's (what would be the current then ?),and still he had no voltage drop :D
Brad

Because the current produced by the inductor is equal and opposite to that being provided by the ideal voltage source,and so no current flows,but it dose rise to an infinite amount.
I have asked this question beforewhat happens when an unstoppable force meets an unmovable object?.
Brad
Like I said before...
The Big Bang. Then, it all starts all over again. An endless cycle.
Of course, this is just a theory.
Bill

tinman,I've had a great find.
Have a look at ElectronicsTutorials w.s.
With all the crap you've been coming out with lately you're like me and
really need to start from the very beginning.
There's a lot about inductors and it's very basic and even I've been able
to inch forward, albeit very slowly.
Good luck,John.

Because the current produced by the inductor is equal and opposite to that being provided by the ideal voltage source,and so no current flows,but it dose rise to an infinite amount.
I have asked this question beforewhat happens when an unstoppable force meets an unmovable object?.
Brad
right, in that scenario it could be said there was never any current at all. infinite is the same thing as none at all in practice.
however, if we were to assume this infinitely rising current without any curvature continues to infinity, that in itself is a current change, therefore how can it be function as an ideal coil? In the very nature of how its operation is described it seems as if it should be infinitely resting with 0 current flow or infinite current flow.. which as I say is the same thing so we might as well just say none at all.

@Magneticitist
I don't see why we should have to take it further than "perfectly and absolutely resists current change". This automatically means no EMF or counter EMF doesn't it?
An ElectroMotive Force relates to electric fields and a MagnetoMotive Force relates to the Magnetic fields. The Electric field is the cause or the source of the force which initially acts on the charges causing them to move... a current. If a charge tries to move it instantly produces a magnetic field which opposes it's motion which we call a Cemf. We call it a Cemf because the conductor contains billions of (+) and () charges and the moment one tries to move the rest oppose it because they are linked by electric fields.
So yes the Emf is present as an electric field of force... before... anything moves such as a current. The Emf is the cause of the current not vice versa.
@Picowatt
This would more so describe an ideal inductor that also has an infinite amount of inductance. As such, when connected across a voltage source, no current would ever flow as the time constant would also be infinite. An ideal inductor with infinite inductance would appear to be a continuous open circuit when connected in parallel with a voltage source.
First we should ask what is inductance?, Inductance: the property of an electrical conductor by which a change in current through it induces an electromotive force in both the conductor itself and in any nearby conductors. Inductance is not something but a property of something which is actually selfinductance which relates to a Cemf. So when we say Inductance what we really mean is selfinductance or the generation of a Cemf which opposes the source Emf.
In this case we are speaking of an Ideal Inductor, a perfect inductor with no losses of any kind hence the term "ideal". Now if a charge tries to move it must instantaneously invoke an equal and opposite Cemf because we have defined action and reaction occurring under ideal conditions, no losses. Thus any inductance (selfinductance) of any kind in an ideal inductor qualifies as infinite (however not an infinitely large field) because the term inductance relates solely to the Cemf which must oppose the source Emf. An ideal closed loop superconductor produces a continuous current at zero voltage and an ideal closed loop superinductor should produce a continuous electric field opposition or Emf/Cemf at zero current. Can you see the symmetry here?.
I know this seems difficult to understand because this leads to questions which many find deeply disturbing. What is an Electric Field, what is a Magnetic Field fundamentally?. You see it's like trying to describe a bike without knowing what a wheel is. This is why everyone tends to throw about terminology never knowing what the terms actually mean or what they are supposed to describe in reality.
AC

@Magneticitist
An ElectroMotive Force relates to electric fields and a MagnetoMotive Force relates to the Magnetic fields. The Electric field is the cause or the source of the force which initially acts on the charges causing them to move... a current. If a charge tries to move it instantly produces a magnetic field which opposes it's motion which we call a Cemf. We call it a Cemf because the conductor contains billions of (+) and () charges and the moment one tries to move the rest oppose it because they are linked by electric fields.
So yes the Emf is present as an electric field of force... before... anything moves such as a current. The Emf is the cause of the current not vice versa.
@Picowatt
First we should ask what is inductance?, Inductance: the property of an electrical conductor by which a change in current through it induces an electromotive force in both the conductor itself and in any nearby conductors. Inductance is not something but a property of something which is actually selfinductance which relates to a Cemf. So when we say Inductance what we really mean is selfinductance or the generation of a Cemf which opposes the source Emf.
In this case we are speaking of an Ideal Inductor, a perfect inductor with no losses of any kind hence the term "ideal". Now if a charge tries to move it must instantaneously invoke an equal and opposite Cemf because we have defined action and reaction occurring under ideal conditions, no losses. Thus any inductance (selfinductance) of any kind in an ideal inductor qualifies as infinite (however not an infinitely large field) because the term inductance relates solely to the Cemf which must oppose the source Emf. An ideal closed loop superconductor produces a continuous current at zero voltage and an ideal closed loop superinductor should produce a continuous electric field opposition or Emf/Cemf at zero current. Can you see the symmetry here?.
I know this seems difficult to understand because this leads to questions which many find deeply disturbing. What is an Electric Field, what is a Magnetic Field fundamentally?. You see it's like trying to describe a bike without knowing what a wheel is. This is why everyone tends to throw about terminology never knowing what the terms actually mean or what they are supposed to describe in reality.
AC
thanks for the explanation.. but I'm not in disagreement that EMF is the cause of the current. I'm saying I don't see how there would be any current, EMF, CEMF, or any kind of interaction I would consider a Lenz Law interaction that would require more current to overpower an equal opposition.
That would be the scenario I see played out were current actually flowing into this inductor with infinite rise time and no curvature.
but I personally can't get past the sheer logical aspect of the argument. I don't see the need to have to get into the complex details of charges and opposing forces when the ideal coil is said to no dissipate energy and perfectly resist change in current. I don't see how that doesn't imply no current at all, unless we are approaching this imaginary scenario adding the parameter that this ideal inductor is already partially charged and just accept that as imaginary scientific parameters.

There's a lot about inductors and it's very basic and even I've been able to inch forward, albeit very slowly.
Did you discover yet, that inductance is the amount of magnetic flux that is generated by a given electric current ?
In other words, Henry = Weber / Ampere.

Thanks for joining PW.
I have to say that i dont agree with the ideal inductor needing to have an infinite inductance value.
An ideal inductor does not have to have an infinite inductance.
What I said was your description of an inductor whose EMF and CEMF are in perfect balance would be descriptive of an inductor with infinite inductance. An inductor with infinite inductance, when connected across a V source, would forever appear as an open circuit.
It is hard for some to understand what !ideal! mean's,but think about it long enough,and you begin to put all the pieces together.
A straight length of ideal conductor, such as a straight wire with no resistance or capacitance, would be an ideal inductor. Just like its less ideal real world counterpart, a finite amount of time is required for current to flow thru the wire because of the wire's inductance.
Wrapping that same straight length of ideal (or normal) wire into a helix increases the coupling between sections (turns) of the wire and increases its inductance.
I do not understand why you believe that an ideal inductor would cause an instantaneous and infinite amount of current to flow when connected across a voltage source. As I said, that would be more descriptive of an ideal capacitance.
Gotta' go...
PW

Because the current produced by the inductor is equal and opposite to that being provided by the ideal voltage source, and so no current flows, but it dose rise to an infinite amount.
It is true only for an ideal inductor that has an infinite inductance.
In an ideal inductor having a finite inductance, in series with an ideal voltage source, the current will be able to flow and it will increase linearly in time without a limit.
Anyway, that statement above is so awkwardly worded.
First you write about two currents flowing and at the end you write about currents not flowing  that sounds contradictory.
I think you wanted to write about two currents, that would flow if they were not opposing each other.
Specifically, one current, that would flow due to the nature of a shorted voltage source and a second current, that would flow in a shorted ideal inductor (shorted by the voltage source).
That's all, that I was able to decode, so far.

An ideal inductor does not have to have an infinite inductance.
True. An infinite ideal inductor is a special case of an ideal inductor.
What I said was your description of an inductor whose EMF and CEMF are in perfect balance would be descriptive of an inductor with infinite inductance. An inductor with infinite inductance, when connected across a V source, would forever appear as an open circuit.
I agree and I think Tinman was stating the same thing but in terms of opposing hypothetical currents instead of opposing EMFs.

until some form of 'distance' or 'length' is brought into the equation, which has a direct relation with resistance in the real world (I think?) isn't 'time' out of the window altogether because without anything to slow electron drift velocity (pardon if that makes no sense I just read some stuff and thought I made some sense of it) wouldn't the current technically travel at light speed? what about collisions? would that be meeting a resistance? if we have nothing with which to truly factor a Tau
into the equation then why can't 'current infinitely rising' be synonymous with 'no current flow at all'.
the real problem here is nobody want's to come out and say Ohm's law needs to be thrown out
of the window in this situation isn't it? and we are supposed to be trained to rely on Ohms Law.
Brad was trying to show extremely high amps calculated at extremely low resistance as a metaphor for approaching "infinite" using Ohms Law. But as we have been discussing, and maybe do not all agree on it, but infinite=0.

Ohm's Law applies to resistors, it does not apply to inductors or capacitors. I am keeping it simple and with that in mind forget about Ohm's Law, we are discussing an inductor.
So you are correct, for this discussion we can throw out Ohm's Law.

The inductor is impeding current flow but when I look up impedance it
refers to ac circuits and reactance doesn't seem to fit the bill either.
It's all good fun and the whole thing proves that not many of us know
that much!
You've got to hand it to the Henrys and Faradays and Maxwells for
figuring it out so well in the first place.
John.

Ohm's Law applies to resistors, it does not apply to inductors or capacitors. I am keeping it simple and with that in mind forget about Ohm's Law, we are discussing an inductor.
So you are correct, for this discussion we can throw out Ohm's Law.
I believe the twin brother to the issue of Ohms law being a factor or not, is the possibility
that an absolute 0 resistance removes the characteristics of an inductor/capacitor/conductor/resistor altogether so they might as well all be considered ideal conductors at 0 resistance. Does that sound like insane mumbo jumbo cause by freaking golly it makes a degree of sense to me.

It's mostly mumbo jumbo talk.
But here is the clue: Even if "the resistance is zero," in other words there is no resistance in the circuit, it does not necessarily mean that something isn't impeding the current flow.

It's mostly mumbo jumbo talk.
But here is the clue: Even if "the resistance is zero," in other words there is no resistance in the circuit, it does not necessarily mean that something isn't impeding the current flow.
:D :D :D

...an absolute 0 resistance removes the characteristics of an inductor/capacitor/conductor/resistor altogether so they might as well all be considered ideal conductors at 0 resistance.
"Absolute 0 resistance" removes the characteristic of resistance but it does not remove "inductive reactance (https://en.wikipedia.org/wiki/Electrical_reactance#Inductive_reactance)" of inductors nor "capacitive reactance (https://en.wikipedia.org/wiki/Electrical_reactance#Capacitive_reactance)" of capacitors (despite all three being measured in Ohms).
Impedance = Resistance + Reactance.

"Absolute 0 resistance" removes the characteristic of resistance but it does not remove "inductive reactance (https://en.wikipedia.org/wiki/Electrical_reactance#Inductive_reactance)" of inductors nor "capacitive reactance (https://en.wikipedia.org/wiki/Electrical_reactance#Capacitive_reactance)" of capacitors (despite all three being measured in Ohms).
Impedance = Resistance + Reactance.
But can we say for sure that our understanding of those 2 forms of reactance is not derived from
some existing level of resistance? were we in some alternate universe where we never
even had a variable to call R in the first place, can we say all our circuit theory would be exactly
the same using the same math? and sheesh it's like, definitely ultra confusing, when we have different variables measures in the same unit but they are somehow working entirely independent
from that unit. not trying to be difficult it just this type of thought pattern seems to be expected
when it's almost inherent in the forum name.

It is true only for an ideal inductor that has an infinite inductance.
In an ideal inductor having a finite inductance, in series with an ideal voltage source, the current will be able to flow and it will increase linearly in time without a limit.
Specifically, one current, that would flow due to the nature of a shorted voltage source and a second current, that would flow in a shorted ideal inductor (shorted by the voltage source).
That's all, that I was able to decode, so far.
Anyway, that statement above is so awkwardly worded.
First you write about two currents flowing and at the end you write about currents not flowing  that sounds contradictory.
No,i said currents being produced,not flowing.
I think you wanted to write about two currents, that would flow if they were not opposing each other.
And that is what i wrote.
The current produced by the EMF of our ideal voltage,would meet the equal and opposite from the CEMF produced by the ideal inductor.
When dealing with ideals,we deal with absolutes,and there for the CEMF is ideal,meaning that it is equal and opposite to that of the EMF.
Brad

author=verpies link=topic=16589.msg483812#msg483812 date=1462987219]
It is true only for an ideal inductor that has an infinite inductance.
In an ideal inductor having a finite inductance, in series with an ideal voltage source, the current will be able to flow and it will increase linearly in time without a limit.
And how would it do that if the CEMF is equal to the EMF that created it?
Brad

And how would it do that if the CEMF is equal to the EMF that created it?
The current would not increase if CEMF = EMF, but they would be so only in the case when the ideal inductor had infinite inductance.
Note that I was explicitly writing about the current increase through an ideal inductor with finite inductance.

But can we say for sure that our understanding of those 2 forms of reactance is not derived from
some existing level of resistance?
No, the resistance behaves significantly different from inductive reactance. The former converts current to heat and the latter converts current to magnetic flux.
...and sheesh it's like, definitely ultra confusing, when we have different variables measures in the same unit but they are somehow working entirely independent.
They both have the same units, but mathematically they are in different complex dimensions which are orthogonal.
You don't seem to have a problem with distances in different orthogonal dimensions, that are not affecting each other despite being expressed by the same units: meters or feet.

What a silly thing this has turned out to be. Ideal seems to have caused total confusion.
The whole thing is just a matter of a bit of maths,which I can't do!
I would love to do the experiment, a good quality choke and a couple of forklift LA. cells
would make an ideal power supply. I can see smoke in the equation.
I don't see superconductors as the equivalent zero resistance.
Another point I'd like to make is the continual appearance of things like "You lied MH."
to me lying is deliberate fraud whereas misinformation due to incomprehension
isn't lying.
I lied when I said I knew the answer when in reality I didn't.
Have a great day you all,warmest regards, John.

Note that I was explicitly writing about the current increase through an ideal inductor with finite inductance.
The current would not increase if CEMF = EMF, but they would be so only in the case when the ideal inductor had infinite inductance.
And so the reverse would apply.
The moment an EMF is placed across the ideal inductor,and a current started to flow,then the CEMF produced would be the same as the EMF that created it. That being the case,then the inductor would act as though it had infinite inductance.
Just thoughts :)
Brad

author=minnie link=topic=16589.msg483887#msg483887 date=1463045835]
I don't see superconductors as the equivalent zero resistance.
I lied when I said I knew the answer when in reality I didn't.
Have a great day you all,warmest regards, John.
I would love to do the experiment, a good quality choke and a couple of forklift LA. cells
would make an ideal power supply. I can see smoke in the equation.
Of course you will get smoke,as you will have resistance,and be far from an ideal inductor.It is the resistance that causes the heat loss,no resistance=no heat.
Remember,the ideal inductor dose not dissipate power. The ideal inductor is wound from ideal conducting wire.
I don't see superconductors as the equivalent zero resistance.
https://en.wikipedia.org/wiki/Meissner_effect
In Type II superconductors, raising the applied field past a critical value Hc1 leads to a mixed state (also known as the vortex state) in which an increasing amount of magnetic flux penetrates the material, but there remains no resistance to the flow of electric current as long as the current is not too large. ;)
Another point I'd like to make is the continual appearance of things like "You lied MH."
to me lying is deliberate fraud whereas misinformation due to incomprehension
isn't lying.
You mean like making an absolute claim about there being no resonant systems in or around an ICE in any way,shape ,or formonly to later say>well i dont know that much about ICEs :o
So you have given only two option's,fraud,or misinformation.
As much as myself and MH may not get along at times,due to nothing more that a difference in opinion,i would think morea mistake would be a better word to usewhich we have all made.
Brad

Perhaps we can actually carry out this experiment :o
Will insulated copper wire become super conductive if submerged in liquid nitrogen ?.
If not,what kind of wire would we need to achieve our super conductive windings at liquid nitrogen temperatures.
Brad

Or this one: https://www.youtube.com/watch?v=PIvZJ9xGutI (https://www.youtube.com/watch?v=PIvZJ9xGutI)
Hear the: Sorry about the light, and welcome in mine Labrborattorry (Nicky''''''s, kitchen table)!?

Or this one: https://www.youtube.com/watch?v=PIvZJ9xGutI (https://www.youtube.com/watch?v=PIvZJ9xGutI)
Hear the: Sorry about the light, and welcome in mine Labrborattorry (Nicky''''''s, kitchen table)!?
That almost sounds like TK lol.
Brad

That almost sounds like TK lol.
Brad
Hole Dbase with different Cookie's remarks collected!
What do you think about:
Always, Independent 3e party validation, all less is don't pay attention, except for: Nasa, Mit, Lewin sometimes, .............
Sweet dreams, and go to bed!

The current would not increase if CEMF = EMF, but they would be so only in the case when the ideal inductor had infinite inductance.
Note that I was explicitly writing about the current increase through an ideal inductor with finite inductance.
Verpies,
Can you have a fixed finite inductance from an inductor that has resistance=0 and capacitance=0

PW
This would more so describe an ideal inductor that also has an infinite amount of inductance. As such, when connected across a voltage source, no current would ever flow as the time constant would also be infinite.
The time constant is infinite.
Tau=L/R. There is no R,as it's an ideal inductor.
Tau=L/0 = infinity.
Brad

the ball, shopping cart, whatever.. floats through the perfect vacuum unimpeded. ok.
well how do we know this if there is absolutely nothing else in this vacuum to compare
its relative motion to. the presence of something else, anything else, could
be considered a resistance. a way to set a Tau. For all intents and purposes, it's at rest.
To say otherwise would be to simply assume it in test parameters. Newtonian Physics
really don't much belong in quantum mechanics anyhow do they? This seems more of
a time paradox. I understand I provide nothing to the convo without being able
to at least properly explain it, but maybe possibly someone else can see where I'm coming
from here. The 'ground' as Webby put it imo seems like light speed. Resistance is what keeps
our energy from traveling at light speed right? Whether, electric current, or whatever?

The time constant is infinite.
Tau=L/R. There is no R,as it's an ideal inductor.
Tau=L/0 = infinity.
Brad
Tau has no bearing on whether current can/will flow or not. The effect it has is how "curvy" the rise of current is, relative to the timing of your test. With an infinite tau, the curve is going to be a straight line, not only because the time is infinite, but also because the beginning part of the curve is almost straight anyway.

the ball, shopping cart, whatever.. floats through the perfect vacuum unimpeded. ok.
well how do we know this if there is absolutely nothing else in this vacuum to compare
its relative motion to. the presence of something else, anything else, could
be considered a resistance. a way to set a Tau. For all intents and purposes, it's at rest.
To say otherwise would be to simply assume it in test parameters. Newtonian Physics
really don't much belong in quantum mechanics anyhow do they? This seems more of
a time paradox. I understand I provide nothing to the convo without being able
to at least properly explain it, but maybe possibly someone else can see where I'm coming
from here. The 'ground' as Webby put it imo seems like light speed. Resistance is what keeps
our energy from traveling at light speed right? Whether, electric current, or whatever?
Electrons move at the speed of light even through resistors as far as I know.
Bill

Electrons move at the speed of light even through resistors as far as I know.
Bill
That is where things get even more complicated and the atomic structure of conductors come into play. The EM waves can propagate at typically light speed, or just under, but the electron's have a
velocity we call the electron drift that is actually much much slower.
(in a way it's analogous to how slowly a magnet falls through the copper pipe demonstration)

That is where things get even more complicated and the atomic structure of conductors come into play. The EM waves can propagate at typically light speed, or just under, but the electron's have a
velocity we call the electron drift that is actually much much slower.
(in a way it's analogous to how slowly a magnet falls through the copper pipe demonstration)
The magnet slowing through a copper pipe is a demonstration of Lenz Law due to the fact that copper can be diamagnetic. Electrons move at the speed of light...you can limit the number of them moving through a circuit but not slow them down.
Bill

The magnet slowing through a copper pipe is a demonstration of Lenz Law due to the fact that copper can be diamagnetic. Electrons move at the speed of light...you can limit the number of them moving through a circuit but not slow them down.
Bill
It's definitely not something that can be so thoroughly explained in a short comment, but essentially this is true, electrons are traveling chaotically at near light speed. This is the Fermi velocity. But when we bring current flow into the equation the drift velocity of the electron particle through a conductor can become analogous to the magnet in the copper pipe, simply in the context of the magnet moving much much slower than you would intuitively imagine without understanding Lenz Law and eddy currents.

Tau has no bearing on whether current can/will flow or not. The effect it has is how "curvy" the rise of current is, relative to the timing of your test. With an infinite tau, the curve is going to be a straight line, not only because the time is infinite, but also because the beginning part of the curve is almost straight anyway.
Poynt.
;)
Brad

Tau has no bearing on whether current can/will flow or not. The effect it has is how "curvy" the rise of current is, relative to the timing of your test. With an infinite tau, the curve is going to be a straight line, not only because the time is infinite, but also because the beginning part of the curve is almost straight anyway.
edit

The magnet slowing through a copper pipe is a demonstration of Lenz Law due to the fact that copper can be diamagnetic. Electrons move at the speed of light...you can limit the number of them moving through a circuit but not slow them down.
Bill
No
It's because the shorted single turn copper coil is producing a magnetic field that apposes that which created it.
Brad

yes, with inertia supplied by gravity.. in this case metaphorically, the ideal voltage/infinite current.
without a time constant, we can say this magnet is falling forever and ever and the pipe is an infinite length. (edit, infinite in both directions... with no 'starting point'.. so how is it moving?) Since we have no relative way of knowing if it's moving other than counting the seconds and
using some constitutive equation to solve for current, we are left to assume it's actually moving
based upon the equation and the implication that it's moving because it is being acted upon by
a force. it is, however as I have thought, more complicated than newtonian physics because we
are dealing with a kind of special relativity that we still do not understand the full nature of despite
the amazing things that engineers are capable of doing as this time. R seems like it can mathematically be represented in nature in a variety of ways. Excluding R from all
of eternity and assuming our understanding of Inductance would remain absolute doesn't
seem like something we can prove.

No
It's because the shorted single turn copper coil is producing a magnetic field that apposes that which created it.
Brad
MIT video demonstrating Lenz Law: https://www.youtube.com/watch?v=N7tIi71AjA (https://www.youtube.com/watch?v=N7tIi71AjA)
Lenz Law:
"If an induced current flows, its direction is always such that it will oppose the change which produced it."
You are basically saying the same thing...except for the diamagnetic business.
Bill

MIT video demonstrating Lenz Law: https://www.youtube.com/watch?v=N7tIi71AjA (https://www.youtube.com/watch?v=N7tIi71AjA)
Lenz Law:
"If an induced current flows, its direction is always such that it will oppose the change which produced it."
You are basically saying the same thing...except for the diamagnetic business.
Bill
lol basically, I have no doubt both of you understand Lenz Law. The term Diamagnetic
can be confusing though and I'm not even sure as to whether that's a 'proper' way to
say it although I perfectly understand what you mean because the copper essentially
does become 'diamagnetic' as far as I can tell. seems like the same explanation.
However I also thought diamagnetic materials referred more to materials that naturally
exhibit a strong diamagnetism, much more so than what may be found in normal conductors like
copper. For example Bismuth. still, confusing.

lol basically, I have no doubt both of you understand Lenz Law. The term Diamagnetic
can be confusing though and I'm not even sure as to whether that's a 'proper' way to
say it although I perfectly understand what you mean because the copper essentially
does become 'diamagnetic' as far as I can tell. seems like the same explanation.
However I also thought diamagnetic materials referred more to materials that naturally
exhibit a strong diamagnetism, much more so than what may be found in normal conductors like
copper. For example Bismuth. still, confusing.
Aluminum, another good conductor, is also diamagnetic. I have seen videos of a magnet slowly sliding down an Al plate. Is Bismuth diamagnetic or paramagnetic? I can't remember...it has been a while.
Bill
PS Here is just one of the aluminum vids: https://www.youtube.com/watch?v=7IzlnFH7ww (https://www.youtube.com/watch?v=7IzlnFH7ww)

Aluminum, another good conductor, is also diamagnetic. I have seen videos of a magnet slowly sliding down an Al plate. Is Bismuth diamagnetic or paramagnetic? I can't remember...it has been a while.
Bill
hell it's been a while myself, I'd have to look up 'paramagnetic' but i'd imagine it's the opposite effect.
I'm pretty sure Bismuth is diamagnetic though. It would certainly seem counter intuitive to say
that copper is not diamagnetic when it does seem to behave that way, like aluminum. I have
never heard that word used to describe the Lenz effect however which is why I can imagine
there is some scientific separation. I'm not one to nitpick over terminology as long as I can understand you of course, but if there IS some separation in the physical explanation of Lenz
law and diamagnetism then I suppose Brad would be correct in excluding the word Diamagnetic.
I suppose the difference lies in the changing magnetic field vs a non changing. Bismuth requires no lenz interaction to get the opposition. (at least from the newtonian side of things to the quantum side of things lol)
edit ok so copper and aluminum are diamagnetic like many other metals. However their
naturally occurring diamagnetism is quite negligible when compared to Bismuth.
So you could say, yes, the magnet acts the way it does in the pipe because copper is a diamagnetic. But to give that explanation also requires that we explain the diamagnetic interaction between the magnet and the copper is a direct result of the Lenz effect, the opposing fields.
so long as a changing magnetic field is brought into the explanation it seems totally proper.
It only becomes confusing when someone can think that copper is naturally diamagnetic like
Bismuth and will yield that field opposition without any changing field interacting with it.

hell it's been a while myself, I'd have to look up 'paramagnetic' but i'd imagine it's the opposite effect.
I'm pretty sure Bismuth is diamagnetic though. It would certainly seem counter intuitive to say
that copper is not diamagnetic when it does seem to behave that way, like aluminum. I have
never heard that word used to describe the Lenz effect however which is why I can imagine
there is some scientific separation. I'm not one to nitpick over terminology as long as I can understand you of course, but if there IS some separation in the physical explanation of Lenz
law and diamagnetism then I suppose Brad would be correct in excluding the word Diamagnetic.
I suppose the difference lies in the changing magnetic field vs a non changing. Bismuth requires no lenz interaction to get the opposition. (at least from the newtonian side of things to the quantum side of things lol)
edit ok so copper and aluminum are diamagnetic like many other metals. However their
naturally occurring diamagnetism is quite negligible when compared to Bismuth.
So you could say, yes, the magnet acts the way it does in the pipe because copper is a diamagnetic. But to give that explanation also requires that we explain the diamagnetic interaction between the magnet and the copper is a direct result of the Lenz effect, the opposing fields.
so long as a changing magnetic field is brought into the explanation it seems totally proper.
It only becomes confusing when someone can think that copper is naturally diamagnetic like
Bismuth and will yield that field opposition without any changing field interacting with it.
If the pipe is cooled down enough,so as it became superconductive,would the magnet still fall?
Think about that,and how it relates to the ideal coil,which is also made of many superconductive loops.
Brad

If the pipe is cooled down enough,so as it became superconductive,would the magnet still fall?
Think about that,and how it relates to the ideal coil,which is also made of many superconductive loops.
Brad
hmm good question, I don't see why it would. but even those scenarios involve an observable magnetic field. which means it's still radiating.

If the pipe is cooled down enough,so as it became superconductive,would the magnet still fall?
Think about that,and how it relates to the ideal coil,which is also made of many superconductive loops.
Brad
Brad:
No! No...
Please do not introduce an ideal pipe into the mix, ha ha. My head is about to explode from all of the other ideal items we have already.
Bill
PS No, the magnet would not fall as Lenz would be increased proportionally to the increased conductivity of the copper pipe.

Brad:
No! No...
Please do not introduce an ideal pipe into the mix, ha ha. My head is about to explode from all of the other ideal items we have already.
Bill
PS No, the magnet would not fall as Lenz would be increased proportionally to the increased conductivity of the copper pipe.
almost spit my mountain dew out at that one

ok, now think about what would happen if you dropped a suitably sized magnet
down a pipe of pure bismuth.
now super cool the bismuth.
lmfao

Can you have a fixed finite inductance from an inductor that has resistance=0 and capacitance=0
Yes

If the pipe is cooled down enough,so as it became superconductive,would the magnet still fall?
Think about that,and how it relates to the ideal coil,which is also made of many superconductive loops.
As well, think about the large number of inductors in use everyday all over the world made just that way...
PW

Brad:
No! No...
Please do not introduce an ideal pipe into the mix, ha ha. My head is about to explode from all of the other ideal items we have already.
Bill
PS No, the magnet would not fall as Lenz would be increased proportionally to the increased conductivity of the copper pipe.
Lolsorry Bill :D
What about thisas we are talking about the magnet falling through the copper pipe.
Now lets say the leading field(the one on the bottom of the PM as we place it in the pipe) of the PM is the north field,and the trailing field is the south field. The magnet will induce a field in that copper pipe that apposes that of the magnet,ans so apposes the motion of the magnet. So just below the falling magnet,the copper pipe will be producing a north fieldthat apposes the motion of the magnet. But dose that mean that a north field is also produced just above the falling magnetso just above the south field of the magnet,that wants to also appose the motion of the magnet.
Dose this mean that the copper pipe now has produced a mono magnetic field? :onorth below the falling magnetpushing against the falling magnet,and north above the falling magnetpulling on the falling magnet.
This is the way it would work with a shorted coil.
A north field approaches the shorted coil,and the shorted coil produces a magnetic field that apposes the north field of the magnet. Then if we turn the magnet around,and we pull the south field away from the shorted coil,then that shorted coil will produce a north field that wants to pull the magnet back toward it.
Sorry in advance Bill :D
Brad

almost spit my mountain dew out at that one
At least your not drinking the cool aid ;)
Brad

Koolaid is good, as long as it is cherry flavour. ;)

Koolaid is good, as long as it is cherry flavour. ;)
We dont have cool aid over herewe have lemon aid of all flavours is that the same?
Brad

If it is a flavoured powder you add to sugar and water, then probably same thing. ;)

If it is a flavoured powder you add to sugar and water, then probably same thing. ;)
Oh no
Lemonaid is like coca cola or pepsi.
Sounds more like our Staminadethats a powder you mix with water to get a fizzy soft drink.
Brad

If it is a flavoured powder you add to sugar and water, then probably same thing. ;)
Have you tried our famous vegimite Poynt?
You have it on toast
Brad

The time constant is infinite.
Tau=L/R. There is no R,as it's an ideal inductor.
Tau=L/0 = infinity.
Good God, Brad!  you are doing a mathematical analysis. Your experimental skills mixed with that one will accomplish wonders.
Anyway, you are correct that when R=0 then Tau=∞ , ...however Tau is not inductance.
Infinite Tau does not mean that current through an ideal inductor cannot change or that it is always constant (i.e. zero) ...like it would be with an infinite ideal inductor.
This is because if you insert R=0 into the formula for inductor's current i(t)=(V/R)*( 1 (e^(t*R/L)) ) you do not get a constant current function ...because there are two competing limits in this formula, one for V/R=∞ and another for (1e^0)=0. If you multiply them together you get ∞ * 0 ...:o
Which limit wins out or how the two limits combine requires higher level of math analysis. Do you want me to go into it?

.

Tau has no bearing on whether current can/will flow or not. The effect it has is how "curvy" the rise of current is, relative to the timing of your test. With an infinite tau, the curve is going to be a straight line,
That is correct.
Making the resistance R=0 "turns off" the real component of impedance but leaves the imaginary part (the inductive reactance (X) ) unaffected.
Their magnitudes combine according to SQRT(R^{2}+X^{2})

It's because the shorted single turn copper coil is producing a magnetic field that apposes that which created it.
I agree and this can be observed in this this video (https://www.youtube.com/watch?v=wUaqXk6axOo).
If the pipe is cooled down enough,so as it became superconductive,would the magnet still fall?
Depends how large the inner diamaeter of the pipe is compared to the magnet. If it is small enough then the magnet would bounce forever.
This video (https://www.youtube.com/watch?v=uL4pfisCX14) illustrates how a falling magnet behaves when it is released from the middle of a single shorted turn that has zero resistance.
P.S.
Again, I remind everyone that the magnitude of the current induced in a perfectly conducting ring does not depend on how fast that magnet is moving.
Also copper cooled by liquid nitrogen does not become superconductive, but the YBCO ceramic does.
Niobium metal becomes superconductive when cooled by liquid helium.

@Magneticist
You should look at that post (http://www.overunityresearch.com/index.php?topic=2684.msg43692#msg43692) of mine from another thread.

Aluminum, another good conductor, is also diamagnetic. I have seen videos of a magnet slowly sliding down an Al plate. Is Bismuth diamagnetic or paramagnetic? I can't remember...it has been a while.
Yes Bismuth is one of the best diamagnets.
The difference between Copper or Aluminum pipe/sheet and Bismuth is that Bismuth will always oppose and repel permanent magnets, while Copper/Aluminum will do so only when these magnets are moving.

No, the magnet would not fall as Lenz would be increased proportionally to the increased conductivity of the copper pipe.
If the superconducting ring or plate is "frozen" while the magnet is away then:
the magnet will fall through a superconductive ring that is much larger than the magnet and if the ring's diameter is sufficiently small compared to the diameter of the magnet, then the magnet will bounce.
And if small superconducting ring or a large plate is "frozen" while the magnet is nearby, then:
the magnet will hover over or under the superconducting ring/plate.

@Magneticist
You should look at that post (http://www.overunityresearch.com/index.php?topic=2684.msg43692#msg43692) of mine from another thread.
I have no issue with that representation because it helps to explain a mathematical
way we can calculate charge/discharge using the constant, at the very least. It also
seems like a great presentation so I will definitely give it a look and thank you for
providing it.
I also have no issue with Poynt99's statement :
"Tau has no bearing on whether current can/will flow or not.
The effect it has is how "curvy" the rise of current is, relative to
the timing of your test. With an infinite tau, the curve is going to
be a straight line,"
the issue I have is when trying to imagine what would 'actually'
happen if this 'ideal inductor' question was suddenly made reality.
It gets into the atomic model and the more detailed aspects of ultra
complicated quantum physics where I am so lost in the grand scheme of things
I can't say I really know how to make heads or tails of it..
I see a more paradoxical issue where we just wouldn't be able to
'see' or 'observe' the current in any way in the real world so it might
as well not exist as 'flowing'. My position is almost agreeing that current
will flow, but flow just as much as it wouldn't. I know this doesn't seem
to make any logical sense to others but what am I left to do? beat my
head against the wall until I no longer see it that way? You can't quite
prove to me that current would flow over 0 resistance unless we are
talking mathematical constructs. and even then, you say you can use
that math to prove this theory but with R=0 I don't see how it could
possibly result in any answer that is not 'undefined', even if you claim
we are using formulas that do not even require a relationship with a unit
of resistance. Somewhere along the lines, a unit of resistance has to matter
in quantum mechanics. The only 'constant' which I see available for us to
make the determination that current will flow, is the assumption it would
do so because it does indeed flow with very little resistance.
In all honestly I don't wholeheartedly believe in hardly any of this
electron theory. I give it the respect it deserves by not completely
dismissing it and attempting to gain an understanding of it because
as far as I know it retains a complex level of mathematical continuity
across the board but that doesn't mean I'm ready to completely
and absolutely accept every aspect of these theories.
they are not totally tangible to me and a lot of times
just flat out go against my better judgement and intuition.
if I had a specific capacitor charged to a certain voltage
and said it had discharged to x volts over a period of time,
and someone was to tell me they did the math and that wasn't
possible.. chances are I was wrong somewhere because as I said
it's been figured out to a complex level of mathematical continuity.
but what if I wasn't wrong? what if all my units were correct?
what would be the real reality of my capacitor? would that need
investigation or should it be immediately dismissed as wrong?
what if I so immediately dismissed it as wrong only to overlook
a possibility that the math was not accounting for an additional
variable? what if? an important discovery or revelation could be
made, or a lot of freaking time could be wasted having a severe
brain fart cause a decimal was where it shouldn't be. to each his own.

Which limit wins out or how the two limits combine requires higher level of math analysis. Do you want me to go into it?
I may not understand the math but I would appreciate a walk through of it.
I thought that the selfinductance was a physical property,, as in if a force of acceleration is applied against the charge carrier the charge carrier would then supply a force of opposition to that acceleration.

@Magneticist
You should look at that post (http://www.overunityresearch.com/index.php?topic=2684.msg43692#msg43692) of mine from another thread.
this is great stuff and really the 'gist' of what we should be focusing on regarding
inductor charging.

oops repeat

I thought that the selfinductance was a physical property,
On the most basic level it is a simple ratio that tells you how many Webers you get per Ampere.
L=Φ/i

How about trying to state inductance in something a bit more tangible that people can relate to? It's all basically the inertia of the current flow when passing through an inductor.
If you have one Henry of inductance and you put a voltage of one volt across it, you get one amp of current after one second.
If you have two Henrys of inductance and you put a voltage of one volt across it, you get onehalf amp of current after one second.
If you have one Henry of inductance and you put a voltage of two volts across it, you get two amps of current after one second.
More Henrys = less amps.
More voltage = more amps.
When you think about that hard you realize that one Henry is one voltsecond per amp.
It's perhaps easier to think about the inverse of that, which is one amp per volt per second.
So, one Henry of inductance gives you one amp per volt per second.
Onehalf of a Henry of inductance is less inertia, so you get two amps per volt per second.
Two Henrys of inductance is more inertia, so you get onehalf of an amp per volt per second.
It's really as simple as that. It's like pushing on a bloody shopping cart, the lighter the shopping cart, the faster it will accelerate, the heavier the shopping cart, the slower it will accelerate. The speed of the shopping cart is the current flow. The mass of the shopping cart is the inductance.
It's all very simple and very understandable if you choose friendly concepts like inertia and friendly units like amps per volt per second.
But unfortunately this simplicity was destroyed by a positively insane argument that no current would flow in an ideal inductor. It's a travesty.
Travesty: 1. a false, absurd, or distorted representation of something. 2. represent in a false or distorted way.

On the most basic level it is a simple ratio that tells you how many Webers you get per Ampere.
L=Φ/i
Thank you

How about trying to state inductance in something a bit more tangible that people can relate to? It's all basically the inertia of the current flow when passing through an inductor.
If you have one Henry of inductance and you put a voltage of one volt across it, you get one amp of current after one second.
If you have two Henrys of inductance and you put a voltage of one volt across it, you get onehalf amp of current after one second.
If you have one Henry of inductance and you put a voltage of two volts across it, you get two amps of current after one second.
More Henrys = less amps.
More voltage = more amps.
When you think about that hard you realize that one Henry is one voltsecond per amp.
It's perhaps easier to think about the inverse of that, which is one amp per volt per second.
So, one Henry of inductance gives you one amp per volt per second.
Onehalf of a Henry of inductance is less inertia, so you get two amps per volt per second.
Two Henrys of inductance is more inertia, so you get onehalf of an amp per volt per second.
Travesty: 1. a false, absurd, or distorted representation of something. 2. represent in a false or distorted way.
It's really as simple as that. It's like pushing on a bloody shopping cart, the lighter the shopping cart, the faster it will accelerate, the heavier the shopping cart, the slower it will accelerate. The speed of the shopping cart is the current flow. The mass of the shopping cart is the inductance.
It's all very simple and very understandable if you choose friendly concepts like inertia and friendly units like amps per volt per second.
But unfortunately this simplicity was destroyed by a positively insane argument that no current would flow in an ideal inductor. It's a travesty.
It's really not that simple MH. And the travesty is you have not taken the time to draw out your own circuit,or realize what you have described. Most every other EE guy has just followed your lead,without thinking or doing the same.
It is like i said,you cannot place an ideal voltage from an ideal voltage source across an ideal inductor. The reason you dont understand this,is because you dont understand your own two component circuit.
Verpies was the only one to touch on why you cannot have or place an ideal voltage across an ideal inductor(quote: Since an ideal inductor must have a zero resistance, this means that it must be shorted (if it ain't shorted, it ain't ideal) and it becomes physically impossible to connect any real voltage sources in series with it.),but i think that your circuit may have past him by as well. Your circuit is an oxymorona paradox,and cannot work in reality,as one cancels out the other. If you took the time to draw out your own circuit,and write down all the values of that circuit,and applied all that you have stated in this(and the JT)thread,then you would see the error of your ways.
But as you continue to try and relate ideal coils to non ideal coils,and ideal voltage sources with non ideal sources,you havnt a hope in hell in seeing what your circuit represents.
I can debunk your circuit in just 5 lines of text,but i will give you and the other EE guys here say4 to 8 weeks lol,just kidding,say 4 days to think about it.
Brad

The travesty continues.
Unfortunately Brad you are not understanding what Verpies is stating or you simply are not reading his actual text properly and/or it is not registering in your brain.
He does NOT say that you cannot place an ideal voltage source across an ideal inductor. He clearly states that you cannot place a real voltage source across an ideal inductor. This is the second time I am telling you what he said in an attempt to correct your misunderstanding.
If you are still confused, then please take it up with Verpies.

It's really not that simple
This may be true but please refrain from engaging with MH here on this thread,, at least for a while so it makes it easier for me to follow and find posts that I am using to further my understanding of things as seen from from another POV.
I would appreciate it,,
Thanks :)

I have no issue with that representation because it helps to explain a mathematical
way we can calculate charge/discharge using the constant, at the very least. It also
seems like a great presentation so I will definitely give it a look and thank you for
providing it.
I also have no issue with Poynt99's statement :
"Tau has no bearing on whether current can/will flow or not.
The effect it has is how "curvy" the rise of current is, relative to
the timing of your test. With an infinite tau, the curve is going to
be a straight line,"
beat my
head against the wall until I no longer see it that way? You can't quite
prove to me that current would flow over 0 resistance unless we are
talking mathematical constructs. and even then, you say you can use
that math to prove this theory but with R=0 I don't see how it could
possibly result in any answer that is not 'undefined', even if you claim
we are using formulas that do not even require a relationship with a unit
of resistance. Somewhere along the lines, a unit of resistance has to matter
in quantum mechanics. The only 'constant' which I see available for us to
make the determination that current will flow, is the assumption it would
do so because it does indeed flow with very little resistance.
In all honestly I don't wholeheartedly believe in hardly any of this
electron theory. I give it the respect it deserves by not completely
dismissing it and attempting to gain an understanding of it because
as far as I know it retains a complex level of mathematical continuity
across the board but that doesn't mean I'm ready to completely
and absolutely accept every aspect of these theories.
they are not totally tangible to me and a lot of times
just flat out go against my better judgement and intuition.
if I had a specific capacitor charged to a certain voltage
and said it had discharged to x volts over a period of time,
and someone was to tell me they did the math and that wasn't
possible.. chances are I was wrong somewhere because as I said
it's been figured out to a complex level of mathematical continuity.
but what if I wasn't wrong? what if all my units were correct?
what would be the real reality of my capacitor? would that need
investigation or should it be immediately dismissed as wrong?
what if I so immediately dismissed it as wrong only to overlook
a possibility that the math was not accounting for an additional
variable? what if? an important discovery or revelation could be
made, or a lot of freaking time could be wasted having a severe
brain fart cause a decimal was where it shouldn't be. to each his own.
the issue I have is when trying to imagine what would 'actually'
happen if this 'ideal inductor' question was suddenly made reality.
It gets into the atomic model and the more detailed aspects of ultra
complicated quantum physics where I am so lost in the grand scheme of things
I can't say I really know how to make heads or tails of it..
I see a more paradoxical issue where we just wouldn't be able to
'see' or 'observe' the current in any way in the real world so it might
as well not exist as 'flowing'. My position is almost agreeing that current
will flow, but flow just as much as it wouldn't. I know this doesn't seem
to make any logical sense to others but what am I left to do?
It makes perfect logical sense. It is only those here that are trying to relate real world device with ideal devices,and the transition just dose not exist .
PW mentioned the real world superconductors we use today,like in MRI scanners. The problem with that is,they are not charged by an ideal voltage source,and so we only have half the circuit MH has proposed.
Current may flow through a shorted ideal inductor,but like you saidhow is it measured?,and how can a voltage placed across an ideal shorted inductor induce a current flow through a shorted ideal inductor?. :o !!Hopefully Poynt will stand back,and look at the circuit MH poses,as MH seems to refuse to do that)
I told MH in the JT thread,that i did not know the answer to the question,in the hope that he would move on,and the JT thread would return back to it's original discussion. But noMH insist that i continue on with the question,and says the first part has been answered. Truth is,it has not been answerednon of it.
So i stand by my answer due to MHs insistence.
You cannot place an ideal voltage from an ideal voltage source across an ideal inductor.
Brad

Some things to consider.
An ideal voltage source is a voltage source that supplies constant voltage to a circuit despite the current which the circuit draws.
This means that despite the resistance which a load may be in a circuit, the source will still provide constant and steady voltage.
An ideal voltage source has the following characterstic that allows it to act as a 100% efficient source of voltage: it has zero internal resistance.
An "ideal inductor" has inductance, but no resistance or capacitance, and does not dissipate or radiate energy.
Why can a voltage not exist across a shorted ideal inductor that has current flowing through it>
Because V=IxR,and R=0
The polarity of the voltage across an inductor is determined bythe positive being the terminal the current is flowing into,and negative being the terminal the current is flowing out from. With a shorted ideal inductor,there are no terminals to measure the voltage across,and the current flowing through that ideal inductor is the same at any two points of that inductor.
Brad

I think use of the term "shorted" when referring to an ideal inductor is not good nor accurate nomenclature.
An ideal inductor simply has zero series resistance. It is not "shorted" in any way, and will not present itself as a "short" if/when a voltage source (ideal or not) is connected across it.

Or,,
if you view it as a single turn connected super conductor,,
I know there is a source but for now,,
There could be voltage that you can not measure or interact with because all the way around the loop the voltage would be uniform,, so current flow but no way to measure the voltage.
Part of what I am thinking over is that there is a nexus event that can and is described from several views of the same event.
The konstruct I have for my way of looking at things is not the same as what is accepted,, but the way I look at things I can use other POV's besides my own at the same time,, with that I do need to know more than just the "basics" of this simple thing known as electricity from the accepted POV so I can compare and adjust my konstruct so that it does not violate what is known to happen.
Like right now I am thinking about the actual inertial interaction of masses,, these little parts do have some mass and if the electron is the charge carrier then to change its random course requires an acceleration,, the input for that comes from somewhere, the reactionary force goes somewhere and somehow,, and then there is the binding force. On top of that the proton also would have an acceleration as well as anything that is bound to that.
This brings things for me more into the mechanical realm which I am more comfortable with,, the CP that binds the electron orbit can then kind of loose the field value,, force at a distance,, and can then be substituted for a physical interaction,, a magic one sure, but a fancy spring for me is easy.

I think use of the term "shorted" when referring to an ideal inductor is not good nor accurate nomenclature.
An ideal inductor simply has zero series resistance. It is not "shorted" in any way, and will not present itself as a "short" if/when a voltage source (ideal or not) is connected across it.
You need to think a little beyond what you are Poynt,and draw the circuit as MH said i should,
I have provided the definitions of the ideal voltage from an ideal voltage source,and an ideal inductor.
If an ideal inductor is shorted,so as it becomes an endless loop,can a voltage be measured anywhere across any two points of that(now looped) ideal inductor while a current is flowing through it?.
Brad

author=webby1 link=topic=16589.msg484135#msg484135 date=1463187555]
There could be voltage that you can not measure or interact with because all the way around the loop the voltage would be uniform,, so current flow but no way to measure the voltage.
It is good to see some one is thinking here.
Part of what I am thinking over is that there is a nexus event that can and is described from several views of the same event.
A paradox ;)
No voltage can be measured across a shorted(looped) ideal inductor,even though current may be flowing through that inductor,as V=IxR stands,and the very reason that current can continue to flow through an ideal inductor is because there is no resistance,and hence the fact that an ideal inductor dose not dissipate any power.
The ideal voltage source that supplies our ideal voltage,by true definition has no internal resistance. As it has no internal resistance,then it too will not dissipate any power by way of an internal resistance. Having no internal resistance,the current flow is in no way impeded ,an so that is what makes it an !ideal! voltage source.
At T=0,the ideal voltage source is connected to the ideal inductor>what have you just done?
Brad
Brad

You need to think a little beyond what you are Poynt,and draw the circuit as MH said i should,
I have provided the definitions of the ideal voltage from an ideal voltage source,and an ideal inductor.
If an ideal inductor is shorted,so as it becomes an endless loop,can a voltage be measured anywhere across any two points of that(now looped) ideal inductor while a current is flowing through it?.
Brad
What is there that is beyond a voltage source and an inductor in series/parallel? I can draw it out in my head, what am I missing?
Let's throw in an ideal switch as well. What happens when the switch closes?
The second part of your question sounds like Faraday induction with the coil shorted.

What is there that is beyond a voltage source and an inductor in series/parallel? I can draw it out in my head, what am I missing?
Let's throw in an ideal switch as well. What happens when the switch closes?
The second part of your question sounds like Faraday induction with the coil shorted.
I would like to ask you the question again Poynt.
1If we have an inductor that has no resistance,and the two ends of that inductor are joined,so as the inductor is now just a continual loop of wire(shorted),can we place a voltage across that inductor that has no resistance.
2If there is now a current flowing in that shorted ideal inductor,is there any two points across that ideal inductor where a voltage can be measured?
Brad

I would like to ask you the question again Poynt.
1If we have an inductor that has no resistance,and the two ends of that inductor are joined,so as the inductor is now just a continual loop of wire(shorted),can we place a voltage across that inductor that has no resistance.
2If there is now a current flowing in that shorted ideal inductor,is there any two points across that ideal inductor where a voltage can be measured?
Brad
As long as the current flowing is increasing or decreasing over time, the answer to both is yes.
It is an inductor...
PW

As long as the current flowing is increasing or decreasing over time, the answer to both is yes.
It is an inductor...
PW
1If we have an inductor that has no resistance,and the two ends of that inductor are joined,so as the inductor is now just a continual loop of wire(shorted),can we place a voltage across that inductor that has no resistance.
2If there is now a current flowing in that shorted ideal inductor,is there any two points across that ideal inductor where a voltage can be measured?
As long as the current flowing is increasing or decreasing over time, the answer to both is yes.
It is an inductor...
I would like to see you do that PW.
Th inductor is a continual loop,and so the current would flow in both directions from the positive connection,to the negative.
the magnetic field produced by one half of the windings would be equal and opposite to that of the field produced by the other half of the windings. So there would be no inductive effect. Than now leaves you with only the resistance in which a voltage can be seen across,and the resistance value is 0.
Brad

I think this is an example of the root cause of one of Brad's problems:
<<< Why can a voltage not exist across a shorted ideal inductor that has current flowing through it>
Because V=IxR,and R=0 >>>
He keeps on going back to Ohm's law for an inductor. He hears "an inductor has no resistance when DC current is flowing through it" then he says "R=0" and then he says the current must be infinity. At least sometimes he says that.
So I don't think he can make a distinction between the special condition where an inductor manifests zero resistance and Ohm's law and zero ohmic resistance.
Assuming that I am correct perhaps someone can lead him out of that quagmire.
MileHigh

I think this is an example of the root cause of one of Brad's problems:
<<< Why can a voltage not exist across a shorted ideal inductor that has current flowing through it>
Because V=IxR,and R=0 >>>
He keeps on going back to Ohm's law for an inductor. He hears "an inductor has no resistance when DC current is flowing through it" then he says "R=0" and then he says the current must be infinity. At least sometimes he says that.
So I don't think he can make a distinction between the special condition where an inductor manifests zero resistance and Ohm's law and zero ohmic resistance.
Assuming that I am correct perhaps someone can lead him out of that quagmire.
MileHigh
The question for you MH
1If we have an inductor that has no resistance,and the two ends of that inductor are joined,so as the inductor is now just a continual loop of wire(shorted),can we place a voltage across that inductor that has no resistance.
2If there is now a current flowing in that shorted ideal inductor,is there any two points across that ideal inductor where a voltage can be measured?
Brad

I think this is an example of the root cause of one of Brad's problems:
<<< Why can a voltage not exist across a shorted ideal inductor that has current flowing through it>
Because V=IxR,and R=0 >>>
He keeps on going back to Ohm's law for an inductor. He hears "an inductor has no resistance when DC current is flowing through it" then he says "R=0" and then he says the current must be infinity. At least sometimes he says that.
So I don't think he can make a distinction between the special condition where an inductor manifests zero resistance and Ohm's law and zero ohmic resistance.
Assuming that I am correct perhaps someone can lead him out of that quagmire.
MileHigh
Well, that depends. Is it an ideal Quagmire? If so, it will be bottomless and have infinity adhesion qualities. By definition, no one can ever escape an ideal quagmire...if they could, then it would have not be an ideal one.
Bill
PS Hopefully it is obvious that I am just attempting a bit of humor. It is, of course, not ideal humor, otherwise everyone would be laughing which I am sure they are not. It is Friday and it has been a long, hard week and I am just blowing off a bit of steam over here.

You are making me think of the Flintstones and the moaning dinosaurs trapped in the tar pits.

I would like to see you do that PW.
Th inductor is a continual loop,and so the current would flow in both directions from the positive connection,to the negative.
the magnetic field produced by one half of the windings would be equal and opposite to that of the field produced by the other half of the windings. So there would be no inductive effect. Than now leaves you with only the resistance in which a voltage can be seen across,and the resistance value is 0.
Brad
Do you believe you can you induce a current flow in an "ordinary" inductor with its ends shorted together?
PW

Brad, the short answer is that you can always place a voltage source across an ideal inductor even if it has "zero resistance." The moment you place a voltage source across an inductor it acts as an impedance and the voltage really exists, and the current starts to change.

The question for you MH
1If we have an inductor that has no resistance,and the two ends of that inductor are joined,so as the inductor is now just a continual loop of wire(shorted),can we place a voltage across that inductor that has no resistance.
2If there is now a current flowing in that shorted ideal inductor,is there any two points across that ideal inductor where a voltage can be measured?
Brad
I have made the question very simple MHsee diagram below.
Brad

I have made the question very simple MHsee diagram below.
Brad
Is this loop current steady, or is it continuouslyvarying with time?

I have made the question very simple MHsee diagram below.
Brad
When the supply is first connected or turned on (T=0) the voltage measured across the inductor will essentially be whatever the V supply voltage is.
It is still an inductor with reactance, and it will take time for the current flow to build.
PW

When the supply is first connected or turned on (T=0) the voltage measured across the inductor will essentially be whatever the V supply voltage is.
It is still an inductor with reactance, and it will take time for the current flow to build.
PW
You have not answered the question asked in the diagram below.
Brad

You have not answered the question asked in the diagram below.
Brad
I believe I did answer the question.
However, if you are referring to DC conditions, the answer would be no. But under time varying conditions, as in when you first connect the supply, the answer is yes.
As this relates to MH's original question, at T=0 there is a measurable voltage across the inductor.
PW

Tinman,
I did not notice the short circuit you drew into your diagram. I mistook those arrows for measurement points (only glanced, time is short right now).
Surely you do not believe that is the equivalent circuit for an ideal inductor with zero resistance.
PW

Tinman,
I did not notice the short circuit you drew into your diagram. I mistook those arrows for measurement points (only glanced, time is short right now).
PW
Surely you do not believe that is the equivalent circuit for an ideal inductor with zero resistance.
You forgot to take notice of the rest of the circuit described in MHs question.
This is what i have been trying to explain all along.
That will come soon enough.
Question 2see diagram below.
Brad

You forgot to take notice of the rest of the circuit described in MHs question.
This is what i have been trying to explain all along.
That will come soon enough.
Question 2see diagram below.
Brad
Why do you indicate a wire shorting your coil? You do know that does not represent an ideal inductor don't you?
As you did not specify an ideal wire, just one with no resistance, shall we assume the wire still has inductance (like any wire)?
PW

You forgot to take notice of the rest of the circuit described in MHs question.
This is what i have been trying to explain all along.
That will come soon enough.
Question 2see diagram below.
Brad
My short answer to question number two as stated in the diagram is yes.
PW

@tinman
For your last question, my answer is yes there will always be some form of impulse voltage but nothing that could be considered as gain since the short is bleeding it right away. This would somewhat equal your Lewan test.
@all
OK, then let me ask you guys this crazy question if possible. We will call it a Bizarre Four Stage Coil (BFSC). It is not an ideal coil, just a regular coil with regular copper wire of let's say 16 awg. You can give it the resistance value you want, it should not change the question.
You apply a DC voltage on the BFSC where in the first quarter of the coil the voltage can exist as applied, in the second quarter the voltage cannot exist, in the third quarter the voltage can exist again as applied and in the final quarter the voltage cannot exist again. What is meant by cannot exist just means the voltage will be zero on those quarters of the coil but will not hinder the voltage from continuing to the next quarter as if the previous quarter was not there. This would infer that the voltage would be as applied through the first and third quarters as if the second quarter was not there although the second and third quarters are there but at zero volts. Zero here should not infer infinity.
The inductance of the total coil is 1H. You apply a DC voltage of 10 volts at 5 amps.
The question is what will the current be in each quarter?
Also, to prevent any "that's impossible" drama you can modify this question any way you want as long as you keep the four quarters and their voltage handling attributes.
I will tell you in advance that this is not a trap question but it will lead to a second question that you may not like. This is an example of puzzles I make for myself to verify EE logic. Let's see how objective you guys really are. You can't get more basic the this while still providing a spread that can actually be discussed.
wattsup

Why do you indicate a wire shorting your coil? You do know that does not represent an ideal inductor don't you?
As you did not specify an ideal wire, just one with no resistance,
PW
As we are talking about ideal inductors,i thought you would have nutted that out for your self.\If we are talking about an ideal coil,would it not be wound with ideal wire?that has no resistance?.
shall we assume the wire still has inductance (like any wire)?
Makes no difference to the answer to the question.
This is a simple test that can be carried out on the bench using a real coil,and provide the correct result.
Brad

Is this loop current steady, or is it continuouslyvarying with time?
.
Brad

As we are talking about ideal inductors,i thought you would have nutted that out for your self.\If we are talking about an ideal coil,would it not be wound with ideal wire?that has no resistance?.
The equivalent circuit model for an ideal inductor is not an inductor with a wire shorting across its ends.
You do know that don't you?
PW

The poor tinman is obviously very,very lost here.
I really like him all the same.
One needs to go right back to the basic concept and start from there.
John.

The equivalent circuit model for an ideal inductor is not an inductor with a wire shorting across its ends.
You do know that don't you?
PW
Of course you're right, the wire is ... ?

Why not a small EEBuiltOff:
Copper pipe, with water flowing in, 2 cell's of a lead battery with SuperCaps, ......??
Just with all the ideas from, WITH respect our EE's, to make / compensate almost ideaal components from what we have on material, is that maybe and hopefully possible?
That would make lovers / rosa glasses from ................ yep, also me!!
Please, 5 min, YT?
Regards, Johan

Tinman,
I did not notice the short circuit you drew into your diagram. I mistook those arrows for measurement points (only glanced, time is short right now).
PW
Surely you do not believe that is the equivalent circuit for an ideal inductor with zero resistance.
I believe that is an ideal inductor that a voltage cannot be measured across at any two points when current is flowing through it.
Brad

As you did not specify an ideal wire, just one with no resistance, shall we assume the wire still has inductance (like any wire)?
PW
Why do you indicate a wire shorting your coil? You do know that does not represent an ideal inductor don't you?
Indeed i do PWsomething for you to remember stating when the end result is presented.
Brad

My short answer to question number two as stated in the diagram is yes.
PW
As i said,this i would like to see.
Brad

The equivalent circuit model for an ideal inductor is not an inductor with a wire shorting across its ends.
You do know that don't you?
PW
Yes i do.
Remember you stated that PW. ;)
Brad

The poor tinman is obviously very,very lost here.
I really like him all the same.
One needs to go right back to the basic concept and start from there.
John.
What one needs to do,is look at MHs circuit as described,and understand what it is.
As soon as you do the MH paradox conversion,where you simply dismiss things like infinite and no resistance,because you !think! they have no meaning due to there value,then you will never arrive at a correct conclusion as to what will happen at T=0
The fact that there is no resistance,is the very reason you do not just dismiss it,just because you think inductance is going to make it obsolete>big error.
Are you going to have a go at answering MHs question John,or are you just going to continue to say i am wrong,even though you can answer the question your self.
Much like telling some one that they dont know what the magnetic force is,even though you have no clue of your own as to what it is.
One can only conclude that your statements and remarks toward me ,are nothing more than you riding on the back of MHand others here as well.
Agree with those you think are correctnot because you have an opinion of your own.
Brad

What is the diagram with a shorted inductor?
John.

tinman, did you study Lewin, Kirchoff for the birds, striped shirt about
16 mins YouTube?
John.

.
Brad

Is this loop current steady, or is it continuouslyvarying with time?
To be more accurate,lets say the current is increasing at 800mA a second.
Brad

tinman, did you study Lewin, Kirchoff for the birds, striped shirt about
16 mins YouTube?
John.
I guess you missed the whole thread dedicated to this very subject that Poynt started.
The total sum of the voltage around that loop = 0,and Kirchoffs law hold's.
So now you have to decide whether you believe Lewin or Poynt?
If lewin is correct,and Kirchoffs law dose not hold,then im screwed.
If Poynt is correct,then my statement also remains correct,in that there is no potential difference that can be measured at any two points across that looped ideal coil. Also remember there is no resistance value in the wire,and so the two resistors in Lewins experiment are omitted.
Below is a circuit showing a real world inductor.
Using this circuit,turn it into an ideal inductor that is looped(shorted)
Lets see if you can do that John.
Brad

When you think about that hard you realize that one Henry is one VoltSecond per Amp.
That is true, too. This is a very useful relation for building electronic circuits. But note, that it involves 4 variables.
For this reason, I do not consider the above relation to be most basic, like the relation of one Henry being one Weber per Amp.
Note that it involves 3 variables. It is also less useful for building electronic circuits  albeit not so for building motors.
Also, notice that it is impossible to have Amps without Webers but it is possible to have Volts without Amps.
This lack of difference makes the 3 variable relation more basic and coherent than the 4 variable relation.

An "ideal inductor" has inductance, but no resistance or capacitance, and does not dissipate or radiate energy.
But it stores energy in the form of magnetic flux. Because of this an inductor can have reactance even of it does not have resistance
An ideal voltage source is a voltage source that supplies constant voltage to a circuit despite the current which the circuit draws.
This means that despite the resistance which a load may be in a circuit, the source will still provide constant and steady voltage.
An ideal voltage source has the following characterstic that allows it to act as a 100% efficient source of voltage: it has zero internal resistance.
That's quite true
It is impossible to connect such voltage source across a shorted ideal inductor, because in such case this voltage source would see a load, which does not have any resistance nor reactance
However it is possible to connect such voltage source in series with an ideal inductor. In this case the inductor becomes shorted by the zero internal resistance of the ideal voltage source and this voltage source sees a load, that does not have any resistance but has reactance.
Why a voltage cannot exist across a shorted ideal inductor that has current flowing through it>
Because there is no place to place the voltmeter probes on a shorted ideal inductor ...in practice as well as in theory.
...and the current flowing through that ideal inductor is the same at any two points of that inductor.
And there is no voltage drop anywhere along the ideal shorted inductor even if there is current flowing through it.

There could be voltage that you can not measure or interact with because all the way around the loop the voltage would be uniform, so current flow but no way to measure the voltage.
And in such case voltage becomes just a postulate ...since you think it's there but cannot measure it.

Having no internal resistance,the current flow is in no way impeded ,an so that is what makes it an !ideal! voltage source.
At T=0,the ideal voltage source is connected to the ideal inductor>what have you just done?
Indeed the ideal voltage source has no internal resistance nor reactance.
However an ideal inductor has reactance even if it does not have resistance.
Impedance = Resistance + Reactance
That's why if an ideal voltage source is connected in series with an ideal inductor, then the current will be impeded by its reactance despite not being impeded by its resistance.
This is the reason why a simple Ohm's law i=V/R is not applicable to calculating current in an L circuit, as it totally disregards the reactance (half of the complex impedance).

What is there that is beyond a voltage source and an inductor in series/parallel?
It seems that you are conflating the series connection of the voltage source with an inductor, with a parallel connection (across a shorted inductor) stipulated by Tinman.

As long as the current flowing is increasing or decreasing over time, the answer to both is yes.
I disagree.
Current can vary in an ideal shorted^{*} inductor only when the flux penetrating it is varied externally somehow.
But despite that variance, there is no place to connect a voltmeter and measure the voltage or voltage drop.
...and presto!, you have a voltageless current  finite current flowing in zero resistance outside of the confines of the Ohm's law.
^{*} (shorted by an ideal wire, not by an ideal voltage source)

Brad,
I am curious as to why you wish to apply an ideal voltage to a shorted ideal coil in reference to MH's original question? IMO, it really has no bearing. Perhaps this idea stems from the second condition of the question where zero volts is applied for 2 secs!?
In regards to an ideal voltage source not changing, we must qualify the change. Assume for a moment that I am the creator of an ideal voltage source. I am free from all known laws to set the voltage level at any magnitude I choose. The magnitude I choose however will not change with any attached load but I am still free to change the magnitude at any given time I wish. The output is still unable to change with any load variation. This is the ideal voltage source MH used in his question.
partzman

I disagree.
Current can vary in an ideal shorted^{*} inductor only when the flux penetrating it is varied externally
Which is what I stipulated, that the current flowing had to vary over time to be able to measure a voltage as per the question.
But despite that variance, there is no place to connect a voltmeter and measure the voltage or voltage drop.
^{*} (shorted by an ideal wire, not by an ideal voltage source)
Consider a series string of ideal inductors connected into a loop with a shorting wire. If you induce a time varying current into the loop, would there not be a measurable voltage drop between the inductors due to the reactance of those inductors?
PW

I think this is an example of the root cause of one of Brad's problems:
<<< Why can a voltage not exist across a shorted ideal inductor that has current flowing through it>
Because V=IxR,and R=0 >>>
He keeps on going back to Ohm's law for an inductor.
Yes, the Ohm's law is not applicable to inductors because it totally disregards the reactance of the inductor.
Resistance is only one half of the total Impedance. It actually is the reason why we have all these words to describe it.

To be more accurate,lets say the current is increasing at 800mA a second.
Brad
It doesn't matter really. All inductors, whether ideal or real will have a voltage across them when there is current through them, regardless if the current is changing or not.

It seems that you are conflating the series connection of the voltage source with an inductor, with a parallel connection (across a shorted inductor) stipulated by Tinman.
I'm not referring to tinman's circuit, but the original one stipulated by MH.

Consider a series string of ideal inductors connected into a loop with a shorting wire.
With or without mutual inductance?
If you induce a time varying current into the loop, would there not be a measurable voltage drop between the inductors due to the reactance of those inductors?
If k<1 and the time varying current was caused asymmetrically, e.g. by varying the flux through only one of the inductors, then there would be a measurable voltage across that inductor as it would act as a voltage source in series with the other inductors.

1If we have an inductor that has no resistance,and the two ends of that inductor are joined,so as the inductor is now just a continual loop of wire(shorted),can we place a voltage across that inductor that has no resistance.
2If there is now a current flowing in that shorted ideal inductor,is there any two points across that ideal inductor where a voltage can be measured?
No to 1 & 2.

Do you believe you can you induce a current flow in an "ordinary" inductor with its ends shorted together?
Yes, by varying the flux penetrating that inductor.
And since that inductor has resistance now, then the current flow will cause a voltage drop across any two points on it.

Brad, Mags,
Please see the attached.
This is the ideal varying voltage source MH poses in his question. It is an arbitrary wave form really, constructed of a number of voltage levels and times. That's all, nothing mysterious, other than it can source infinite current, being ideal and all.
Does this make sense now? I trust this is clearly understood.
The question is, what happens to the circuit current?, i.e. draw out the circuit current on this same graph.

Brad, the short answer is that you can always place a voltage source across an ideal inductor even if it has "zero resistance."
I think you are conflating the series connection of the voltage source and the inductor with the parallel connection of the voltage source to a shorted inductor, as stipulated by Tinman,
Picowatt wrote that you had a different connection in mind earlier in the thread (which I did not read) so the meaning of the word "across" might be at the root of your disagreement with Tinman.

I did not notice the short circuit you drew into your diagram. I mistook those arrows for measurement points (only glanced, time is short right now).
That could be the root of the misunderstanding with Tinman. When he writes "shorted inductor" he means "shorted by an ideal wire" not "shorted by an ideal voltage source".
Surely you do not believe that is the equivalent circuit for an ideal inductor with zero resistance.
I think he does and he is correct.
Why do you indicate a wire shorting your coil? You do know that does not represent an ideal inductor don't you?
The opposite to a shorted inductor is an "open inductor" and such inductor behaves as if it did not exist at all. This can be seen when an "open inductor" is subjected to varying magnetic flux. i.e from an approaching permanent magnet.
Most real inductors are somewhere inbetween  between the "shorted inductor" and "open inductor" ...closer to the shorted, though.

Question 2see diagram below.
No.

Indeed the ideal voltage source has no internal resistance nor reactance.
However an ideal inductor has reactance even if it does not have resistance.
Impedance = Resistance + Reactance
This is the reason why a simple Ohm's law i=V/R is not applicable to calculating current in an L circuit, as it totally disregards the reactance (half of the complex impedance).
That's why if an ideal voltage source is connected in series with an ideal inductor, then the current will be impeded by its reactance despite not being impeded by its resistance.
The connection is not just a series connection when there is only two components.
The connection is both series and parallel.
The fact that the ideal voltage source has no internal resistance,and is connected across the ideal inductor,means the inductor is now a loop connection with no resistance to current flow through that now looped inductor. And as we know,a voltage cannot be measured across any two points of a looped(shorted) ideal inductor.
Regardless of the fact that the ideal voltage source can produce a voltage,the fact that the ideal voltage source is now connected across that ideal inductor,means that the current flowing through it is in no way impeded,nor is there any loss to resistance,as the resistance value is 0and so no power is dissipated.
So now you have a situation where we cannot measure a voltage across any two points of the circuit,and yet at the same time,we are trying to place the very same thing across that circuit that cannot be measured.
Brad

That could be the root of the misunderstanding with Tinman. When he writes "shorted inductor" he means "shorted by an ideal wire" not "shorted by an ideal voltage source".
I think he does and he is correct.
The opposite to a shorted inductor is an "open inductor" and such inductor behaves as if it did not exist at all. This can be seen when an "open inductor" is subjected to varying magnetic flux. i.e from an approaching permanent magnet.
Most real inductors are somewhere inbetween  between the "shorted inductor" and "open inductor" ...closer to the shorted, though.
What are you saying exactly?
That an ideal inductor is one that has a short across its terminals? Please clarify your point.

That could be the root of the misunderstanding with Tinman. When he writes "shorted inductor" he means "shorted by an ideal wire" not "shorted by an ideal voltage source".
I think he does and he is correct.
The opposite to a shorted inductor is an "open inductor" and such inductor behaves as if it did not exist at all. This can be seen when an "open inductor" is subjected to varying magnetic flux. i.e from an approaching permanent magnet.
Most real inductors are somewhere inbetween  between the "shorted inductor" and "open inductor" ...closer to the shorted, though.
Verpies,
I disagree here. The coil resistance should be in series with the inductance not in parallel. If this were a multiple winding arrangement, then the reflected resistance/impedance would be in parallel with the referenced winding.
partzman

Brad,
I am curious as to why you wish to apply an ideal voltage to a shorted ideal coil in reference to MH's original question? IMO, it really has no bearing. Perhaps this idea stems from the second condition of the question where zero volts is applied for 2 secs!?
In regards to an ideal voltage source not changing, we must qualify the change. Assume for a moment that I am the creator of an ideal voltage source. I am free from all known laws to set the voltage level at any magnitude I choose. The magnitude I choose however will not change with any attached load but I am still free to change the magnitude at any given time I wish. The output is still unable to change with any load variation. This is the ideal voltage source MH used in his question.
partzman
Because an ideal voltage source has no internal resistance,and that is what makes it ideal.
The ideal voltage source is a series/parallel connection,as there is only two components in the circuit. As that ideal voltage source provides the very same link across the inductor as the piece of non resistant wire dose,then as soon as you hook the ideal voltage source across that ideal inductor,you have just shorted(looped) that ideal inductor.
If the voltage was reduced to 0 volts on the ideal voltage source,the current flow would continue through the loop that now exist in the ideal coil.=,as the voltage source has no internal resistance to impede the current flow.
Ask your self this.
When MH turns his voltage source down to a value of 0 volts,will the current flow continue on?
If not,then explain as to why notwhat will impede that current flow,when the complete loop from the ideal inductor across the ideal voltage supply has no resistance ?
Who here can draw the complete circuit,along with the resistance values of that circuit?.
Brad

I think he does and he is correct.
Are you stating that you believe the equivalent circuit for an ideal inductor is an inductor in parallel with (shorted) by an ideal wire?
Do you also believe that the equivalent circuit for a normal inductor has its wire resistance in parallel with the inductor?
PW

The equivalent circuit model for an ideal inductor is not an inductor with a wire shorting across its ends.
Just because most of the world does it wrong does not mean that we have to.
An energized capacitor can be correctly modeled in an open state, the inductor  just the opposite.
Are you stating that you believe the equivalent circuit for an ideal inductor is an inductor in parallel with (shorted) by an ideal wire?
Yes, with an addendum that it can also be shorted by an ideal voltage source.
I know that SPICE does not draw an inductor this way, but internally it calculates it that way.
Do you also believe that the equivalent circuit for a normal inductor has its wire resistance in parallel with the inductor?
No, I believe that when the parasitic capacitance is disregarded then the real inductor's equivalent circuit has its wire resistance in series with its inductance and that entire circuit is closed by an ideal wire just like with an ideal inductor devoid of resistance.

Which is what I stipulated, that the current flowing had to vary over time to be able to measure a voltage as per the question.
PW
Consider a series string of ideal inductors connected into a loop with a shorting wire. If you induce a time varying current into the loop, would there not be a measurable voltage drop between the inductors due to the reactance of those inductors?
A MH paradox added?
There is only one inductornot a series of them.
My question was very clear.
Brad

Yes, the Ohm's law is not applicable to inductors because it totally disregards the reactance of the inductor.
Resistance is only one half of the total Impedance. It actually is the reason why we have all these words to describe it.
But it is applicable when that ideal inductor is shorted(becomes a continual loop)
Even when a current is flowing through that looped ideal inductor,ohms law states that V=IxR,and as there is no R,then there is no voltage across that looped inductoras we know.
Brad

I'm not referring to tinman's circuit, but the original one stipulated by MH.
You were answering my question posted on my circuit diagram Poynt.
How can you say that you were referring to MHs question?
Brad

No to 1 & 2.
Thank you verpies
As we know,you are correct.
Brad

A MH paradox added?
There is only one inductornot a series of them.
My question was very clear.
Brad
Not a paradox, a model/tool for visualization purposes.
Every inductor can be thought of as being equivalent to a series string of inductors.
Would not measuring between several turns of your single inductor be equivalent to measuring between several series connected inductors?
PW

What are you saying exactly?
That an ideal inductor is one that has a short across its terminals? Please clarify your point.
No
An ideal inductor that is looped(has a short placed across it by an ideal wire)is a shorted ideal inductor. If there is no wire across the ideal inductor terminals,then it is just an open circuit.
Brad

The connection is not just a series connection when there is only two components.
The connection is both series and parallel.
Seemingly yes, if you do not count the ideal wire that shorts the inductor.
Pay attention to this issue and the associated terminology because in my opinion this is the reason why you and MH cannot agree.
The fact that the ideal voltage source has no internal resistance,and is connected across the ideal inductor,means the inductor is now a loop connection with no resistance to current flow through that now looped inductor.
Yes, there is no resistance to current flow but there is impedance to current flow. That is why the current does not become infinite immediately.
Impedance has two components. Please consider both of them.

Because an ideal voltage source has no internal resistance,and that is what makes it ideal.
The ideal voltage source is a series/parallel connection,as there is only two components in the circuit. As that ideal voltage source provides the very same link across the inductor as the piece of non resistant wire dose,then as soon as you hook the ideal voltage source across that ideal inductor,you have just shorted(looped) that ideal inductor.
If the voltage was reduced to 0 volts on the ideal voltage source,the current flow would continue through the loop that now exist in the ideal coil.=,as the voltage source has no internal resistance to impede the current flow.
Ask your self this.
When MH turns his voltage source down to a value of 0 volts,will the current flow continue on?
If not,then explain as to why notwhat will impede that current flow,when the complete loop from the ideal inductor across the ideal voltage supply has no resistance ?
Who here can draw the complete circuit,along with the resistance values of that circuit?.
Brad
Let me first answer your question. When the ideal voltage source goes to zero volts, the current in the ideal coil will remain at the level it reached with the previous condition and will stay that way until the ideal voltage source is changed to a different magnitude.
What was confusing is that it appeared you wished to apply an ideal voltage across an ideal coil with an ideal short. This is not equivalent to the circuit as MH proposed.
The circuit diagram is simply an ideal source in parallel with an ideal inductor with no resistance.
partzman

No, I believe that when the parasitic capacitance is disregarded then the inductors equivalent circuit has it wire resistance in series with its inductance and that entire circuit is closed by an ideal wire just like with an ideal inductor devoid of resistance.
I was agreeing with this as far as the wire resistance of a normal inductor being in series with the inductor, but when you got to the point where you stated "that entire circuit is closed by an ideal wire just like with an ideal inductor devoid of resistance" you lost me.
Are you referring to some circuit in particular or are you stating that the equivalent model for every inductor includes a short circuit across its terminals?
The model for a normal inductor has the wire resistance in series with the inductor. The model for an ideal inductor removes that series resistor (or places its value at zero). There is no short circuit across either inductor.
PW

You were answering my question posted on my circuit diagram Poynt.
How can you say that you were referring to MHs question?
Brad
Well, actually I wasn't. What I was doing which led to my response was the following:
I think use of the term "shorted" when referring to an ideal inductor is not good nor accurate nomenclature.
An ideal inductor simply has zero series resistance. It is not "shorted" in any way, and will not present itself as a "short" if/when a voltage source (ideal or not) is connected across it.
Here I was responding to a post from verpies. Referring to an ideal inductor as being "shorted" brings no practical utility to the discussion other than to add confusion.
You need to think a little beyond what you are Poynt,and draw the circuit as MH said i should,
I have provided the definitions of the ideal voltage from an ideal voltage source,and an ideal inductor.
If an ideal inductor is shorted,so as it becomes an endless loop,can a voltage be measured anywhere across any two points of that(now looped) ideal inductor while a current is flowing through it?.
Brad
What is there that is beyond a voltage source and an inductor in series/parallel? I can draw it out in my head, what am I missing?
Let's throw in an ideal switch as well. What happens when the switch closes?
The second part of your question sounds like Faraday induction with the coil shorted.
Here I was responding to your response. Clearly I am referring to the original circuit, an ideal inductor with an ideal voltage source connected across it. Then I introduce an ideal switch. You said it yourself, "draw the circuit as MH asked you to", to paraphrase. There is nothing more to the circuit, specifically, no short across the inductor.
It seems that you are conflating the series connection of the voltage source with an inductor, with a parallel connection (across a shorted inductor) stipulated by Tinman.
Here verpies misinterprets my post as referring to your circuit diagram with the short. As I explained, that is not what I was referring to.
I would strongly suggest that all focus on the original question and purpose of this thread. I have now given you a clear illustration of the voltage source contained within the question, and reworded it to get directly to the point of the question.
So Brad, or Mags, will you give it a go? That is, plot out the current trace right over top of the voltage trace I posted?

What was confusing is that it appeared you wished to apply an ideal voltage across an ideal coil with an ideal short. This is not equivalent to the circuit as MH proposed.
That's classic Brad. From what I can gather the schematic that he posted did not match his written description from a few posts earlier, creating confusion.

Seemingly yes, if you do not count the ideal wire that shorts the inductor.
Pay attention to this issue and the associated terminology because in my opinion this is the reason why you and MH cannot agree.
Yes, there is no resistance to current flow but there is impedance to current flow. That is why the current does not become infinite immediately.
Impedance has two components. Please consider both of them.
Quote
Ideal inductors and capacitors have a purely imaginary reactive impedance:
Verpies
Can you define what they mean by !imaginary!,as this seems to only be associated with ideal inductors,where as non ideal(real world)inductors are always stated as having impedance,without the use of the word !imaginary! .
2nd
If the ideal voltage source is reduced to a value of 0 volts instantly,but the non resistive link of the ideal voltage source is still across the ideal inductor,will the current continue to flow through the circuit,as there is nothing to impede this current flow.
Brad

Seemingly yes, if you do not count the ideal wire that shorts the inductor.
Pay attention to this issue and the associated terminology because in my opinion this is the reason why you and MH cannot agree.
That is why the current does not become infinite immediately.
Impedance has two components. Please consider both of them.
Yes, there is no resistance to current flow but there is impedance to current flow.
But that current flow must still be the same throughout the circuit/loop,regardless of the impedance created by the inductor.
As we have both stated,a voltage cannot be measured at any two points across a shorted ideal inductor. Adding the ideal voltage source to the open inductor,also shorts that inductor,due to the fact that the ideal voltage source also has no internal resistancethe loop is complete ,void of any resistance to current flow,regardless of the value of that current flow.
Brad

That's classic Brad. From what I can gather the schematic that he posted did not match his written description from a few posts earlier, creating confusion.
The schematic i posted represents your circuit MH
Would it be easier if i placed a + and  sign in there for you ?
Brad

That's classic Brad. From what I can gather the schematic that he posted did not match his written description from a few posts earlier, creating confusion.
Here is your circuit MH
Make it any easier to understand now?
The MH paradox?
Brad

But that current flow must still be the same throughout the circuit/loop,regardless of the impedance created by the inductor.
As we have both stated,a voltage cannot be measured at any two points across a shorted ideal inductor. Adding the ideal voltage source to the open inductor,also shorts that inductor,due to the fact that the ideal voltage source also has no internal resistancethe loop is complete ,void of any resistance to current flow,regardless of the value of that current flow.
Brad
As the circuit is prescribed, there is always a voltage measured across the terminals of the inductor, and that voltage precisely follows what the voltage source is set to at any point in time.

Most of you lot are very like lavatory brushes, clean round the bend!!
Luckily we've got poynt to point us in the right direction.
It's really good entertainment though.
John.

Here is your circuit MH
Make it any easier to understand now?
The MH paradox?
Brad
Brad,
An ideal voltage source is not properly illustrated as a short circuit.
You need to draw in a voltage source, or function generator there.

Here verpies misinterprets my post as referring to your circuit diagram with the short. As I explained, that is not what I was referring to.
Yes, and I stand corrected. I thought that you were referring to the circuit stipulated by Tinman and depicted in Fig.3.

Yes, and I stand corrected. I thought that you were referring to the circuit stipulated by Tinman and depicted in Fig.3.
Please draw out the current trace of your figure no. 3 then if V=1V and L=1H (both ideal).
If it is anything but an infinite amplitude, what is the point of illustrating it with a short circuit, other than to add confusion?

Brad,
An ideal voltage source is not properly illustrated as a short circuit.
You need to draw in a voltage source, or function generator there.
If it has no internal resistance through that ideal voltage source,how can it not be a short circuit?
Please post a diagram you !your! ideal voltage source.
This is the paradox i am referring to,and why an ideal voltage source cannot exist,and/or be placed across an ideal inductor.
What would happen if we reduced the internal series resistance of a charged cap(say 1000uF with 10 volts across it) to a value of 0(added),and then used that as our ideal voltage source,and placed the ideal inductor across it ?.
Brad

Please draw out the current trace of your figure no. 3 then if V=1V and L=1H (both ideal).
If it is anything but an infinite amplitude, what is the point of illustrating it with a short circuit, other than to add confusion?
Figure 3 represents MHs circuit.
Brad

Figure 2 represents my circuit.

Figure 2 represents my circuit.
No
figure 3 represents your circuit.
Replace your ideal voltage source with an ideal capacitor,and you have the same.
an ideal capacitor is your voltage source,where that ideal capacitor has no series resistance.
You charge your capacitor to 4 volts,and you place it across your ideal coil.
Can you now measure a voltage anywhere around that circuit,even though current is now flowing through it?.
Brad

Please draw out the current trace of your figure no. 3 then if V=1V and L=1H (both ideal).
If it is anything but an infinite amplitude, what is the point of illustrating it with a short circuit, other than to add confusion?
It is to show you that you cannot place a voltage across a shorted(looped) ideal inductor.
Brad

Yes, and I stand corrected. I thought that you were referring to the circuit stipulated by Tinman and depicted in Fig.3.
Verpies, this is mass confusion day. I find your figure 3 confusing. If we view a shorted ideal coil like a classic toroidal inductor, then aren't we talking about putting a voltage source across the toroid such that it ends up looking like two separate coils in parallel?
If that is the case, I fail to see any utility in that.
(sorry for the chicken scratchings drawing.)

Please draw out the current trace of your figure no. 3 then if V=1V and L=1H (both ideal).
In that case:
Current (i) flowing through the inductor equals i=ΔΦ/1H and the current flowing trough the voltage source is infinite.

No
figure 3 represents your circuit.
Replace your ideal voltage source with an ideal capacitor,and you have the same.
an ideal capacitor is your voltage source,where that ideal capacitor has no series resistance.
You charge your capacitor to 4 volts,and you place it across your ideal coil.
Can you now measure a voltage anywhere around that circuit,even though current is now flowing through it?.
Brad
No I assure you that figure 2 represents my circuit.

If it has no internal resistance through that ideal voltage source,how can it not be a short circuit?
If it has 0V across it, then yes it is equivalent to a short circuit. However, this is a voltage source that starts out with +4V across it, therefore it must be drawn as an ideal voltage source or function generator.
Please post a diagram you !your! ideal voltage source.
It is a voltage source (set to voltage X) with a series resistor equaling 0 Ohms.
This is the paradox i am referring to,and why an ideal voltage source cannot exist,and/or be placed across an ideal inductor.
There is no paradox. You are misinterpreting an ideal voltage source as a zero Ohm wire. That is not the case unless as I said, the voltage is set to 0V.
What would happen if we reduced the internal resistance of a charged cap(say 1000uF with 10 volts across it) to a value of 0 ?.
You would have a good quality capacitor. What would happen? Nothing if it wasn't connected to anything. If connected to some load it would happily discharge.

Verpies, this is mass confusion day. I find your figure 3 confusing. If we view a shorted ideal coil like a classic toroidal inductor, then aren't we talking about putting a voltage source across the toroid such that it ends up looking like two separate coils in parallel?
If that is the case, I fail to see any utility in that.
(sorry for the chicken scratchings drawing.)
Well they wont be two seperate coils in parallel MH,because one winding will be CW and the other CCW.
So they will be in series,not parallel
Brad

Verpies, this is mass confusion day.
I kind of like it because it stimulates us to think.
I think this discussion is beneficial to all involved as long as Ad Hominem remarks are absent.
I find your figure 3 confusing. If we view a shorted ideal coil like a classic toroidal inductor, then aren't we talking about putting a voltage source across the toroid such that it ends up looking like two separate coils in parallel?
The confusion will disappear immediately when you consider the coupling coefficient (k) between these two "separate coils" connected "in parallel".
In an ideal toroid the flux is completly shared and k=1, thus in fact these coils are connected in antiparallel when flux direction is considered. Thus they do not posses any inductance collectively. Consequently their antiparallel combination possesses zero reactance and zero resistance, leading to zero impedance and the current flowing through the voltage source rises immediately to infinity.
If that is the case, I fail to see any utility in that.
It is pretty useless from an engineering standpoint, but it has a great educational value.
Anyway, it is what Tinman was referring to all along while you were apparently analyzing Fig.2.

In that case:
Current (i) flowing through the inductor equals i=ΔΦ/1H and the current flowing trough the voltage source is infinite.
Now there is your paradox Brad.
When you place an ideal voltage source across an ideal short, who wins? The voltage source or the ideal wire? verpies seems to indicate that the voltage source wins, as the voltage holds and the inductor still gets some current.

Well they wont be two seperate coils in parallel MH,because one winding will be CW and the other CCW.
So they will be in series,not parallel
Brad
If it really was a toroid, then there will be two junction points where the voltage source makes contact with the toroid as shown in the diagram. Current will flow say from top to bottom in each half coil.
There will be some flux cancellation at the two junction points. But excluding that, for all practical intents and purposes each half of the toroid will be a separate and independent coil. Because they are separate and independent, the winding direction is meaningless.
The two halftoroid coils will look like two separate coils in parallel across the voltage source.
MileHigh
.....................................................
This is an addendum.....
Okay so Verpies is talking about an ideal toroidal inductor where there is total flux coupling from coil turn to coil turn no matter what the physical configuration. It's almost like there is a "magic ideal core" to the toroid directing the flux.
In this case if you connect up the voltage source as per my diagram, the CW and the CCW turns due to the current splitting up in different directions will cause total flux cancellation and it will look like a direct short.
Note as per the toroidal inductor diagram, if the connections are not 180 degrees apart, you get a kind of "dial up an ideal inductor value" configuration.
I can get this but Verpies I think it had to be described better. I hope I am right, and please correct me if I am wrong.

No
figure 3 represents your circuit.
Brad
I interpreted MH's setup more like figure 2,, but it has a one way clutch in the darn thing because once current starts to flow there is nothing that will stop it,, turn down the source voltage to zero and there would be no reduction in current flow,, but then if you go and reverse the polarity of the source you must have and instant stoppage of current flow,, but you can not have that because the source will force the flow the other way.
Now it looks like we would have inertia in play and would that make the ideal not ideal in all facets??

author=poynt99 link=topic=16589.msg484272#msg484272 date=1463239770]
If it has 0V across it, then yes it is equivalent to a short circuit. However, this is a voltage source that starts out with +4V across it, therefore it must be drawn as an ideal voltage source or function generator.
And who states that this is the case?
Regardless of whether there is a voltage or not,the fact that there is no internal series resistance means that the current can still flow through that voltage source unimpeded.
When we drop down to 0 volts (as stated in MHs question),the current will continue to flow,and at a steady state. As verpies stated,you cannot measure a voltage across a shorted(looped) ideal inductor.
It is a voltage source (set to voltage X) with a series resistor equaling 0 Ohms.
There is no paradox. You are misinterpreting an ideal voltage source as a zero Ohm wire. That is not the case unless as I said, the voltage is set to 0V.
And MHs question requires a o voltage at one point in the test. At that point,the current still flow's,as there is no resistance that impedes that current flow through the now looped(shorted )circuit. Now,as virpies stated,you cannot place a voltage across a shorted ideal inductor,and you have just stated that at 0 volts,then yes,the ideal voltage source is like a 0 ohm wire. So now how do you once again increase the voltage after 0 volt's,as you are now trying to place a voltage across an ideal shorted(looped) inductor?
See the paradox now?.
Brad
You would have a good quality capacitor. What would happen? Nothing if it wasn't connected to anything. If connected to some load it would happily discharge.
Now we place that ideal inductor across that ideal capacitorwhat happens?
Can you measure a voltage across any part of that circuit?.

Figure 3 represents MHs circuit.
Brad
Figure 2 represents MH's question, and is an accurate representation of an ideal voltage source and inductor.
An ideal inductor is "theoretically" made from ideal wire having no resistance or capacitance. It is not "shorted" by a wire having no resistance.
A straight length of ideal wire still has inductance and behaves accordingly. Coiled up, that same length of ideal wire has increased inductance. At no time is there any resistance in the ideal wire and at no time is there a requirement to short the inductor with another ideal wire to model it.
PW

An ideal voltage source is not properly illustrated as a short circuit.
So what is the resistance of a 0V voltage source?
....or if you connect an ideal capacitor, charged to 11V, to an ideal 1V voltage source, then what current will flow through the capacitor at t_{0}?

What would happen if we reduced the internal series resistance of a charged cap(say 1000uF with 10 volts across it) to a value of 0(added),and then used that as our ideal voltage source,and placed the ideal inductor across it ?.
An ideal LC circuit will be formed and sinusoidal current will appear.
An infinite current will not flow because the ideal inductor will still possess nonzero reactance and impedance despite having zero resistance.

If it really was a toroid, then there will be two junction points where the voltage source makes contact with the toroid as shown in the diagram. Current will flow say from top to bottom in each half coil.
There will be some flux cancellation at the two junction points. But excluding that, for all practical intents and purposes each half of the toroid will be a separate and independent coil. Because they are separate and independent, the winding direction is meaningless.
The two halftoroid coils will look like two separate coils in parallel across the voltage source.
MileHigh
And here we have another MH paradox added,so as he is not found to be wrong again.
Correct answer.
The two coils will be in seriesnot parallel.
The top of one coil (coil A) will produce say a north field,and the top of the other coil(coil B) will produce a south field. The opposite end of each coil will of course produce the opposite field. As it is a toroid,we can follow a circular path. We will have a field configuration that represents a series connectionNorth to south to north to south. If it were parallel,then we would have north/north at one end,and south /south at the other.
You only have to replace the two coils with batteries,where north may represent positive,and south represent negative,to see that it is a series arrangement ,and not a parallel arrangement.
Brad

And here we have another MH paradox added,so as he is not found to be wrong again.
Correct answer.
The two coils will be in seriesnot parallel.
The top of one coil (coil A) will produce say a north field,and the top of the other coil(coil B) will produce a south field. The opposite end of each coil will of course produce the opposite field. As it is a toroid,we can follow a circular path. We will have a field configuration that represents a series connectionNorth to south to north to south. If it were parallel,then we would have north/north at one end,and south /south at the other.
You only have to replace the two coils with batteries,where north may represent positive,and south represent negative,to see that it is a series arrangement ,and not a parallel arrangement.
Brad
I did an addendum to my posting #382 to account for Verpies' explanation of his configuration.

And who states that this is the case?
That an ideal 0V source is equivalent to an ideal conductor? By deduction, that is what it is. It is common sense. It is impractical of course (except in simulators where it can be used to measure current), but it is a precise equivalent. Doesn't an ideal noninductive conductor always have 0V across it? Well, so does an ideal voltage source set to 0V.
Regardless of whether there is a voltage or not,the fact that there is no internal series resistance means that the current can still flow through that voltage source unimpeded.
When we drop down to 0 volts (as stated in MHs question),the current will continue to flow,and at a steady state.
Yep, I'm not arguing against that.
As verpies stated,you cannot measure a voltage across a shorted(looped) ideal inductor.
And MHs question requires a o voltage at one point in the test. At that point,the current still flow's,as there is no resistance that impedes that current flow through the now looped(shorted )circuit. Now,as virpies stated,you cannot place a voltage across a shorted ideal inductor,and you have just stated that at 0 volts,then yes,the ideal voltage source is like a 0 ohm wire. So now how do you once again increase the voltage after 0 volt's,as you are now trying to place a voltage across an ideal shorted(looped) inductor?
I'm not sure what verpies is saying, but the fact is that when the voltage drops to 0V (even after it was at some nonzero level), you will measure 0V across the inductor.
See the paradox now?.
No.
Now we place that ideal inductor across that ideal capacitorwhat happens?
Can you measure a voltage across any part of that circuit?.
Of course you can.

So what is the resistance of a 0V voltage source?
It is whatever the internal resistance of that source is.
....or if you connect an ideal capacitor, charged to 11V, to an ideal 1V voltage source, then what current will flow through the capacitor at t_{0}?
Infinite current of course.

author=poynt99 link=topic=16589.msg484272#msg484272 date=1463239770]
And MHs question requires a o voltage at one point in the test. At that point,the current still flow's,as there is no resistance that impedes that current flow through the now looped(shorted )circuit. Now,as virpies stated,you cannot place a voltage across a shorted ideal inductor,and you have just stated that at 0 volts,then yes,the ideal voltage source is like a 0 ohm wire. So now how do you once again increase the voltage after 0 volt's,as you are now trying to place a voltage across an ideal shorted(looped) inductor?
See the paradox now?.
Brad
If a voltage source has resistance, this resistance is equivalent to being in series with an ideal voltage generating means. The voltage generating means is free to vary while still maintaining the same series resistance. The same exists in an ideal voltage source except the series resistance is zero. So, you are correct that if we short the ideal voltage source, infinite current will flow. You are also correct that if the voltage source is zero when connected to an ideal inductor, any previous current flowing will continue without change.
However, we are allowed to vary the voltage generating means with a zero series resistance connected to an ideal coil which also has a zero series resistance and due to the nature of the self inductance of the coil, we will experience a change in current depending on the circuit values.
partzman
EditChanged to "connected to"

<<< Now,as virpies stated,you cannot place a voltage across a shorted ideal inductor, >>>
Again, I believe I understand what Verpies meant because he responded to my questions about this.
A "shorted ideal inductor" just becomes a dead short circuit by his definition. Half of the ideal turns cancel out the flux of the other half of the ideal turns and you end up with a "ideal zero ohm resistor with no inductance."
By his definition a "shorted ideal inductor" simply disappears from the circuit and looks like an ideal wire with zero inductance.

It is not "shorted" by a wire having no resistance.
An open inductor makes no more sense than a shorted capacitor and these components are complete conjugates of each other.
Open conductor cannot:
 store energy,
 maintain the level of flux that penetrates it,
 obey the Lenz law,
 generate magnetic flux density gradient that can do mechanical work.
In fact an open inductor loses all of the properties of inductors except an empty voltage potential under varying flux conditions, that cannot even be measured without completing the inductor's circuit.
A straight length of ideal wire still has inductance and behaves accordingly.
Yes, but what does it matter if k (https://en.wikipedia.org/wiki/Inductance#Coupled_inductors_and_mutual_inductance)=1 ?
At no time is there any resistance in the ideal wire and at no time is there a requirement to short the inductor with another ideal wire to model it.
The requirement to close the inductor's circuit is there every time you want an inductor to act as an inductor and do more than nothing.
That's why some good simulators will not even allow you to have an open inductor in a simulation and out of those they don't have problems with open capacitors.
I remind you that inductors and capacitors are complete conjugates (col: opposites) of each other.

That an ideal 0V source is equivalent to an ideal conductor? By deduction, that is what it is. It is common sense. It is impractical of course (except in simulators where it can be used to measure current), but it is a precise equivalent. Doesn't an ideal noninductive conductor always have 0V across it? Well, so does an ideal voltage source set to 0V.
No.
Of course you can.
Regardless of whether there is a voltage or not,the fact that there is no internal series resistance means that the current can still flow through that voltage source unimpeded.
When we drop down to 0 volts (as stated in MHs question),the current will continue to flow,and at a steady state.
Yep, I'm not arguing against that.
I'm not sure what verpies is saying, but the fact is that when the voltage drops to 0V (even after it was at some nonzero level), you will measure 0V across the inductor.
But we agree that the inductor is now a shorted loop,and current is flowing through this circuit loop that includes the ideal voltage source(not to be confused with the ideal voltage,which now has a value of 0 volts)
To quote verpies
It is impossible to connect such voltage source across a shorted ideal inductor, because in such case this voltage source would see a load, which does not have any resistance nor reactance
So now i ask
At 0 volts we both agree that we are in a shorted(looped) condition,and a steady current is flowing through this shorted ideal inductor loop,and due to the 0 resistance value,it will continue to flow infinitely. How is this short removed just by turning up the voltage on the ideal voltage source?
If it is not removed,then how are you placing a voltage across this !now shorted! ideal inductor?
The paradox beingquote verpies>It is impossible to connect such voltage source across a shorted ideal inductor.
Is there some sort of two kinds of !shorted!?.
A short is a shortis it not?.
Brad

I wonder if MH would have said that the current continued to flow through his circuit during the 0 volt time of his question?
I think not,as it would have been something he would never have seen,due to not understanding the meaning of Ideal.
But we will never know,as he never would answer his own question ::)
Brad

I wonder if MH would have said that the current continued to flow through his circuit during the 0 volt time of his question?
I think not,as it would have been something he would never have seen,due to not understanding the meaning of Ideal.
But we will never know,as he never would answer his own question ::)
Brad
You think wrong. I intentionally stated that the voltage becomes zero volts to see if you guys would get that.
You're just being ridiculous when you say I do not understand the meaning of "ideal."
You guys apparently are slowly but surely getting up the learning curve.
Do you now accept the fact that an ideal voltage source can output a voltage waveform that changes in time?

I kind of like it because it stimulates us to think.
I think this discussion is beneficial to all involved as long as Ad Hominem remarks are absent.
The confusion will disappear immediately when you consider the coupling coefficient (k) between these two "separate coils" connected "in parallel".
In an ideal toroid the flux is completly shared and k=1, thus in fact these coils are connected in antiparallel when flux direction is considered. Thus they do not posses any inductance collectively. Consequently their antiparallel combination possesses zero reactance and zero resistance, leading to zero impedance and the current flowing through the voltage source rises immediately to infinity.
It is pretty useless from an engineering standpoint, but it has a great educational value.
Anyway, it is what Tinman was referring to all along while you were apparently analyzing Fig.2.
Great thinking. ;) It is the reverse of the ideal inductor idea of no current can flow, where in your point here is there is no reactance, and the single inductor has 100% reactance. ;) In an ideal world.
Mags

Attached is an equivalent circuit of MH's question using ideal components. The questions are, can we vary Vg and can we use delta I = E*t/L to analyze this circuit? If not, why? The main objection has been the short circuited this and that. I don't see any short circuits. What an I missing here?
partzman

But we will never know,as he never would answer his own question ::)
And isn't it a good thing that I refused to answer the easy question and only answered the hard question, because slowly but surely you are actually learning something now.

But we agree that the inductor is now a shorted loop,and current is flowing through this circuit loop that includes the ideal voltage source(not to be confused with the ideal voltage,which now has a value of 0 volts)
To quote verpies
So now i ask
At 0 volts we both agree that we are in a shorted(looped) condition,and a steady current is flowing through this shorted ideal inductor loop,and due to the 0 resistance value,it will continue to flow infinitely.
As I said, I am not arguing against that.
How is this short removed just by turning up the voltage on the ideal voltage source?
If it is not removed,then how are you placing a voltage across this !now shorted! ideal inductor?
The short is not removed. The voltage source itself is the short (if you will), but it doesn't short itself out!
The paradox beingquote verpies>It is impossible to connect such voltage source across a shorted ideal inductor.
There is no paradox, and verpies is wrong because the inductor does not represent a short the moment it is connected to something, even an ideal voltage source. The only true paradox I've seen so far is verpies' application of an ideal voltage source across an ideal short. Which one wins? That is your paradox Brad.
As I said, talk of such abstract theories as being posed is not helping the understanding here in any way, it is only hindering it.

As I said, talk of such abstract theories as being posed is not helping the understanding here in any way, it is only hindering it.
I agree, Verpies is too far out sometimes. His comments on the thread from a couple of days ago were barely comprehensible.
Verpies: I suggest that you step it down a notch and add some more description at times so your message is more readily understood by both the "ordinaries" and the "gurus."

Attached is an equivalent circuit of MH's question using ideal components. The questions are, can we vary Vg and can we use delta I = E*t/L to analyze this circuit? If not, why? The main objection has been the short circuited this and that. I don't see any short circuits. What an I missing here?
partzman
You are absolutely 100% dead on. Will the train see the light at the end of the tunnel and actually emerge from the tunnel in one piece? That is the question of the hour.

As I said, I am not arguing against that.
The short is not removed. The voltage source itself is the short (if you will), but it doesn't short itself out!
There is no paradox, and verpies is wrong because the inductor does not represent a short the moment it is connected to something, even an ideal voltage source. The only true paradox I've seen so far is verpies' application of an ideal voltage source across an ideal short. Which one wins? That is your paradox Brad.
As I said, talk of such abstract theories as being posed is not helping the understanding here in any way, it is only hindering it.
I agree with you on the no current flow at t/0 of an ideal inductor.
But I dont think that if we're to study the ideals if an ideal inductor that we should ignore the mechanism that makes it do what it does. If resistance is zero, and no losses, then that underlying ideal inductor mechanism should be lossless and 100% efficient also. And if that mechanism is lossless then the the inductor should continuously impede an emf presented at the input. ;) PW says that a straight wire has inductance, and I agree, no matter how tiny the inductance is. So it may be that the ideal straight wire may not be able to pass current if the inductance mechanism is 100% efficient. Does that make any sense at all? If not then where might we find reference that tells us otherwise so we can examine that if possible?
In the real world we have resistance mostly no matter what. So all those losses, voltage drops no matter how tiny would definitely affect the efficiency of that mechanism to be less than 100% efficient, thus there could not be a 100% impediment to the input and the inductor would now work as we know them.
Mags

I agree with you on the no current flow at t/0 of an ideal inductor.
But I dont think that if we're to study the ideals if an ideal inductor that we should ignore the mechanism that makes it do what it does. If resistance is zero, and no losses, then that underlying ideal inductor mechanism should be lossless and 100% efficient also. And if that mechanism is lossless then the the inductor should continuously impede an emf presented at the input. ;) PW says that a straight wire has inductance, and I agree, no matter how tiny the inductance is. So it may be that the ideal straight wire may not be able to pass current if the inductance mechanism is 100% efficient. Does that make any sense at all? If not then where might we find reference that tells us otherwise so we can examine that if possible?
In the real world we have resistance mostly no mater what. So all those losses, voltage drops no matter how tiny would definitely affect the efficiency of that mechanism to be less than 100% efficient, thus there could not be a 100% impediment to the input and the inductor would now work as we know them.
Mags
See, something MH said bothered me a bit. He said that a resistance of .000001ohm, 1uohm was virtually seemless to being an ideal inductor. Would you agree with that statement? Not trying to pit you against him. But it would be nice if that statement were to be considered true by you or not and give us your understanding as to why your answer is what it is.
I say it is very far from seamless, pretend world or not. It was said by Carl Segan that if we cut an apple pie in half, then cut 1 half in half, then cut 1/4 in half, and keep going, I think the number was about 70 or 90 cuts to get to a single atom. It was less than 100 cuts. But we could go further and further, somehow. When does it end? So the '1uohm is seamless with an ideal component' is not an accurate statement and we cannot accept that as fact here. .5uohm has half the resistance of his idealized 1uohm. What about .001uohm? .001 pico ohm? .000000001pico ohm? So I think that should be resolved on that seamless bit. It is not a good representation, of which happens quite often.
Mags

See, something MH said bothered me a bit. He said that a resistance of .000001ohm, 1uohm was virtually seemless to being an ideal inductor. Would you agree with that statement? Not trying to pit you against him. But it would be nice if that statement were to be considered true by you or not and give us your understanding as to why your answer is what it is.
I say it is very far from seamless, pretend world or not. It was said by Carl Segan that if we cut an apple pie in half, then cut 1 half in half, then cut 1/4 in half, and keep going, I think the number was about 70 or 90 cuts to get to a single atom. It was less than 100 cuts. But we could go further and further, somehow. When does it end? So the '1uohm is seamless with an ideal component' is not an accurate statement and we cannot accept that as fact here. .5uohm has half the resistance of his idealized 1uohm. What about .001uohm? .001 pico ohm? .000000001pico ohm? So I think that should be resolved on that seamless bit. It is not a good representation, of which happens quite often.
Mags
Cut that 1uohm in half, 90 times. Are we done yet? Are we on the verge of an ideal component with zero losses yet?
Mags

I agree with you on the no current flow at t/0 of an ideal inductor.
But I dont think that if we're to study the ideals if an ideal inductor that we should ignore the mechanism that makes it do what it does. If resistance is zero, and no losses, then that underlying ideal inductor mechanism should be lossless and 100% efficient also. And if that mechanism is lossless then the the inductor should continuously impede an emf presented at the input. ;) PW says that a straight wire has inductance, and I agree, no matter how tiny the inductance is. So it may be that the ideal straight wire may not be able to pass current if the inductance mechanism is 100% efficient. Does that make any sense at all? If not then where might we find reference that tells us otherwise so we can examine that if possible?
In the real world we have resistance mostly no matter what. So all those losses, voltage drops no matter how tiny would definitely affect the efficiency of that mechanism to be less than 100% efficient, thus there could not be a 100% impediment to the input and the inductor would now work as we know them.
Mags
Mags,
I am not the one you asked for an answer on this but I will go ahead and offer my opinion. It appears from your comment highlighted above, you assume that if an inductance is 100% efficient, it will not allow any current to flow. At 100% efficiency, the emf is totally cancelled by the cemf thus resulting in zero current flow or infinite inductance. It must have even the smallest amount of resistance to perform like a real inductor.
When we apply the formula delta I = E*t/L or rearrange the formula to solve for inductance of emf, we assume ideal conditions such as an ideal inductance. The answer we get from using this formula on an inductance with resistance is very close to what we would measure on the bench. The less dc resistance the coil has, the closer the answer is. From this we can deduce that an ideal 5h inductance is just that, 5h with no impediment from any resistance.
partzman

Mags,
I am not the one you asked for an answer on this but I will go ahead and offer my opinion. It appears from your comment highlighted above, you assume that if an inductance is 100% efficient, it will not allow any current to flow. At 100% efficiency, the emf is totally cancelled by the cemf thus resulting in zero current flow or infinite inductance. It must have even the smallest amount of resistance to perform like a real inductor.
When we apply the formula delta I = E*t/L or rearrange the formula to solve for inductance of emf, we assume ideal conditions such as an ideal inductance. The answer we get from using this formula on an inductance with resistance is very close to what we would measure on the bench. The less dc resistance the coil has, the closer the answer is. From this we can deduce that an ideal 5h inductance is just that, 5h with no impediment from any resistance.
partzman
Yes you do understand my point. Thanks.
As for the rest of it, how can we deduce that the CEMF that counters the input isnt equal to the input? Maybe there is something a miss here. Why isnt the cemf ideal also when it comes to the ideal inductor? Where is the calculation of some loss that keeps the cemf lower than the input?
Again, the sim says what the sim says. Look at the graph again and expand that time out for hours or even days. There will always be a curve in the real world no matter the level of resistance. Who is to say that with absolute zero resistance that the mechanism that is at work in the inductor that impedes the input still has some sort of loss in order for it to be that the cemf is not equaled to the input applied, all of the time? How does that formula account for that? This is the question. ;) Maybe that possibility is ignored some how? Like how would we know for sure without actual hands on testing of such a device? Yet we talk about it as if it is just standard thinking without going any deeper. Just think a little deeper. How deep can we see beyond just some formula dropped on us and think thats all we need to know?
Mags

You are dealing with the same problems in understanding, and the same "rules" that are seemingly made up on the fly using incorrect logic, and the same lack of understanding in how an inductor actually works.
When you connect a real or ideal voltage source to a real or an ideal inductor, or to a resistor, then the EMF imposed on any of those three devices results in a CEMF in the device that is always equal and opposite to the EMF.
The idea that the CEMF must be a bit lower than the EMF for current to flow is 100% wrong, it's a "rule" that has been made up because it "sounds right."
The resistor responds to the EMF with a resultant current flow and that power is burnt off as heat.
The real or ideal inductor responds to the EMF with a resultant current flow and that power is stored in the magnetic field and if there is a resistance then a small amount of that power is burned off as heat.
Why does the inductor do what it does? Why does the shopping cart move when you push on it? Read about an inductor and understand its dynamic response to changing voltage conditions. This is a device whose response is determined by differential and integral equations. Learn intuitively how an inductor responds and learn and understand the equations that back up the intuitive understanding of how the inductor responds. A shopping cart's response is intuitive, you have to get to the point where the inductor's response is intuitive.
I will repeat that this notion that "if the CEMF is the same as the EMF then current will not flow" is a fallacy. It does not apply to a resistor, and it does not apply to an inductor.
What about a capacitor? If the CEMF from the capacitor is the same as the EMF applied to it, does any current flow? The answer is no, there is no current flow. The capacitor is simply not the same as the resistor or the inductor. That's just the way it is and if you want to understand electronics then you have to understand these things.
What about "ideal CEMF?"
You guys define "ideal CEMF" first, and then perhaps we can discuss it. What does that actually mean? Or is that just a meaningless term that has been made up on the fly? Lay your cards on the table about this "ideal CEMF" business.

Who is to say that with absolute zero resistance that the mechanism that is at work in the inductor that impedes the input still has some sort of loss in order for it to be that the cemf is not equaled to the input applied, all of the time? How does that formula account for that? This is the question. ;)
Mags
That's a fair enough question. Let's see if we can logically work out an answer. Obviously the formula(s) don't seem to supply a suitable answer because they assume ideal components. So, maybe we can look at one factor that determines the inductance L of a real air coil for example. This would be the geometry of the coil such as turns, spacing, length of winding and diameter of windings. Apart from any outside influence, this geometry would determine the effect of the self induced emf or cemf produced by the time rate change of current or dI/dt on the applied emf which is basically the inductance of the coil. If we now reduce the resistance of the coil windings to zero, we still maintain the same geometry. So are we to say that the laws of induction due to this particular geometry are going to change because of zero resistance? Or will this example coil become an ideal inductor with pure inductance as determined by it's geometry?
Regarding simulation of an ideal inductor, our current limited versions only offer us the ability to compare a reasonable resistance to a very small resistance. I was able to get LtSpice to simulate the 5H inductor with a dcr of 1e320 without blowing up. The point of this is, the current slope is straighter or has less droop over time with every decrease in dcr. So, could we assume that a trend is established that says if we continue to decrease the dcr to zero, we will experience an infinite straight current line with a perfect dI/dt representing the ideal inductance?
partzman
Edit

MH, ;)

Working on my bike. The quick release lever shaft broke and the nut to nut fasteners on one end of the axle came loose and tightened the axle solid. Couldnt even push it back as the threads of the axle were digging into the axle holder slots. Was on my way back from the beach to get a little sun and see the beautiful women in bikinis everywhere. Then tragedy. About 3 miles from my shop or home. Some guys in a truck offered me a ride. ;D So now I am in fix the bike mode.
The bike is electric. About 80lb. So walking it back was a hard option without the wheel working in any way that it should.
I looked over what both of you posted. Ill go over it again after. Thanks for the explanations. That is all Ive been asking for. Lets see if it all fits.
Back later.
Mags

There is no paradox, and verpies is wrong because the inductor does not represent a short the moment it is connected to something, even an ideal voltage source.
Indeed I would be wrong if I wrote this about an inductor connected to a voltage source as in Fig.2 (http://overunity.com/16589/mhsidealcoilandvoltagequestion/dlattach/attach/157668/) , but I was writing about the circuit stipulated by Tinman, which is the circuit depicted in Fig.3 (http://overunity.com/16589/mhsidealcoilandvoltagequestion/dlattach/attach/157668/).
The latter circuit is equivalent to the diametrically connected (http://overunity.com/16589/mhsidealcoilandvoltagequestion/dlattach/attach/157670/) toroidal inductor that MH invented as an example and was analyzing in collaboration with me.

Verpies: I suggest that you step it down a notch and add some more description at times so your message is more readily understood by both the "ordinaries" and the "gurus."
I don't do it deliberately.
I even consider it my failing if I am not understood.
I need detailed feedback so I can adjust my wording. Without it I am blind to such problems.

See, something MH said bothered me a bit. He said that a resistance of .000001ohm, 1uohm was virtually seemless to being an ideal inductor. Would you agree with that statement? Not trying to pit you against him. But it would be nice if that statement were to be considered true by you or not and give us your understanding as to why your answer is what it is.
You can not really go by the absolute resistance alone to intuit if that will make the inductor act close to "ideal" or not. It really comes down to the ratio of the inductance to resistance, i.e. tau.
Yes, I would agree that 1u Ohm with 5H is sufficient to get good results when compared to an ideal inductor.
Let's explore if tau could be a good indicator of what inductance to resistance ratio is acceptable as an "ideal" inductor for our particular application. So let's look at MH's example, and use 1u Ohm as the inductor's series resistance: 5H/1u = 5Ms. That's a little less than 3/4 of a year. That is a ratio of 5M. Maybe this is excessive? We can also do some tests to see what might be reasonable: With R=10m Ohm, the end current after 3s is 2.389A, which is pretty close to 2.4. The L/R ratio for this case is 500. So we could establish this as a minimum ratio to achieve nearideal results. If your L/R ratio is 500 or greater, you will achieve very close to ideal results.
Another example: if your inductance is 100mH, then you can have a maximum of 200u Ohm before it won't be so close to ideal when using the previous ratio of 500:1.
For interest sake, let's see how far we would be off with a ratio of 100:1. So with L=5H, that would be 50m Ohm. That gives us an average current of 2.34A, so the error is 2.5% or so? What is your desired tolerance? If 5% is acceptable, then perhaps we could go to a ratio of 50:1. Let's see: L=5H, R=100m Ohm. This gives us an average current of 2.28A, which is about a 2.28, exactly 5% error. If that is close enough, then we need only ensure a 50:1 L/R ratio to get results within 5% of ideal.

Indeed I would be wrong if I wrote this about an inductor connected to a voltage source as in Fig.2 (http://overunity.com/16589/mhsidealcoilandvoltagequestion/dlattach/attach/157668/) , but I was writing about the circuit stipulated by Tinman, which is the circuit depicted in Fig.3 (http://overunity.com/16589/mhsidealcoilandvoltagequestion/dlattach/attach/157668/).
The latter circuit is equivalent to the diametrically connected (http://overunity.com/16589/mhsidealcoilandvoltagequestion/dlattach/attach/157670/) toroidal inductor that MH invented as an example and was analyzing in collaboration with me.
Figure 3 is essentially nonsense, and does not aid in the understanding of the problem at hand.

I agree with you on the no current flow at t/0 of an ideal inductor.
So do I and I even came to the same conclusion in a stepbystep analysis of a resistive inductor here (http://www.overunityresearch.com/index.php?topic=2684.msg43692#msg43692).
If resistance is zero, and no losses, then that underlying ideal inductor mechanism should be lossless and 100% efficient also. And if that mechanism is lossless then the the inductor should continuously impede an emf presented at the input. ;)
That's a reasonable thinking but note that constant and continuous impediment does not mean complete impediment. The lack of complete impediment means that the current is not frozen and can increase with time.
That's why when an ideal voltage source is suddenly inserted into an ideal but finite deenergized inductor (as in Fig.2 (http://overunity.com/16589/mhsidealcoilandvoltagequestion/dlattach/attach/157668/)), the current through the inductor increases linearly from zero to infinite current in infinite time  not instantaneously.
PW says that a straight wire has inductance, and I agree, no matter how tiny the inductance is.
I agree with that too.
So it may be that the ideal straight wire may not be able to pass current if the inductance mechanism is 100% efficient.
but when this wire shares the inductor's flux and when the coupling coefficient (k) approaches unity, then this wire, and its inductance, becomes a part of the inductor and closes/completes it losslessly.

It appears from your comment highlighted above, you assume that if an inductance is 100% efficient, it will not allow any current to flow.
His comment referred to current at t_{0} only.
He never wrote that the current will be zero at t_{1}.
The current will be totally impeded at all times (including t_{1}), only if the ideal inductor has infinite inductance.
An ideal inductor must have zero resistance but it does not need to have an infinite inductance (nor infinite reactance).
At 100% efficiency, the emf is totally cancelled by the cemf thus resulting in zero current flow or infinite inductance. It must have even the smallest amount of resistance to perform like a real inductor.
No, an inductor does not need any resistance to perform its magic. Ideal inductors impede current flow due to their reactance, not due to their resistance.
Also, efficiency is inversely proportional to the square of the resistance because only resistance irreversibly converts energy to heat. Inductive reactance does not irreversibly convert energy to heat  it converts it to magnetic field. That conversion is lossless and reversible.
When we apply the formula delta I = E*t/L
The proper formula for current in a series RL circuit stimulated by a positive voltage step is:
i(t)= (V/R)*( 1e^(t*R/L) )
If you take it to the limit R>0. then you get a linear current increase. (from zero if the inductor was not already energized at t_{0})

Why isnt the CEMF ideal also when it comes to the ideal inductor?
Because the CEMF is a function of reactance, not of resistance.
An ideal inductor must have zero resistance but it does not need to have an infinite inductance (nor infinite reactance).

His comment referred to current at t_{0} only.
He never wrote that the current will be zero at t_{1}.
Mags agreed that I was correct with my assumptions so he will have to confirm what he really meant.
[/quote]
The current will be totally impeded at all times (including t_{1}), only if the ideal inductor has infinite inductance.
An ideal inductor must have zero resistance but it does not need to have an infinite inductance (nor infinite reactance).
No, an inductor does not need any resistance to perform its magic. Ideal inductors impede current flow due to their reactance, not due to their resistance.
[/quote]
Yes I am fully aware of this. Let me make it clear that all of my first paragraph is stating what I assume Mags believes at this point in time and is not what I hold to believe. I apologize for my wording being confusing.
[/quote]
Also, efficiency is inversely proportional to the square of the resistance because only resistance irreversibly converts energy to heat. Inductive reactance does not irreversibly convert energy to heat  it converts it to magnetic field. That conversion is lossless and reversible.
The proper formula for current in a series RL circuit stimulated by a positive voltage step is:
i(t)= (V/R)*( 1e^(t*R/L) )
[/quote]
Yes I agree but my reason for using the formula as stated is because no resistance is included so therefore it is assuming an ideal inductor which was my point.
[/quote]
If you take it to the limit R>0. then you get a linear current increase. (from zero if the inductor was not already energized at t_{0})
[/quote]
Yes I again agree.
partzman

I was agreeing with this as far as the wire resistance of a normal inductor being in series with the inductor, but when you got to the point where you stated "that entire circuit is closed by an ideal wire just like with an ideal inductor devoid of resistance" you lost me.
Are you referring to some circuit in particular or are you stating that the equivalent model for every inductor includes a short circuit across its terminals?
The model for a normal inductor has the wire resistance in series with the inductor. The model for an ideal inductor removes that series resistor (or places its value at zero). There is no short circuit across either inductor.
PW
Is this not the model below?
Regarding MHs question,will there not be an alternating current flow?
If we take into account that MH thinks DC means a steady state flow of current in one direction,then we would have to use the AC model to satisfy MH :D
Brad

Now there is your paradox Brad.
When you place an ideal voltage source across an ideal short, who wins? The voltage source or the ideal wire? verpies seems to indicate that the voltage source wins, as the voltage holds and the inductor still gets some current.
What you would get is a big explosion the unstoppable force meets the unmovable object.
If there is a dead short across the ideal voltage supply,the current would simply build in the ideal voltage supply until either the short exploded,or the ideal voltage supply exploded. This would depend on which one of the two could contain the most energy before it failed>or they(the shorted ideal wire and ideal voltage source) would continue to store the energy for an infinite time.
Brad

You have an ideal voltage source and an ideal coil of 5 Henrys. At time t=0 seconds the coil connects to the ideal voltage source. For three seconds the voltage is 4 volts. Then for the next two seconds the voltage is zero volts. Then for two seconds the voltage is negative three volts, and then for the next six seconds the voltage is 0.5 volts. Then after that the voltage is zero volts.
What happens from T=0 when the ideal voltage is connected to the ideal coil?.
After reading all of what has been posted from everyone here>
At T=5 seconds,MHs device explodes.
Brad

If we take into account that MH thinks DC means a steady state flow of current in one direction,then we would have to use the AC model to satisfy MH :D
Brad
I don't really think that you want to act like a little imp with respect to me when it comes to this discussion. I assure you that people that understand inductors knew exactly what I meant by saying "there is no voltage across an ideal inductor inductor with DC current flowing through it." I did not have to say "constant DC." You are the person that shockingly mixed up "constant DC" and "changing DC" and you used that ignorance in a vain attempt to "set me up" and it totally backfired on you to an extreme level.
An ideal voltage source does not "contain energy" and likewise an ideal short does not "contain energy." You still need to work on that angle.
No "little imp" please. Try to answer the question, that's your real goal.
MileHigh

Why is that Brad?

I don't really think that you want to act like a little imp with respect to me when it comes to this discussion. I assure you that people that understand inductors knew exactly what I meant by saying "there is no voltage across an ideal inductor inductor with DC current flowing through it." I did not have to say "constant DC." You are the person that shockingly mixed up "constant DC" and "changing DC" and you used that ignorance in a vain attempt to "set me up" and it totally backfired on you to an extreme level.
No "little imp" please. Try to answer the question, that's your real goal.
MileHigh
An ideal voltage source does not "contain energy" and likewise an ideal short does not "contain energy." You still need to work on that angle.
When we are talking about your circuit MH,then while your voltage value from your ideal voltage source is 0v,then yes,the ideal inductor dose contain/store energy,and that energy can be recovered when the shorted(looped) ideal inductor becomes open.
Regarding the DC current flow thing,it was you that told me to be specific when defining terms,when those terms have multiple meaningssuch as a DC current dose.
Get off your ass,and stop being lazy,and be specific as you have told me to befor those less in the know,as you described.
Brad

Why is that Brad?
Very simple Poynt.
As we have agreed on,while the ideal voltage source has a value of 0 volts,the current will continue to flow through this ideal inductor loop.
At T=5 seconds,the voltage source wants to instantly reverse that current flow direction.
So we have an ideal 5 Henry coil,with current flowing in say a CW direction around the loop.
At T= 5 seconds,the coil is instantly supplied negative 3 volts from an ideal voltage source.
At that instant,you have to infinite current values trying to flow in opposite directions.
Lets use MHs water flow in pipes analogy.
We have water flowing through a pipe at a set rate. We then try to change that flow of water to the opposite direction in an instant. The instantaneous pressure would be infinite if nothing gave outbut something will.
Here in this situation that is !ideal!,and no energy can be dissipated either by the ideal inductor,or the ideal voltage source,you have to wonder what will happen at the instant of the collision of the two meeting apposing current flows.
In order for the current to stop flowing,it must be dissipated into a loadone that can dissipate the energy of that current flow. Normally this can be done just through the resistance of the winding wire it self,but here we have ideal wire that cannot do this.
So at T=5 second's,something has got to give.
Brad

At T=0,the ideal voltage source is placed across the ideal inductor.
At T=3s,the current flowing through that inductor/ideal voltage source loop will be 2.4 amps.
From T=3s to T=5s(the 0 volt phase),the current flowing through that ideal loop will remain at 2.4 amps.
At T=5s,the ideal voltage source(being one that will supply 3 volts across a load,no matter what the load)wants to supply a voltage that will cause a current that apposes that which is already flowing through the ideal inductor/voltage source loop.
Being that both the inductor and voltage source is ideal,the energy stored in the ideal loop from T=3s to T=5s cannot be dissipated in order for a current to start flowing in the opposite direction,and the ideal voltage source says that it will supply 3 volts in opposition to the present current flow no matter what .
At T=5s,MHs circuit explodes.
Brad

@tinman
I did this one in 2013.
https://www.youtube.com/watch?v=iAeoktG6hkg
Could say more but.
wattsup

At T=5s,MHs circuit explodes.
Well, the circuit doesn't explode. So think it through some more and try to come up with a better answer or perhaps someone else will chime in.

Well, the circuit doesn't explode. So think it through some more and try to come up with a better answer or perhaps someone else will chime in.
Oh,of course MH.
The MH paradox kicks in,and the stored energy before T=5s,just disappears(it has to,as there is nothing in the circuit that can dissipate this energy),so as the ideal voltage source can now induce a reversed state of energy into the inductor loop.
This MH paradox is truly fantasticit makes everything work just the way you want it to.
Throw out resistance,throw out time constant's,make DC mean what ever you want it to meanwho needs all that crap when you have the MH paradox>you bloody ripper :D
Brad

Oh,of course MH.
The MH paradox kicks in,and the stored energy before T=5s,just disappears(it has to,as there is nothing in the circuit that can dissipate this energy),so as the ideal voltage source can now induce a reversed state of energy into the inductor loop.
This MH paradox is truly fantasticit makes everything work just the way you want it to.
Throw out resistance,throw out time constant's,make DC mean what ever you want it to meanwho needs all that crap when you have the MH paradox>you bloody ripper :D
Brad
You have to think Plan B because Plan A is a nogo.

You have to think Plan B because Plan A is a nogo.
Ok,i will give it some more thought.
Brad

Ok,i will give it some more thought.
Brad
No
No matter how i try and find a way for the stored energy to be dissipated before the opposite potential of that stored energy is released into the system,there is just no where for it to go.
Even using your flywheel analogy MH,it would mean the the flywheel would have to be stopped instantly,and sent spinning in the opposite direction. If there is no where to burn off(dissipate)that stored energy in the spinning flywheel,then the flywheel cannot be stoppedbut were going to do that anyway with the ideal voltage source.
The homopolar motor would be a very close representation of this situation.
You place a voltage across the round magnet,and the magnet starts spinning in one direction. You remove the voltage ,and the magnet will keep spinning due to inertia/energy stored within that spinning mass. We then swap the voltage polarity over,and apply it to that spinning magnet,and now the magnet wants to spin the other way. The spinning magnet now becomes a generator that is producing a current flow opposite to that of the now inverted voltage. As we know,at this point in time,we will now have a very high current value between the spinning magnet,and the voltage source,and a lot of heat will be produced because of this. In this case,the energy stored in the spinning magnet is dissipated as heat,and the magnet will slow to a stop,and start spinning the other way. But in the ideal case,there is no resistance,and there for no way to dissipate the stored energy as heat. One current source want to go one way,and the other current source want the current to start flowing the other way. The result is an extreme instant current between the stored energy and the ideal voltage source.
I can see no where in the ideal circuit,where the stored energy can be dissipated,and it cant be stored within the system,as it is the opposite to that of what the ideal voltage source wishes to place in the system.
Brad

Hi Tinman, "I can see no where in the ideal circuit,where the stored energy can be dissipated,and it cant be stored within the system,as it is the opposite to that of what the ideal voltage source wishes to place in the system."
Maybe ,and is what I think is, that all systems do store opposites at the same time occupying the same space.
artv

Hi Tinman, "I can see no where in the ideal circuit,where the stored energy can be dissipated,and it cant be stored within the system,as it is the opposite to that of what the ideal voltage source wishes to place in the system."
Maybe ,and is what I think is, that all systems do store opposites at the same time occupying the same space.
artv
To put it simple,the flywheel cannot spin CW and CCW at the same time.
The energy is stored as a magnetic field in this case,and the magnetic field must decline to a 0 value before the opposite field can be induced. The only way the original magnetic field can collapse,is if the energy stored in that field has some where to go,or can be dissipated. As it has no where to go,due to it being in a closed loop,and there is no way of dissipating it's stored energy,due to there being no resistance in the loop,then it must remain.
Brad

To put it simple,the flywheel cannot spin CW and CCW at the same time.
The energy is stored as a magnetic field in this case,and the magnetic field must decline to a 0 value before the opposite field can be induced. The only way the original magnetic field can collapse,is if the energy stored in that field has some where to go,or can be dissipated. As it has no where to go,due to it being in a closed loop,and there is no way of dissipating it's stored energy,due to there being no resistance in the loop,then it must remain.
Brad
Not trying to get into a debate, not trying to learn anything or teach anyone anything. These are my opinions, they do not require the approval anyone reading this.
A flywheel cannot spin in two directions at once, however, one can spin two in opposite directions at the same time and arrange them in a manner which will allow them to constructively or destructively influence one another.
In my opinion its not wise to compare flywheel energy storage with magnetic energy storage, there are many reasons for this, for the sake of this post I only refer to two or three. The speed at which a magnetic energy storage can be charged and discharged is orders of magnitude faster than a flywheel energy storage is able to charge and discharge. The reactive cross section of the magnetic energy storage can be manipulated on the fly where as the flywheel energy storage cannot. There's the issue with the magnetic field of a magnetic energy storage system inverting when the current responsible for its manifestation is removed. The potential which is induced by an expanding magnetic field (magnetic energy storage being charged) is not the same polarity as the potential which is induced by a collapsing magnetic field (magnetic energy storage being discharged). This latter bit is difficult pill for some to digest, I take comfort in Faraday's law, from it one is informed of the polarity of the inducing field via the polarity of the EMF it induces.
When we close the switch on the circuit which includes an inductor "something" is motivated to move in a very specific manner through and about that circuit. Whatever this "something" is, it tends to want to continue in this very same direction after the circuit is opened, whatever this "something" is, its crystal clear (to me) it's not magnetism, the magnetism inverted, that which remains nameless was motivated by the magnetism, it continues to move in the direction dictated by the conditions established when the magnetic energy storage was charging....
I don't need anyone to tell me what I can and more than likely have read over and over again on my own. What I have read, what we've all read, cannot explain what I am seeing on the bench, so please save the lessons in electronics, and "accepted" component behavior I am not interested. Please don't take this the wrong way, I am simply growing tired of coming here and seeing nothing but this game of who has the biggest pair. It should be clear by now that the guy with the biggest pair is the one who was kicked the hardest.
Fact is....we are all looking for something, most are looking for the same thing, it's clear from a casual review of the various topics that no one here has it, all the more reason to work together and not waste time nit picking bullshit. Stop trying to teach and correct each other, it's not working, its counter productive. If we work together and find it, but don't know what we have because we don't know what we are looking at....BIG DEAL, we will loose it, let them loose it! We will be forced to retrace our steps, and this second, third or 100th time around we will be mindful, motivated in by the fact that we did it, but lost it, and also motivated by the whispers of "I told you so....I told you so..."
Regards

Not trying to get into a debate, not trying to learn anything or teach anyone anything. These are my opinions, they do not require the approval anyone reading this.
A flywheel cannot spin in two directions at once, however, one can spin two in opposite directions at the same time and arrange them in a manner which will allow them to constructively or destructively influence one another.
In my opinion its not wise to compare flywheel energy storage with magnetic energy storage, there are many reasons for this, for the sake of this post I only refer to two or three. The speed at which a magnetic energy storage can be charged and discharged is orders of magnitude faster than a flywheel energy storage is able to charge and discharge. The reactive cross section of the magnetic energy storage can be manipulated on the fly where as the flywheel energy storage cannot. There's the issue with the magnetic field of a magnetic energy storage system inverting when the current responsible for its manifestation is removed.
I don't need anyone to tell me what I can and more than likely have read over and over again on my own. What I have read, what we've all read, cannot explain what I am seeing on the bench, so please save the lessons in electronics, and "accepted" component behavior I am not interested. Please don't take this the wrong way, I am simply growing tired of coming here and seeing nothing but this game of who has the biggest pair. It should be clear by now that the guy with the biggest pair is the one who was kicked the hardest.
Fact is....we are all looking for something, most are looking for the same thing, it's clear from a casual review of the various topics that no one here has it, all the more reason to work together and not waste time nit picking bullshit. Stop trying to teach and correct each other, it's not working, its counter productive. If we work together and find it, but don't know what we have because we don't know what we are looking at....BIG DEAL, we will loose it, let them loose it! We will be forced to retrace our steps, and this second, third or 100th time around we will be mindful, motivated in by the fact that we did it, but lost it, and also motivated by the whispers of "I told you so....I told you so..."
Regards
The potential which is induced by an expanding magnetic field (magnetic energy storage being charged) is not the same polarity as the potential which is induced by a collapsing magnetic field (magnetic energy storage being discharged). This latter bit is difficult pill for some to digest, I take comfort in Faraday's law, from it one is informed of the polarity of the inducing field via the polarity of the EMF it induces.
Well as we are not talking about an EMF,then i dont know what your post is about?.
We are talking current flow,and the direction of the current flow that created the magnetic field will continue to be the same when that field collapses. The circuit is a closed ideal loop,and so there is no open circuit for an EMF to be produced.
When we close the switch on the circuit which includes an inductor "something" is motivated to move in a very specific manner through and about that circuit. Whatever this "something" is, it tends to want to continue in this very same direction after the circuit is opened, whatever this "something" is, its crystal clear (to me) it's not magnetism, the magnetism inverted, that which remains nameless was motivated by the magnetism, it continues to move in the direction dictated by the conditions established when the magnetic energy storage was charging....
Once again,the circuit is never openit remains a constant loopan ideal constant loop.
Brad

Very simple Poynt.
As we have agreed on,while the ideal voltage source has a value of 0 volts,the current will continue to flow through this ideal inductor loop.
At T=5 seconds,the voltage source wants to instantly reverse that current flow direction.
So we have an ideal 5 Henry coil,with current flowing in say a CW direction around the loop.
At T= 5 seconds,the coil is instantly supplied negative 3 volts from an ideal voltage source.
At that instant,you have to infinite current values trying to flow in opposite directions.
Why do you believe the current will be infinite? In your posts following this one, you seem to be on board with the idea that the current after the first 3s is 2.4A, correct?
You also mentioned that at T=5s, the circuit would explode. This is not the case. Does the circuit explode when the circuit connection is first made? I can tell you that the two scenarios are no different, and the inductor simply adjusts for a change in current. In other words, going from 0V to 4V, and the resulting final value of 2.4A, is no different than going from 0V to 3V with a starting current of 2.4A. The circuit current simply begins adjusting/changing as a result of the new voltage.
Have you drawn out the circuit current as I suggested? I suspect not. Here is a hint: at t=0s, the current ramps up linearly to the final value of 2.4A at t=3s. Between t=3 and t=5, the current remains at 2.4A. Now, what do you suppose happens to the current between t=5 and t=7? Between t=7 and t=13? From t=13 onward?

Brad,
Here is an inverted color version of the graph in case you want to print it out and finish the current plot with a pen a ruler.

Why do you believe the current will be infinite? In your posts following this one, you seem to be on board with the idea that the current after the first 3s is 2.4A, correct?
You also mentioned that at T=5s, the circuit would explode. This is not the case. Does the circuit explode when the circuit connection is first made? I can tell you that the two scenarios are no different, and the inductor simply adjusts for a change in current. In other words, going from 0V to 4V, and the resulting final value of 2.4A, is no different than going from 0V to 3V with a starting current of 2.4A. The circuit current simply begins adjusting/changing as a result of the new voltage.
Have you drawn out the circuit current as I suggested? I suspect not. Here is a hint: at t=0s, the current ramps up linearly to the final value of 2.4A at t=3s. Between t=3 and t=5, the current remains at 2.4A. Now, what do you suppose happens to the current between t=5 and t=7? Between t=7 and t=13? From t=13 onward?
I do not agree with this. The energy stored in the magnetic field cannot just disappear. That energy has to be dissipated before the opposite magnetic field can be inducedwhich is what the3 volts is trying to do. The startup current is dampened by the induction process due to the CEMF. But there is now a steady current flow through the looped inductor,and a magnetic field that is storing energy. The 3 volts is applied,and the current being produced is trying to collapse that already built magnetic field,which cannot be collapsed due to the steady state current that is flowing that keeps it built. In a real world situation,that energy would be dissipated as heat,but as we have an ideal inductor loop,then the energy cannot be dissipated.
Brad.

The energy in the inductor does not "disappear", some is returned to the source during that period.

I honestly disagree with the handholding at this point. There are others that were interested in this and I suggested they brainstorm. Magluvin may also give it a shot when he comes back.
The big "breakthrough" is apparently Brad now accepts that the current ramps up to 2.4 amps after three seconds. What about the other people that disagreed vehemently about this, what do you have to say?

I wasn't planning on going much further MH.
There are enough clues now to fully and correctly answer the question, as long as the "infinite current" and "shorted inductor" barriers are no longer in the way.

I do not agree with this. The energy stored in the magnetic field cannot just disappear. That energy has to be dissipated before the opposite magnetic field can be induced
Yes the energy stored in the magnetic field cannot just disappear but it can be transferred elsewhere without dissipation.
P99's current graph is correct so far.

Just got to sit and go over it all, and there is more I see.
Have to run out for a while. Life.
Just a quicky....
In an ideal world, Im sure that the windings of the coil would be made to perfection as would be the conductors themselves. Zero flaws. Made in a coil shape, not just bent into shape. Ill go further on that when I get back. Would a single ideal wire allow current to flow? Until I get a grip an all that has been posted since yesterday, Im still looking at the possibility that current wouldnt flow. Just thoughts, not saying you guys are wrong.
Mags

The energy in the inductor does not "disappear", some is returned to the source during that period.
If the source is ideal,and the voltage is fixed at 3v,then the flow of energy is in opposition to that stored in the inductor loop/magnetic field.
So this energy that is stored cannot return to the source,as the energy from the source is flowing in the wrong direction. This would be like trying to force a water flow through a pump in opposition to the water already flowing out of the pump.
Brad

There are others that were interested in this and I suggested they brainstorm. Magluvin may also give it a shot when he comes back.
I honestly disagree with the handholding at this point.
Lets all be honest here MH.
You would never have stated in your answer(if you gave one),that the current would have continued to flow during the 0 volt portion of your question. You would have depicted a sloped drop of current,and then onto the negative voltage protion of the question.
This is why you never answered your questionyou needed others input.
But of course,we will never know,as you did not answer your own question.
The big "breakthrough" is apparently Brad now accepts that the current ramps up to 2.4 amps after three seconds. What about the other people that disagreed vehemently about this, what do you have to say?
No i do not.
I am going through the process using your understanding.
The coil is ideal,and so should be the CEMF.
I am yet to see any reason posted why the CEMF is also not ideal.
Brad

Something that keeps coming to mind as i think....
When we see the demo of a magnet being suspended above a superconductor, It is said that as we put the magnet in place, I suspect a particular height above the super conductive object, that as the field of the magnet sets up lenz currents in the superconductor that are set in motion indefinitely and it sets up an opposing field against the mags field and it floats. So in that case the lenz field is even stronger than the mags field at float as to hold up the weight of the magnets mass, let alone just be equal to the field of the magnet. So is it not possible that this balancing act could be inherent in the ideal perfect inductor? Still thinking on it all. Which is good I suppose. Have some work to do tonight. Will see what I figure on it all
Mags

Lets all be honest here MH.
You would never have stated in your answer(if you gave one),that the current would have continued to flow during the 0 volt portion of your question. You would have depicted a sloped drop of current,and then onto the negative voltage protion of the question.
This is why you never answered your questionyou needed others input.
But of course,we will never know,as you did not answer your own question.
Brad, you really are being silly with what you said there. I know you and MH don't seem to be getting along well, but he deserves more benefit of doubt than you are affording him here on this electronics question. MH did know the answer to the question, and could easily have plotted out the current trace as I have encouraged you to do, but he wanted you and others to work it through yourselves.
MH's question is thoughtprovoking and well thought out, although the wording could have been geared more towards the experimenters here rather than the techies. At any rate, I can assure you that MH knew all along exactly how the circuit currents behave, given the input he specified. After all, he is the one that formulated the question.

If the source is ideal,and the voltage is fixed at 3v,then the flow of energy is in opposition to that stored in the inductor loop/magnetic field.
So this energy that is stored cannot return to the source,as the energy from the source is flowing in the wrong direction.
Sounds to me like you are agreeing with me.
The source is trying to drive the opposite polarity energy into the inductor. When the source is at 3V, it offers up a place for the inductor to dump some of its stored energy. Given more time at 3V, the inductor would fully lose its "positive" energy, and begin to energize in the "negative" direction.
Since it isn't long enough for that to occur, the energy in the inductor simply decreases to a lower "positive" level, and there is only one place for that lost energy to go. You can even see it in the sim.

Had a thought while drilling holes for this sub box that I wanted to bring up.
While thinking about a physical ideal inductor, I visualized current through the conductors. If the conductor were very small, very fine wire, would there be a limit to how much 'current' by definition could flow through that ideal wire? Make the wire half the dia. Any limits there? Cut the dia in half again. Any limits? Lets think that there may be limits in the superconductive ideal conductors. Like if we kept increasing current, would there end up being some of it in skin effect?
If we have enough current through a given short piece of wire that ALL the electrons of the conductor(the ones that can)are in motion(Bill says they move at the same speed when they move, and I agree) with the direction of current. All moving in sync. Can more electrons be packed in to consider having more current flow than full up flow?
Just something to think on. You guys may not find these things as interesting possibilities. Maybe some of you do. Along side of what you may think are crazy, loony thoughts, well we are talking artificial, pretend environments. Im just looking at possible details that may alter what we can think when it comes to the legitimacy of the ideal world components. because we are taking them pretty seriously here, sooo. Example. If the ideal cap follows every capacitor rule except for resistance and inductance, then we could not design a perfect inductor in that environment. And likewise the other way around..
Anyway.. back to making holes. Its a bass reflex box using 2 Kicker 15" for my boss's daughter. Tuned at 30hz.
Mags

@tinman
I put that video here to show what happens when any factor goes above the physical limits of the wire.
The only thing about ideal voltage is that it will not sag like if an ideal car battery is shorted, the voltage will not drop for a few seconds, it will stay at 12 volts and in the case of my video and the copper wire would have popped immediately and not after a few seconds.
But I'll play the game in my OU logic terms....... the math I will leave to others.
Ideal voltage holds the voltage steady while the current CAN rise to infinity, but only if infinity is called for by the inductor. So for current infinity to happen, the inductor would also need to be infinite in either or both wire length or thickness and with or without the resistance the ideal voltage would not care.
But here you have a great disconnect in the logic because current from the ideal voltage that could rise to infinity is now countermanded by an ideal inductor that has no resistance but has an inductance limit to 5H. So the two limiting factors are 4 volts with as much current as required and 5H with no resistance. Figure also that coil impedance will be minimal because most wires have resistance and little reactance to start with so if there is no resistance then only the reactance = impedance = peanuts.
Since the ideal voltage is ideal, it cannot vary but like any DC source, you can add to it, in or out of phase will not matter, but again because its ideal you cannot subtract from it having any less on those two ideal power terminals. Given this is ideal DC, the positive terminal will always be 4 volts (p4) and the negative terminal will always be 0 volts (n0).
During the exercise when you get to the zero volts reading this does not mean there is zero across the inductor since the voltmeter is also a differential reading it just sees the same voltage on each side of the inductor as would a scope that would then produce a flat line. If it shows zero with an ideal voltage, then you have to have a combination present at the two terminals like p4/n4 or add 1 from the inductor to p4 to get p5 the inductor would then have to add 5 to the n0 would mean you have p5/n5 which is still zero. At the 3 volts reading the p4 and n0 still cannot change but the inductor can add a 7 to the n side so you can have p4/n7 and show 3 on the voltmeter. This does not mean the ideal voltage is dropping in voltage but that since it is DC, you can add to it with a positive or a negative value and your volt meter will see this change at the terminals but again it does not imply the ideal voltage changed.
A debate could be had here in terms of can an ideal voltage which also has zero resistance can also receive return voltage from the inductor as I had read here someone implying that at the 3 mark, power is being returned to the voltage source. If that's just another "acceptance" to be able to play the game then OK but I would think not since the ideal part of the voltage should prohibit this. Besides, take a 9 volt and 12 volts battery and connect them in parallel and what do you get. This does not mean the 12 volt battery is feeding anything to the 9 volts battery. This just means you will see 21 volts on your volt meter.
What all parties, especially @MH and company need to understand is that when you ask a question literally, it is taken literally. When you say ideal voltage, then indicate 4 volts, there is no turning back from that. Regardless if the voltage has zero resistance or 300% resistance or any other other factor you wish to equate, that ideal voltage is ideal for only one reason, to stay stable, unchanging, otherwise you just need to call it a standard DC power supply (PS) and then we could have all gone home a long time ago. If it was a regular PS we all know we can adjust the voltage where we want and the current will fix itself as per the inductive requirement. A regular PS is not an ideal voltage because depending on what you run with such a PS, yes you can see on the instrument voltage and current meters that voltage and current will vary with the pulsing or relaxation or inductive kick back that the DUT provides. But as an ideal voltage source that is not the case. Being ideal, one should consider there is a little Jennie inside the power source that magically holds the applied voltage steady under all circumstances. Come hell or high water she shall not flinch.
So to start you need to calculate the first 3 seconds of 4 volts hitting a 5H inductor that measures 4 volts after three seconds. These guys already calculated it to arrive at 2.4 amps of current in the coil. What voltage will that produce in the inductor? Take that apply it to your p4/n0 and go from there.
Anyways I am sure if I asked 1000 students who just graduated from university how they perceived the ideal voltage ideal inductor question, the grand majority will answer "Don't really know, I just faked it and passed". hahahahahaha
But let me make at least one thing very clear with all of you. Regardless of what each holds as truth to a fictitious ideal question, do not have the slightest presumption that this explains how a coil works. This will not tell you how a coil works, it only gives you a presently accepted and measured depiction of how a coil responds. This is far from knowing how it works and for me as an OUer looking to develop a better understanding of the logic behind how a coil actually WORKs, all EE discourse becomes purely presumptive. Now once you have all eaten all this ideal stuff and have washed the dishes and cleaned up the kitchen and then if you really want hard questions of logic, let me know. I have a case full that could keep you busy for a few years. One of them was the last question I asked here but again, better you wait till the kitchen is cleaned.
Went on long enough. As usual, sorry for long post. I even cut it down to half. hahaha
wattsup

Besides, take a 9 volt and 12 volts battery and connect them in parallel and what do you get.
Very high current. Equivalent to a 3V shorted battery.
However the same does not apply to a 9V battery inserted into an ideal inductor, because the inductor is not a voltage source  it is a current source.
The internal impedance of an ideal current source is infinite, while the internal impedance of an ideal voltage source is zero.
That's why a current source connected to a voltage source does not result in any infinities.

MH did know the answer to the question, and could easily have plotted out the current trace as I have encouraged you to do, but he wanted you and others to work it through yourselves.
MH's question is thoughtprovoking and well thought out, although the wording could have been geared more towards the experimenters here rather than the techies. At any rate, I can assure you that MH knew all along exactly how the circuit currents behave, given the input he specified. After all, he is the one that formulated the question.
Brad, you really are being silly with what you said there. I know you and MH don't seem to be getting along well, but he deserves more benefit of doubt than you are affording him here on this electronics question.
You mean like the benefit of doubt that he gave me regarding the ICE having resonant systems?. Guessed you missed all that. This is the very same situation. He told me i had no idea what i was talking about,when Internal combustion engines are my fortemy area of expertise.
The difference is,i backed up my knowledge with provided fact's,and this is something no one here can do with an actual test,as we are talking about ideals we do not have. What we are doing is placing a theory based around !best guesses!.
So i have given MH no more than he has given me,and in fact,i have never used the foul language he has toward me.
Not once did i see you,or any other EE guy here tell MH to calm it down when the roll was reversed,but i see you are quick to jump on me when i do the same that has been done to me.
I have seen this very same thing with other members that disagree with MHs analogy.
It's an !agree with me! or your wrong attitude MH hasplane and simple.
As i said in the other thread,i will now treat him as he treats me.
As i said,there is a pattern that is followed on this forum,and that is the EE guys stick togetherbar one,that being (as i have always said) verpies. I would also put vortex1 in there with verpies,but he dosnt frequent this forum muchsadly.
As verpies said in reply to this question,Quote : The equivalent circuit model for an ideal inductor is not an inductor with a wire shorting across its ends.
verpiesJust because most of the world does it wrong does not mean that we have to.
Brad

Sounds to me like you are agreeing with me.
The source is trying to drive the opposite polarity energy into the inductor. When the source is at 3V, it offers up a place for the inductor to dump some of its stored energy. Given more time at 3V, the inductor would fully lose its "positive" energy, and begin to energize in the "negative" direction.
Since it isn't long enough for that to occur, the energy in the inductor simply decreases to a lower "positive" level, and there is only one place for that lost energy to go. You can even see it in the sim.
Well this is how i see it Poynt.
We have a loop of water pipe that represents our ideal inductor loop. In that loop of water pipe we have a pump. Our pump is our voltage source,and the water in that pipe is our current. We start the pump,and this puts pressure(our voltage) on the water(our current) the water starts to flow around our loop. We have a bypass valve in the pump,so as when we switch the pump off(0 volts),the water can still flow in the direction it was(our current is now flowing with the pump off). While the water is flowing,we start the pump up so as it spins in the opposite direction. The pump(our now reversed voltage) wants to now push the water in the opposite direction to that of which it is already flowing. The energy in the moving water is not added to,or stored in the pumpit is working against the pump,and the pump will draw more current to stop the flow of water that is flowing in the wrong direction, before it can start to move the water(current) in the right direction to that of what the pump wants to move it in.
That is how i see it.
Brad

@Tinman
That's a mechanical/hydraulic analogy. This type of mechanical thinking is a long lost skill among physicists and will lead you to the correct results but the analogy must be precise. Unfortunately yours is not precise enough.
We have a loop of water pipe that represents our ideal inductor loop. In that loop of water pipe we have a pump.
Let's make it a positive displacement pump like a Lobe Pump.
I chose the Lobe Pump for this analogy because in such pump, the direction and speed of the impeller has 1:1 correspondence to the direction and speed of the water current.
Our pump is our voltage source and the water in that pipe is our current.
Corrections:
The water in the pipe symbolizes electric charge. The motion of this charge symbolizes electric current.
BTW: The mass/inertia of the water symbolizes inductance.
The force/torque applied to the pump's impeller symbolizes the voltage.
It is important not to conflate the pump with the voltage, because the pump itself is not the force  it is only a mechanism to transfer the force to the water/charge.
We start the pump,and this puts pressure(our voltage) on the water(our current) the water starts to flow around our loop.
Generaly, I agree.
I would write: "...this puts force/pressure on the water causing its acceleration and motion (current)"
We have a bypass valve in the pump,so as when we switch the pump off (0 volts), the water can still flow in the direction it was (our current is now flowing with the pump off).
Yes, the water can flow even when the force (torque) applied to the impeller is zero, but a bypass valve is not necessary for this flow to continue, as e.g. the impeller of a lobe pump will continue to rotate under the current of water already flowing through this pump.
This "bypass valve" is an extraneous component that should have tipped you off, that the analogy is not precise enough.
While the water is flowing,we start the pump up so as it spins in the opposite direction. The pump(our now reversed voltage) wants to now push the water in the opposite direction to that of which it is already flowing.
Here is where you reap the bad fruit of your analogy ;(
Precisely the cause of the conceptual error is in the phrase: "we start the pump up so as it spins in the opposite direction".
The correct phrase should have been: "we apply a torque to the pump's impeller in the opposite direction".
Note, that the application of opposite force/torque to the impeller, does not immediately result in the reversal of the impeller's direction (nor water's direction). I remind you of the 1:1 correspondence between the direction of the impeller and the direction of the water, (their speed also).
I am sure that now you have the tools to complete the rest of the analogy by yourself.

@Tinman
That's a mechanical/hydraulic analogy. This type of mechanical thinking is a long lost skill among physicists and will lead you to the correct results but the analogy must be precise. Unfortunately yours is not precise enough.
Let's make it a positive displacement pump like a Lobe Pump.
I chose the Lobe Pump for this analogy because in such pump, the direction and speed of the impeller has 1:1 correspondence to the direction and speed of the water current.
Corrections:
The water in the pipe symbolizes electric charge. The motion of this charge symbolizes electric current.
BTW: The mass/inertia of the water symbolizes inductance.
The force/torque applied to the pump's impeller symbolizes the voltage.
It is important not to conflate the pump with the voltage, because the pump itself is not the force  it is only a mechanism to transfer the force to the water/charge.
Generaly, I agree.
I would write: "...this puts force/pressure on the water causing its acceleration and motion (current)"
This "bypass valve" is an extraneous component that should have tipped you off, that the analogy is not precise enough.
Note that the application of opposite force/torque to the impeller does not immediately result in the reversal of the impeller's direction (nor water's direction). I remind you of the 1:1 correspondence between the direction of the impeller and the direction of the water, (their speed also).
I am sure now you have the tools to complete the rest of the analogy by yourself.
Here is where you reap the bad fruit of your analogy ;)
Precisely the cause of the error is in the phrase "we start the pump up so as it spins in the opposite direction".
The correct phrase should have been: "we apply a torque to the pump's impeller in the opposite direction".
I see no difference in what i said,and what you have stated above.
The motor that drives the positive displacement pump you have used,will still encounter the force of the water moving in the opposite direction to that of the applied torque to the pump.
You have simply separated the pump from the motor. But any force placed upon the pump will be transferred to the motor.
Yes, the water can flow even when the force (torque) applied to the impeller is zero, but a bypass valve is not necessary for this flow to continue, as e.g. the impeller of a lobe pump will continue to rotate under the current of water already flowing through this pump.
As the motion of the pump/motor combo i was using to represent the voltage(our force),then i included the bypass valve to represent no motion of the motor/pumpthe equivalent of the 0 volt level in the question. Having the impeller still moving,would be seen as a resistance against the flow of water,and in our ideal loop,we have no resistance to the flow of the current,and so the bypass valve was included for that reason also.
I believe that this is a close representation to what we have,than having the impeller being rotated by the water,resulting in a resistance to that flow of water.
Brad

Have you had a gander at the Waterloo induction analogy?
J.

Something that keeps coming to mind as i think....
When we see the demo of a magnet being suspended above a superconductor, It is said that as we put the magnet in place, I suspect a particular height above the super conductive object, that as the field of the magnet sets up lenz currents in the superconductor that are set in motion indefinitely and it sets up an opposing field against the mags field and it floats. So in that case the lenz field is even stronger than the mags field at float as to hold up the weight of the magnets mass, let alone just be equal to the field of the magnet. So is it not possible that this balancing act could be inherent in the ideal perfect inductor? Still thinking on it all. Which is good I suppose. Have some work to do tonight. Will see what I figure on it all
Mags
Mags,
Let me say again that I'm no expert on superconductivity however, I would like to comment on your thoughts above.
In a type I superconductor, all the magnetic flux is outside the superconducting material. If you approach the type I with a magnet, the magnet will be repelled into space because the magnetic field can not penetrate the type I superconductor.
In a type II superconductor however, the levitating magnet is held in place via flux pinning. In a type II superconductor, magnetic flux penetration is possible and it produces quantum current vortexes or tubes. These penetrating flux tubes loop around the magnet and hold it in fixed position above the superconductor. Wikipedia shows a good representation of this plus a good explanation.
partzman

You mean like the benefit of doubt that he gave me regarding the ICE having resonant systems?. Guessed you missed all that. This is the very same situation. He told me i had no idea what i was talking about,when Internal combustion engines are my fortemy area of expertise.
The difference is,i backed up my knowledge with provided fact's,and this is something no one here can do with an actual test,as we are talking about ideals we do not have. What we are doing is placing a theory based around !best guesses!.
So i have given MH no more than he has given me,and in fact,i have never used the foul language he has toward me.
Not once did i see you,or any other EE guy here tell MH to calm it down when the roll was reversed,but i see you are quick to jump on me when i do the same that has been done to me.
I have seen this very same thing with other members that disagree with MHs analogy.
It's an !agree with me! or your wrong attitude MH hasplane and simple.
As i said in the other thread,i will now treat him as he treats me.
As i said,there is a pattern that is followed on this forum,and that is the EE guys stick togetherbar one,that being (as i have always said) verpies. I would also put vortex1 in there with verpies,but he dosnt frequent this forum muchsadly.
As verpies said in reply to this question,Quote : The equivalent circuit model for an ideal inductor is not an inductor with a wire shorting across its ends.
verpiesJust because most of the world does it wrong does not mean that we have to.
Brad
All I will say Brad is that posts like you made inferring that MH did not know the answer to his question are just not good, for you, nor the forum, especially since it is a question having to do with electrical fundamentals.
And you are right, MH is not infallible, nor am I, TK, nor anyone I know on this forum. But you are wrong about the techies always sticking together. We do sometimes disagree, and that is ok. Like you, if I see something that I feel is incorrect or nonsense, I will point it out, regardless of who posted it. And I expect the same of others too, in regards to what I post.
Brad, if anyone can build a 50:1 L/R ratio inductor, I feel it is you. I would encourage you to try to get one together and give this thing a go so you can see for yourself what the circuit current does. For the supply voltages, you could build a 4V and 3V, and 0.5V supply with reasonable ease, and just use a rotary switch to change the voltages manually while monitoring a stop watch. It won't be exactly precise, but close enough to get the gist of what will happen at each transition.
Going with 5H and a 50:1 ratio however would be difficult to achieve without liquid nitrogen cooling I would think, so perhaps the ratio could be reduced and the resulting error taken into account?
What do you say?
Guitar pickups are on the order of 5H, but their R is on the order of 8k Ohms :(

All I will say Brad is that posts like you made inferring that MH did not know the answer to his question are just not good, for you, nor the forum, especially since it is a question having to do with electrical fundamentals.
The first thing i will say ishave you said the same thing to MH,when he said i did not know what i was talking about and was wrong,when it came to the ICE,which is my speaciality ?
Secondlycan you say without doubt tha MH would have been able to answer the question correctly,without having proof to back it up.
Thirdly Regardless of any answer being considered to be correct,it is just theory based around real world examples. As we cannot make such an ideal voltage source to test the theory,then it remains just thata theory. Lust as verpies saidjust because th erest of the world dose it wrong,dose not mean we have to.
Brad, if anyone can build a 50:1 L/R ratio inductor, I feel it is you. I would encourage you to try to get one together and give this thing a go so you can see for yourself what the circuit current does. For the supply voltages, you could build a 4V and 3V, and 0.5V supply with reasonable ease, and just use a rotary switch to change the voltages manually while monitoring a stop watch. It won't be exactly precise, but close enough to get the gist of what will happen at each transition.
And there in lies a problem. What about that ideal voltage source,where the current is free to flow during the 0 volt cycle. A rotary switch would mean an open circuit in between contacts,and there go's your current flow.
Why not simulate it Poynt,using the lowest value resistance you can.
We have to remember,it is going to be 3 volt's,not a build up of voltage as the current drop's from 2.8 amp's down to the lower value,the ideal voltage source will place 3 volts across that ideal loop regardless of the current. So lets see if you can get your sim to do that.
Going with 5H and a 50:1 ratio however would be difficult to achieve without liquid nitrogen cooling I would think, so perhaps the ratio could be reduced and the resulting error taken into account?
What do you say?
Guitar pickups are on the order of 5H, but their R is on the order of 8k Ohms :(
Why would it have to be a set value of inductance. Could we not see the effect with any decent inductor?.
Brad

Have you had a gander at the Waterloo induction analogy?
J.
And as you can see from that very example minnie,that my prediction is correct. The water wheel being the inductor will keep pumping the water against the pump when the direction of the pump is reversed. This works against the pump,it dose not store it's energy in the pump at Poynt said.
Brad

still reading this but have been wrapped up tinkering hardcore with some stuff someone sent me.
just wanted to leave this here for Mags and Bill. interesting stuff regarding electron movement.
http://hyperphysics.phyastr.gsu.edu/hbase/electric/ohmmic.html

And there in lies a problem. What about that ideal voltage source,where the current is free to flow during the 0 volt cycle. A rotary switch would mean an open circuit in between contacts,and there go's your current flow.
Why not simulate it Poynt,using the lowest value resistance you can.
We have to remember,it is going to be 3 volt's,not a build up of voltage as the current drop's from 2.8 amp's down to the lower value,the ideal voltage source will place 3 volts across that ideal loop regardless of the current. So lets see if you can get your sim to do that.
Why would it have to be a set value of inductance. Could we not see the effect with any decent inductor?.
Brad
Brad,
Trust me when I say that a sim using an ideal voltage source and an inductor with a really, really small resistance will properly demonstrate the point that MH and others are trying to get across. The question is, would you believe it and accept it? To understand this may open up new horizons for you in your quest for OU.
I am not a 'EE' or an 'E' or even an 'e', but I have studied greatly to show myself approved. This is a difficult way to go but one can get an education either formally or informally but we all must be willing to learn. For example, I could raise a rucus here by stating that Distinti's magnetic theory says Maxwell's equations for electromagnetic waves are incomplete. I would guess that the EEs here would/will take issue with this. The point is, we must all be willing to learn.
Poynt or myself could post a sim showing the answer to the question but I chose not to do so out of respect to MH because he asked me not to.
partzman

As I understand it in the analogy the torque of the pump is the voltage.
Am I correct?
Having a bad day "at the office".
J.

The first thing i will say ishave you said the same thing to MH,when he said i did not know what i was talking about and was wrong,when it came to the ICE,which is my speaciality ?
Yes, in private I have talked to MH about keeping his tone in check, and yes I would agree that he still goes a little too far sometimes.
Secondlycan you say without doubt tha MH would have been able to answer the question correctly,without having proof to back it up.
Absolutely. It perplexes me why you would even question this.
Thirdly Regardless of any answer being considered to be correct,it is just theory based around real world examples. As we cannot make such an ideal voltage source to test the theory,then it remains just thata theory. Lust as verpies saidjust because th erest of the world dose it wrong,dose not mean we have to.
Not sure what you mean by "theory based around real world examples", but as I've already explained, and MH did also a while back, we don't need a perfectly ideal source nor inductor to prove out the equation that MH posted, which determines the final value of current in the inductor, based on the L and t, and initial current Io.
And there in lies a problem. What about that ideal voltage source,where the current is free to flow during the 0 volt cycle. A rotary switch would mean an open circuit in between contacts,and there go's your current flow.
Most voltage sources are close enough to ideal for this test. Yes, a rotary switch may cause some loss due to an imperfect transition.
Why not simulate it Poynt,using the lowest value resistance you can.
I have already done so. A number of posts back I gave a recommendation as to the L/R ratio that would still provide results "close to ideal", at least close enough for our purposes. I established this level (50:1) by using the sim with various values of R to get 5% error on the final current. Did you miss that post? It was done with the simulation. I have already posted a partial plot of the current, did you miss that also?
We have to remember,it is going to be 3 volt's,not a build up of voltage as the current drop's from 2.8 amp's down to the lower value,the ideal voltage source will place 3 volts across that ideal loop regardless of the current. So lets see if you can get your sim to do that.
As I said above, I have already fully simulated the circuit, as has Partzman. I am sure our results are identical.
Why would it have to be a set value of inductance. Could we not see the effect with any decent inductor?.
If you followed my posts on establishing the L/R ratio for a "close to ideal" inductor, you would know that any value of inductance can work, as long as we are aware that a 50:1 L/R ratio will present a 5% error in the current values. The current will be 5% (or thereabouts) less than the value predicted by MH's equation.

I already discussed simulating this in the real world on another thread. Say you had a 2x1000watt car audio amp. I am pretty sure they have huge voltage inverters in them so the run off a bipolar supply. They are giant ideal voltage sources within certain bandwidth and IV limits. If you could hack into one so that you insert a current viewing resistor at the output such that the voltage sense is on the far side of the current viewing resistor then you should be good to go. You just need a scope channel for the CVR and I am assuming that the car audio amp is being powered by a car battery. You connect up your arbitrary waveform generator (or iPhone) to the amplifier.
Then perhaps go to the hardware store and buy a bunch of looped hollow copper pipe for your inductor. There is probably a better way of doing the coil but that's the one that comes to my mind.
All that you have to do is shorten the timing and lower the voltages and you should be able to do a setup that is a very good facsimile of what we are discussing here. As long as you don't exceed the IV capabilities of the beefy car audio amplifier you should be fine.
But of course, measurements like these were probably made thousands of times already. What's of more value is the intellectual understanding and solving the problem on paper. That's what we are trying to do here.
MileHigh

As I understand it in the analogy the torque of the pump is the voltage.
Am I correct?
Yes.
It is interesting to consider, which way the energy flows when the pump rotates against that torque.

If the question in this thread is ever answered properly or not, I strongly advise anybody interested to go back and read it again from the beginning. You can decide for yourselves.
As an example, I somehow doubt we are going to get any more challenges about the concept of a timevariable ideal voltage source. Nor will we get any admissions that this was all just silliness and a useless and nonsensical waste of time and energy from the main players.
I will also repeat the main goals for this thread:
1. Brad gets up the learning curve and understands the original question and then answers it correctly all by himself and clearly demonstrates that he understands the concepts and understands what he is doing.
2. Brad admits that he was wrong when he stated that my response to the harder question was wrong.
I would really hope that Brad successfully achieves those two goals.

I see no difference in what i said,and what you have stated above.
The motor that drives the positive displacement pump you have used,will still encounter the force of the water moving in the opposite direction to that of the applied torque to the pump.
You have simply separated the pump from the motor. But any force placed upon the pump will be transferred to the motor.
The difference is huge because now the torque is not synonymous with the rotational direction of the impeller and the pump can rotate with or against the torque.
Thus, we can have a typical 4quadrant operation:
1) The pump moving CW with a CW torque.
2) The pump moving CW against a CCW torque.
3) The pump moving CCW with a CCW torque.
4) The pump moving CCW against a CW torque.
In two of these cases, the energy is transferred to the pump, and in two other cases, the energy is transferred from the pump.
Can you assign the direction of energy flow in these four cases?
As the motion of the pump/motor combo i was using to represent the voltage(our force),then i included the bypass valve to represent no motion of the motor/pumpthe equivalent of the 0 volt level in the question. Having the impeller still moving,would be seen as a resistance against the flow of water,and in our ideal loop,we have no resistance to the flow of the current,and so the bypass valve was included for that reason also.
In an ideal pump, as well as in an ideal pipe, there is no resistance to the water's motion and lately we've been considering an analogy of an ideal L circuit here, so all the analogical hydraulic components must be ideal, too.
Thus, the water can move the impeller without any friction ...and the impeller can move the water without any friction, too.

Yes.
It is interesting to consider, which way the energy flows when the pump rotates against that torque.
I am just going to offer up simplified analogies and perhaps they will help.
The water pump in this case never impedes the water flow, but what it does do is create an increase in pressure in that water itself. So you don't necessarily have to think about a tangible water pump, just what it does. So if the pressure is 10 psi on one side of the pump, then the pressure is 14 psi on the other side of the pump. Or perhaps the pump drives the pressure the other way: the pressure is 10 psi on one side of the pump, and 7 psi on the other side of the pump.
That's the type of water pump we are talking about in this example, a constantpressure pump. You don't care about the water flow rate or even the direction of the flow, the only thing that counts is that the pump establishes a difference in water pressure from one side to the other side. You can even block the water flow, and the pump still does the same thing.
However there is another type of water pump you can also imagine, a constantflow pump. It's easy to imagine perhaps a large diesel engine with a transmission that lowers the shaft output speed that drives some small pistons that pump water at a certain flow rate.
In this pump, you don't care at all what the water pressure is on either side of the pump. The only thing you care about is that the pump pumps say six gallons of water per minute. The pressure at the output side of the pump could be 100 psi or 1000 psi, it doesn't matter and the pump will pump at a constant water flow rate.

If the question in this thread is ever answered properly or not, I strongly advise anybody interested to go back and read it again from the beginning. You can decide for yourselves.
As an example, I somehow doubt we are going to get any more challenges about the concept of a timevariable ideal voltage source. Nor will we get any admissions that this was all just silliness and a useless and nonsensical waste of time and energy from the main players.
I will also repeat the main goals for this thread:
1. Brad gets up the learning curve and understands the original question and then answers it correctly all by himself and clearly demonstrates that he understands the concepts and understands what he is doing.
2. Brad admits that he was wrong when he stated that my response to the harder question was wrong.
I would really hope that Brad successfully achieves those two goals.
Where did the bold words in your post come from? Seems this is the first time this is used in front of the words "ideal voltage source" since this was not in your initial question of ideal voltage on an ideal inductor. If so then why is that graph showing a straight line at 4 volts for 3 seconds? Why would you equate a time variable with ideal voltage, when ideal voltages do not change?
Hmmmmmmmmm (with my wet index finger pointing upwards) did the wind change directions? Or do I sense a hurricane is about to form.
wattsup

Here we go again. Who said that ideal voltages have to be fixed and can't change in time? It's an ideal voltage, I can say it does whatever I want it to do. In my second example I state that the ideal voltage v = 20*t^2. So the ideal voltage increases in value proportional to the square of the time. There is nothing stopping me from saying that.
What about your function generator when you set it to a triangle wave and you connect it to your circuit? Supposing that the triangle wave is somewhat attenuated especially at the peaks because of the load on it from the circuit. Well, what if I said that my "ideal voltage source" had an identical waveform to the somewhat attenuated triangle wave output from your function generator? Under those static load conditions then the ideal voltage source that clones your function generator output and the function generator output are indistinguishable. That means you can look at the output from your function generator as being a timevariable ideal voltage source as long as the load condition remains static and unchanging.

Why would you equate a time variable with ideal voltage, when ideal voltages do not change?
I don't know what an "ideal voltage" is and I think that MH does not know either.
Perhaps you had an "ideal voltage source" in mind.
If "yes, then consider these questions:
Does an "ideal voltage source" need to output a constant voltage all the time?
Can a voltage source still be ideal if it outputs an alternating voltage ?

@MH  @verpies
Common guys. We have been through this how many times.
Here is a common definition.
"An ideal voltage source is a voltage source that supplies constant voltage to a circuit despite the current which the circuit draws."
In this definition, one can summarize that there is no variability in the voltage source. If you adjust your voltage at 4 volts, it stays at 4 volts regardless of any current draw.
In the video I posted on my shorting a wire over a battery, the battery is not an ideal voltage source so the voltage will drop as I short it with that wire. But in that case it is still backed up with 100 amps so that wire went zap in a few seconds. This for @tinman and myself and many others was the complete basis of why the question of ideal voltage on an ideal inductor was more then moot, it approached ludicrous because for us, like I said in my post, come hell or high water, that voltage should never change. This for us signifies a DC voltage holding a fixed straight line as in @partzman's graph showing the first three seconds as a straight 4 volts. Where is the variable in time there for the voltage. None. Yes it will show a variable current rise from t0 onward.
This is why I was stating that for any voltage to change across the terminals of the ideal voltage, it has to come from the inductor.
This was the whole basis of my prior post and of most posts from @tinman and the root of all this commotion that started with an innocent JT circuit.
A time variable ideal voltage source, for me, is not a DC fixed ideal source but either an AC ideal source or a DC ramped ideal source. Totally different animal. But in either of these three, using the term IDEAL will still signify that once they are set, they are set for life. No change possible and thus they become the fixed anchor to other variables that one can experiment with.
Hmmmmmmm.
wattsup

I don't know what an "ideal voltage" is and I think that MH does not know either.
Perhaps you had an "ideal voltage source" in mind.
This is what MH actually said:
As an example, I somehow doubt we are going to get any more challenges about the concept of a timevariable ideal voltage source.
It's quite clear to me what he is referring to.
If "yes, then consider these questions:
Does an "ideal voltage source" need to output a constant voltage all the time?
Can a voltage source still be ideal if it outputs an alternating voltage ?
It is perplexing to me to see these questions ???
To clarify for all interested, one simple answer to cover all questions as to what an ideal voltage source is:
An ideal voltage source is a source that outputs a voltage according to what it is set to, no matter what load is connected to it. This does not preclude ideal function generators.
I trust you can now answer both of your questions.

@MH  @verpies
This for us signifies a DC voltage holding a fixed straight line as in @partzman's graph showing the first three seconds as a straight 4 volts. Where is the variable in time there for the voltage. None. Yes it will show a variable current rise from t0 onward.
wattsup
Wattsup,
Yes, I created that ideal voltage source and applied it to the ideal inductor for three seconds and that voltage source was faithful to it's duty and it did not change for that duration. I could however, create a new ideal voltage source of 0 volts in magnitude and replace the original 4 volt ideal voltage source and it too will do it's job. In a sim, I can do this in few ns of time and maintain a duration of my choosing.
I can continue to replace each ideal voltage source with another as I choose since there is no law I am aware of which prevents me from doing so. Each ideal voltage source I create has the ability to maintain it's magnitude no matter the load. Proceeding in this manner allows us to be able to solve MH's problem.
partzman
Edit

wattsup was just having an "off day". That's all that was up!

@partzman
I know that. Your are on a sim and that sim is not simulating an ideal voltage as per the definition otherwise it would not change. Don't you get it. If your sim could simulate an ideal voltage it would have stayed at 4 volts all the time according to the definition. There is no ideal voltage in your sim. It is just simulating an applied voltage given the other parameters and it will change in time.
Actually there is no point in using a sim with an ideal voltage because you will always have the same voltage straight line.
The idea of the ideal voltage is to maintain a fixed voltage source while you can vary other parameters that cannot vary the ideal voltage but will provide you with how the circuit works when you, let's say, vary the resistance, or inductance, or impedance or capacitance while that ideal voltage always stays the same. The require gain or reduction in current will be automatic but the voltage will never change. You can then measure across different components in the circuit to see those changing values but the ideal voltage will never change. Otherwise what is the point in having an ideal voltage when it is just simple voltage.
@all
Common guys. There is process, there is teachings but there is also logic and logic has to win every time. At least this is what has been paying my great salaries for years. Nothing beats logic.
Either that or come forward and convene that this question does not require an ideal voltage source to propose the teaching you wish to promote, because it just does not click at all. It should at least send alarms bells ringing when other @members here cannot adjust to your line of thinking that maybe the teaching requires some "fine tuning".
wattsup

Wattsup:
I would call that the classic ironic twist of the free energy forums. You are supposed to be "out of the box" experimenters. You are supposed to be "free thinkers not held back by the constraints of dogmatic science." You mock the scientists and the engineers and say they "only know what is in books" or, even worse, they "are programmed by the Powers That Be not to think."
And then the subject of a timevariable ideal voltage source comes up. And what do you do? You look for a definition in a book and you stick to it like glue. The definition says nothing at all about whether or not the voltage can vary in time but the concept is seemingly "too radical" and you can't process it so you make up a "new rule" to "fit in" to how you think things should be.
It is so ironic, it's almost surreal. For myself, and Poynt, Verpies, Partzman and countless others the concept of a timevariable ideal voltage source was presented to us, we processed it, and didn't flinch for a second. The simple fact is that it makes perfect sense. The logic that describes it is perfect.
It's almost shocking that so much time and energy has been wasted on this nonissue.
MileHigh
P.S.: I will just repeat that a very powerful car audio amplifier could make for a pretty beastly variable ideal voltage source.

Surreal is right.
Your thoughtexperiment ideal voltage source has a virtual knob on it that says "Voltage output" right? And you can virtually reach out and turn that knob up and down, can't you? Thus making a simple timevarying ideal voltage source. Replace the "knob" with whatever virtual control system is needed to vary the output on whatever time schedule you like. What is so difficult to understand about that?

Yep, you could even imagine my "wrist" had a bandwidth of about 30 kHz and could make a variable ideal voltage source that sounded like two lead guitars trading licks....
https://www.youtube.com/watch?v=oZUp1gUQLyg

Absolutely. It perplexes me why you would even question this.
300 years ago,89% of the worlds population believed in God. Today only 26% of the worlds population believe in Godthe rest have seen the light.
Not sure what you mean by "theory based around real world examples", but as I've already explained, and MH did also a while back, we don't need a perfectly ideal source nor inductor to prove out the equation that MH posted, which determines the final value of current in the inductor, based on the L and t, and initial current Io.
Your theory being based around real world applications,is based on the fact that there exist a resistanceno matter how small. In our ideal situation,that resistance is no existent,and no energy can be dissipated anywhere in that ideal circuit loop. The energy now stored in the ideal loop at the 0v phase ,must be dissipated before the ideal voltage can set a current flowing in the opposite direction.
Most voltage sources are close enough to ideal for this test. Yes, a rotary switch may cause some loss due to an imperfect transition.
I have already done so. A number of posts back I gave a recommendation as to the L/R ratio that would still provide results "close to ideal", at least close enough for our purposes. I established this level (50:1) by using the sim with various values of R to get 5% error on the final current. Did you miss that post? It was done with the simulation. I have already posted a partial plot of the current, did you miss that also?
No i did not miss the post.
First,i do not think you grasp the difference between a very low resistance value,and no resistancethe difference is infinite,and so the outcome has an infinite difference.
Secondthe voltage in this ideal circuit is seen as being in series with the inductor,but as far as the current go's,the connection is both a series and parallel connection. So example 3 that verpies posted some time back,represents MHs circuit as far as the current flow go's,that being the inductor has a short across it's terminals as far as the current is concerned.
As I said above, I have already fully simulated the circuit, as has Partzman. I am sure our results are identical.
If you followed my posts on establishing the L/R ratio for a "close to ideal" inductor, you would know that any value of inductance can work, as long as we are aware that a 50:1 L/R ratio will present a 5% error in the current values. The current will be 5% (or thereabouts) less than the value predicted by MH's equation.
When that L/R =0,what is your ratio then?. 50:0 ?
As you can see,we went from a real world obtainable calculation,to one that dose not exist.
This is what i mean by trying to use real word values against values that just are not there,and expecting to arrive at an accurate outcome.
I have spent countless hours,and thousands of dollars on experimentsnot to mention time.
I have helped those purchase equipment that they otherwise could never hope to obtain. I have spent time organizing fun competitions ,and donated prizes to those judged as being the most innovative at what they built. And yet here we have MH,sitting in his rocking chair,not lifting one finger(other than to type)to help the cause.
I will leave you with one thing to think about Poynt>
Are the books ever going to explain the TPU?. Is what you know going to line up all perfect to that of the operation of the TPU.
Perhaps this is why it(the TPU) still alludes all those trying to replicate it's effectyour trying to place known real world values against values that are infinitethe energy being delivered by the TPU seems to have that valueinfinite.
I love that saying by verpies>just because the rest of the world got it wrong,dose not mean we have to.
Maybe we can learn something from that quote.
Brad

I am pretty sure they have huge voltage inverters in them so the run off a bipolar supply. They are giant ideal voltage sources within certain bandwidth and IV limits. If you could hack into one so that you insert a current viewing resistor at the output such that the voltage sense is on the far side of the current viewing resistor then you should be good to go. You just need a scope channel for the CVR and I am assuming that the car audio amp is being powered by a car battery. You connect up your arbitrary waveform generator (or iPhone) to the amplifier.
Then perhaps go to the hardware store and buy a bunch of looped hollow copper pipe for your inductor. There is probably a better way of doing the coil but that's the one that comes to my mind.
All that you have to do is shorten the timing and lower the voltages and you should be able to do a setup that is a very good facsimile of what we are discussing here. As long as you don't exceed the IV capabilities of the beefy car audio amplifier you should be fine.
But of course, measurements like these were probably made thousands of times already. What's of more value is the intellectual understanding and solving the problem on paper. That's what we are trying to do here.
MileHigh
I already discussed simulating this in the real world on another thread. Say you had a 2x1000watt car audio amp.
And that is where your ideal world all falls apart.
Please explain as to why you need 2000 watts of power to simulate a circuit that dose not dissipate any power?.
Answer that,and then you may begin to understand the difference between a small resistance value,and no resistance at all.
Your circuit MH,should use no power at all,as everything put into it,should be what you get back out of itthere are no losses in an ideal circuit such as your's.
This is based on the fact that what Poynt said is true,and that you can store the energy from the loop,back into the ideal source.
So your 2000 watts are for what?.
Brad

It is perplexing to me to see these questions ???
Oh, C'mon!  these were rhetorical questions. I was not seeking answers to them. I was provoking some thinking.
To clarify for all interested, one simple answer to cover all questions as to what an ideal voltage source is:
An ideal voltage source is a source that outputs a voltage according to what it is set to, no matter what load is connected to it. This does not preclude ideal function generators.
Of course, but this needs to be stated explicitly, since apparently some people are under the impression, that an ideal voltage source is a constant voltage source :o

Please explain as to why you need 2000 watts of power to simulate a circuit that dose not dissipate any power?.
The servo amp with high power output gives you the ability to output high currents (and sink high currents). The amp pumps real power into the coil. The longer you impose a constant voltage across the coil, the higher and higher the power demands get. A highpower car audio amp gives you more headroom and flexibility to carry out various experiments.
Your circuit MH,should use no power at all.
It would seem that only in your mind can things like that be stated in all seriousness.

Oh, C'mon!  these were rhetorical questions. I was not seeking answers to them. I was provoking some thinking.
Of course, but this needs to be stated explicitly, since apparently some people are under the impression, that an ideal voltage source is a constant voltage source :o
C'mon yourself. Your question about whether MH knows what an ideal voltage source is, is absolute bunk.
After that it seemed you were the one that was lost. I encourage you to avoid the socalled rhetorical questions, and instead try to help Brad understand why his thinking on this affair is a little off the tracks.

So, just to be clear here...
I can have an ideal voltage source that puts out 4 volts @ 1 amp but then, after 2 seconds, puts out 10 volts at .5 amps
and later, after 20 seconds puts out 2,000 volts @ 500 amps?
Do we need ideal circuit breakers that do not blow when working with this ideal voltage source?
Honestly, this just completely blows my mind that you (meaning anyone) can just make up some hypothetical device and then assign it ANY operational parameters that you want and then, do an "experiment" with this made up device and then claim that something real was learned?
This sounds like the same crap that Mythbusters might use. I just listened to an audio book that mentioned the Mythbuster's show where they "proved" that Archimedes did not have a death ray. They tried it, it did not work so this was PROOF!
Well, some boys over at MIT did a replication and....boom! It worked as it was supposed to. Those idiots at Mythbusters were ALWAYS changing parameters and then claiming devices to be "impossible". So, they could not find a mirror large enough so, they substituted a bar of soap instead...didn't work...see? Impossible! (Of course I exaggerate here a bit, but not too much)
I apologize...I just can not get my head around making crap up to fit your theory, and when your theory works claiming a victory.
I must be getting old.
Bill

I don't know what an "ideal voltage" is and I think that MH does not know either.
Perhaps you had an "ideal voltage source" in mind.
If "yes, then consider these questions:
Does an "ideal voltage source" need to output a constant voltage all the time?
Can a voltage source still be ideal if it outputs an alternating voltage ?
I think that I can solve the mystery for this one. It took me a few hours to get it.
Verpies is splitting hairs on semantics. What does "ideal voltage" mean vs. "ideal voltage source?" i.e.; How do you define an "ideal beauty?" Is there an "ideal voltage?"
So Verpies was not suggesting that I did not know what an ideal voltage source is, even though I myself thought the same at first. His posting is driven by looking at Wattsup's prose.
Does an "ideal voltage source" need to output a constant voltage all the time?
Can a voltage source still be ideal if it outputs an alternating voltage ?
NO and YES. But how many ideal voltage sources can dance on the head of a pin? ;)
Coles Notes on this subject will be available in six weeks. I looked them up and if you are not Canadian you may have to look them up too. The Americanized version is now called CliffsNotes.

Honestly, this just completely blows my mind that you (meaning anyone) can just make up some hypothetical device and then assign it ANY operational parameters that you want and then, do an "experiment" with this made up device and then claim that something real was learned?
But how did they get to the moon? Answer: Lunar orbit rendezvous.
There is a great documentary, I think it's a Nova. I did a quick check but I don't think I saw it on YouTube. It's all about how the NASA scientists in the late 1950s and early 1960s were struggling to design a system that would get men to the moon and back. Typically, the solutions relied on massive rockets that would go to the moon, land, and then blast off and then go back to Earth for a standard reentry. The launch system to do that had to be humongous, say five times the size of a Saturn V rocket.
There was a lowerlevel scientist that had the idea for lunar orbit rendezvous. He could not get anybody's attention and was ignored for a few years, nobody took his ideas seriously. Out of frustration he sent off a letter and bypassed three levels of management to make his case and get the attention his idea deserved. That's how they ended up choosing the system for getting to the moon.
What's the point? The point is the scientist conceived of the lunar orbit rendezvous plan all by himself, in his own head. He did an entire successful moon mission in his head based on hardware and software that didn't even exist. Then he took out his napkin and started doing some basic number crunching to confirm that what he had conceptualized all by himself was doable. It's a great story about a true unsung hero.
What is the escape velocity for Earth? You can figure that out on a blackboard by writing down a few equations. You don't have to build a rocket and find out by trial and error.
MileHigh

C'mon yourself. Your question about whether MH knows what an ideal voltage source is, is absolute bunk.
I never posed a question whether MH knows "what an ideal voltage source is".
After that it seemed you were the one that was lost.
Just go back and read exactly what I wrote. Read every word, understand it and come back to me with an apology.
I still don't think that MH knows what an "ideal voltage" is. Do you?

I never posed a question whether MH knows "what an ideal voltage source is".
Just go back and read exactly what I wrote. Read every word, understand it and come back to me with an apology.
I still don't think that MH knows what an "ideal voltage" is. Do you?
I certainly don't. But I do know why Mona Lisa is smiling. ;D

@MH
You know what. Your only fall back is to attack me with something that is totally irrelevant to the discussion. You want to discuss constructs then open a new thread and i will be there in my corner of the boxing ring. No problem. If you want we can play knowledge court where you will defend EE and I will accuse it of fraud. That would make a good weekly. Hmmmmm
You had a chance to teach something and you failed. Not me. For me a teacher is like a brain farmer. The farmer knows exactly how to treat the crop, feed it at proper intervals, make sure its thirst is quenched and then if the crop is harvested, he gets paid. Teachers should follow the same modus. No crop no pay. You failed not me. The onus is always on the teacher to read the possible misunderstandings and quench them before they grow into weeds. You failed because you were to busy with your big mouth and fast action posts, not really thinking things through, never judging where and how to properly help the crop grow. You failed not me.
You presented a problem, gave it some parameters and "assumed" that your students will not take it literally. That is your fault and not ours. So you know what. You learned something today. You learned that you can be one hell of a jackass always relying on your hing legs to protect yourself should things go bad. So defer the real question on who's responsibility it is to make sure a question is asked in its proper context. Surely not mine. I only found the fault. I only had to read endless pages of crap to then realize the fault, point it out so you can then chastise me instead of congratulating me for seeing it. Man, just great. So no need to thank me man. Maybe excuse yourself to @tinman for stringing him along while you whipped him good.
I even tried to make your bumble up work but no, not even one comment on that. Not even a "nice try wattsup, interesting way of analyzing the problem". Just your usual selfcentered boy worrying about how the world will see him. Tell me what take more guts, pointing out the problem or following the pack. You should all be ashamed of yourselves. One stupid little question that could have been answered in one page and none of you saw the flaw. Oh, but I did, didn't I and I am the sureal. hahaha
I'll put my logic against anything you want to throw at me boy. You muster up any EE concept you wish, throw it at me and I will cut it down to flea grass. That's because now that I know how and why electricity conveys in our wires and coils, I know which questions to ask you that will prove it to yourself. That's what I am now good at. And actually I have you to that thank for being the typical pompous ass scientist that needs no more outlook on nature then a worm needs more earth. I asked you guys a question a few pages back and no answer, not even a remark because I now know the EE modus. Ignore whatever can eat away at your present comfort zone. Funny thing is it is not outofthebox. It is just normal logic taking its logical course like water eventually taming the stone.
wattsup

But how did they get to the moon? Answer: Lunar orbit rendezvous.
There is a great documentary, I think it's a Nova. I did a quick check but I don't think I saw it on YouTube. It's all about how the NASA scientists in the late 1950s and early 1960s were struggling to design a system that would get men to the moon and back. Typically, the solutions relied on massive rockets that would go to the moon, land, and then blast off and then go back to Earth for a standard reentry. The launch system to do that had to be humongous, say five times the size of a Saturn V rocket.
There was a lowerlevel scientist that had the idea for lunar orbit rendezvous. He could not get anybody's attention and was ignored for a few years, nobody took his ideas seriously. Out of frustration he sent off a letter and bypassed three levels of management to make his case and get the attention his idea deserved. That's how they ended up choosing the system for getting to the moon.
What's the point? The point is the scientist conceived of the lunar orbit rendezvous plan all by himself, in his own head. He did an entire successful moon mission in his head based on hardware and software that didn't even exist. Then he took out his napkin and started doing some basic number crunching to confirm that what he had conceptualized all by himself was doable. It's a great story about a true unsung hero.
What is the escape velocity for Earth? You can figure that out on a blackboard by writing down a few equations. You don't have to build a rocket and find out by trial and error.
MileHigh
I believe you are speaking of Max Fagat...he is the guy that favored not only lunar orbit rendezvous, but came up with the ridiculous idea of throwing away the used portions of the spacecraft. Main booster tanks empty? Throw it away. 2nd stage burned out? Throw it away. Landed on the moon with the lander? Throw away the base. Back in the command module? throw away the lander...and etc., etc. He was a true genius.
Von Braun favored earth orbit rendezvous. He said that if we did not do that, the moon would be the last place we went...and he was right.
But, MH, I am not talking about criticizing brainstorming or creative thinking and analysis here. What if old Max used an ideal booster in his thinking? One that required no fuel and weighed nothing? How far would his thought experiments have gone then? Reality must overrule in any thought experiment with any components of any kind in my opinion. What if Max considered a rocket that would accelerate to 1.5 lightspeed? He would have then designed a spacecraft needing only 3 minutes of oxygen for the crew to get to the moon. Yes, he could have pictured a booster doing this but what good would it do?
I know were are on different wavelengths here and I mean no animosity by it. I have said this before, and I will say it again...you, and the other trained EE guys here have forgotten more about electronics than I will ever learn. But, this is not my field, but a hobby I have picked up in order to learn. I appreciate everyone's input...I just am missing something here with the made up "ideal" component business.
Bill

I never posed a question whether MH knows "what an ideal voltage source is".
Just go back and read exactly what I wrote. Read every word, understand it and come back to me with an apology.
I still don't think that MH knows what an "ideal voltage" is. Do you?
I read every word. How was that post helpful to anyone?
I encourage you to avoid the socalled rhetorical questions, and instead try to help Brad understand why his thinking on this affair is a little off the tracks.

So, just to be clear here...
I can have an ideal voltage source that puts out 4 volts @ 1 amp but then, after 2 seconds, puts out 10 volts at .5 amps
and later, after 20 seconds puts out 2,000 volts @ 500 amps?
You almost got it Bill.
The problem with your guess is that the currents probably aren't going to track that way. The assumption is and always has been that the ideal voltage source is always connected to the same load, regardless of what it is.
Did you see my post on arbitrary wave form generators? Did you see my post of the wave form MH prescribed for the experiment? I don't get what is so difficult to understand about this. Think of it as a special function generator that has zero output impedance and can be programmed to produce any imaginable wave form. Make sense? These aren't imaginary, at least the wave form generator part. The "ideal" part is of course not achievable, but we can get so close that it makes not a bit of difference in the final analysis. This is a hurdle that Brad also seems to be having trouble getting over.
Now I'm beginning to feel like I'm in an episode of the Twilight Zone :P

You almost got it Bill.
The problem with your guess is that the currents probably aren't going to track that way. The assumption is and always has been that the ideal voltage source is always connected to the same load, regardless of what it is.
Did you see my post on arbitrary wave form generators? Did you see my post of the wave form MH prescribed for the experiment? I don't get what is so difficult to understand about this. Think of it as a special function generator that has zero output impedance and can be programmed to produce any imaginable wave form. Make sense? These aren't imaginary, at least the wave form generator part. The "ideal" part is of course not achievable, but we can get so close that it makes not a bit of difference in the final analysis. This is a hurdle that Brad also seems to be having trouble getting over.
Now I'm beginning to feel like I'm in an episode of the Twilight Zone :P
Thanks Darren.
So, is it OK to have more than one "ideal" component in an exercise? Or is it? Or, is it more to have an imaginary component (ideal) replace something you already know in the circuit and are wanting to find out about something else in the circuit? Is this it? Like solving for X when you already know Y?
Picture a place and time when you go to a roadside dinner at 3:00 a.m. and all of the waitresses have one large eye in the center of their head. You have entered...the Twilight Zone.
Bill

@MH
You know what. Your only fall back is to attack me with something that is totally irrelevant to the discussion. You want to discuss constructs then open a new thread and i will be there in my corner of the boxing ring. No problem. If you want we can play knowledge court where you will defend EE and I will accuse it of fraud. That would make a good weekly. Hmmmmm
wattsup
I am not attacking you. In fact the "you" in my posting refers to all those that "don't believe" that an ideal voltage source can vary with respect to time. Like it or not, many people would find just arguing the issue to be strange. It's kind of a mom and apple pie issue and you are of the group saying that you are not allowed to put ice cream on your apple pie. It just makes no sense.
I am not going to address all of the drama in your posting about me nor all of the drama you raise about the question itself. There is no need for all of this, truly. Brad and others are supposed to be trying to answer a simple question and understand all of concepts and the related issues. As far as I am concerned this whole thing should have happened six years ago.
You made reference to a question of yours that was ignored. I don't see a question in post #454. That posting is really hard to digest, BTW. I went back a few more pages and did not see anything so I suppose you are taking about your post #454. If you can tell me what the question is I will try to answer it.
MileHigh

Thanks Darren.
So, is it OK to have more than one "ideal" component in an exercise? Or is it? Or, is it more to have an imaginary component (ideal) replace something you already know in the circuit and are wanting to find out about something else in the circuit? Is this it? Like solving for X when you already know Y?
It is ok to have more than one ideal component in a circuit, yes.
In the sim we can get real close to ideal components (voltage sources are), in real life that is more of a challenge. MH keeps mentioning big audio power amplifiers, well that is to get the output impedance down. Use lots of negative feedback and big hefty transistor outputs, and you will have a nice low Z buffer for your arbitrary wave form generator (assuming it is DC coupled). The inductor is a far greater challenge, but still not insurmountable to get reasonably close, i.e. L/R=50.
Picture a place and time when you go to a roadside dinner at 3:00 a.m. and all of the waitresses have one large eye in the center of their head. You have entered...the Twilight Zone.
Bill
;D

@MH
Maybe you need to understand how an OUer sees the question or anyone else that reads a question and responds to it literally. So just to clarify this without any chance of further misunderstandings, you need to clarify this yourself and I am spelling it out clearly for you to make it easy.
Method #1
1) the voltage stays at 4 volts for 3 seconds
2) then is falls on its own to 0 volts for 2 seconds
3) then it falls on its own to 3 volts for 2 seconds
4) then is rises on its own to 0.5 volts for 7 seconds
5) then it falls on its own to 0 seconds for infinity.
All these events are occurring in succession on their own with only that first 4V setting on a 5H inductor resulting from the actions occurring in the inductor itself. For us, that's how your question was asked and never was there any mention of timevariable anything.
Now with your sudden inclusion of those two words that I knew in advance would generate a hurricane you are saying this.
Method #2
1) the voltage is set manually or programmed to automatically go to 4 volts for 3 seconds
2) the voltage is set manually or programmed to automatically go to 0 volts for 2 seconds
3) the voltage is set manually or programmed to automatically go to 3 volts for 2 seconds
4) the voltage is set manually or programmed to automatically go to 0.5 volts for 7 seconds
5) the voltage is set manually or programmed to automatically go to 0 seconds for infinity.
Are you saying your question was as per Method #2, hence each is a separate event?
Or is there something different besides these two choices that you need to add.
wattsup

Wattsup:
Is there really any difference between the two ways of wording it and the way I worded it? I don't think it is unreasonable to expect people to know what a voltage source is, or what an ideal voltage source is.
How about we go back to TK's metaphor? You are turning a dial to change the voltage on an ideal power supply. You can turn the dial very fast and the resultant voltage output looks like the plot that Poynt posted. The voltage source is active, it's driving the show.
All these events are occurring in succession on their own with only that first 4V setting on a 5H inductor resulting from the actions occurring in the inductor itself.
In my example, the inductor simply is a slave to the actions of the voltage source. That's it, the inductor has no say whatsoever in what voltage exists across its two terminals.
Does this make sense now?
MileHigh

@MH
No. I just need an answer.
I thought spelling it out clearly is better then reading between the lines.
I thrive on precision and not on suppositions. I investigate things until they become clear.
Method #1 or Method #2
And the difference is enormous. 200 pages enormous.
Now do you understand why all this crap has been going on.
I know now just by your sidetracking that your answer is Method #2 and it should not have been asked as it was. It should have been 6 different questions each starting at t0 for their set duration and each starting at their identified voltage setting and each produces one effect.
Instead you led the question to be understood as one voltage setting "transmuting" to the others on their own direct from that one first 4V input. So do you realize the problem. Or now maybe I should ask you to respond to Method #1 and see how you like it while I badger you ever step of the way, telling you, "no no it's a perfectly sound question that I can answer in a jiffy".
Our total argumentation was based on Method #1 and you cannot say they are the same. They are not.
wattsup

I'll pick method number #2 of you want me to pick something.
I don't know what you really mean by "falls on its own" in method #1. You seem to be suggesting that it is a passive action by the voltage source. How could that be? How could the voltage source "know" to fall to 3 volts?
I don't why you focus on things like this. If you have a basic understanding of electronics then the question as originally posed is straightforward, selfevident, and not ambiguous. You are reading into something that is not there.
Instead you led the question to be understood as one voltage setting "transmuting" to the others on their own direct from that one first 4V input. So do you realize the problem.
No, there is no problem at all.
Our total argumentation was based on Method #1 and you cannot say they are the same. They are not.
You are in "how many angels can dance on the head of a pin" territory. People that understand basic electronics don't see any rhyme or reason in what you are saying.

wattsup,tinman and me, what I've learned is if you want to do electronics
you've got to go to college and start from DAY ONE and really pay attention.
I'm managing to get a bit but only by constantly referring to very basic
tutorials.
John.
We must thank the likes of MH. poynt etc. for their patience in inching us along.

Language...
A big part of this problem here is language ,The "taught ", hear and read things thru their learned Language ,which are indeed self evident to _Them_.
that is why one "Taught Man" can send a PDF to build a 747 to another "Taught man" on the other side of the world .
and a 747 can be built with no further contact.
here we try to talk outside the "taught Box" and not only outside of _that _ box but also with many here trying to contribute in tongues and cultures which they are not familiar with.
its a big mish mosh ,and I'm surprised there are not more fights.
And then we throw in Ego's insults and personal "agendas" [never good to do]
and.... Boom
Do we keep rinsing and repeating ??
we can self loop this in perpetuity .....
OR ???
we have a lot of very good men here , dedicated Men !
and plenty of Napkins...

I read every word. How was that post helpful to anyone?
It helped Wattsup to notice that he needs to be more careful with his words.
These are technical posts and every word matters. Making mental shortcuts is sometimes funny but most often it requires "secret decoder rings" from the reader. Those are not good habits for technical communication.
So now we have established that an ideal voltage source merely needs to have:
 zero internal resistance
 zero internal reactance
 an output voltage that is not affected by the load
...but it does not need to output a constant output voltage all the time, and it can be even an ideal alternating voltage source.
We can consider the direction of the current flowing through this ideal voltage source.
Brad will like this:
The current can flow in two directions through the voltage source and the voltage presented by this voltage source can have two polarities, too. So we have 2 variables and each one can have 2 states. This gives us 4 combinations.
Of course this is again the 4quadrant operation, where:
 2 combinations result in energy flowing from the voltage source to the load
 2 other combinations result in energy flowing from the load to the voltage source
Can you enumerate which ones?

If the question in this thread is ever answered properly or not, I strongly advise anybody interested to go back and read it again from the beginning. You can decide for yourselves.
As an example, I somehow doubt we are going to get any more challenges about the concept of a timevariable ideal voltage source. Nor will we get any admissions that this was all just silliness and a useless and nonsensical waste of time and energy from the main players.
I will also repeat the main goals for this thread:
1. Brad gets up the learning curve and understands the original question and then answers it correctly all by himself and clearly demonstrates that he understands the concepts and understands what he is doing.
2. Brad admits that he was wrong when he stated that my response to the harder question was wrong.
I would really hope that Brad successfully achieves those two goals.
MHyou are a true idiot.
Do not come to this thread i started,and belittle me like you have above.
You are no wizz kid,and that much was proven in the JT thread.
To come here nad have the arrogance to think i am the one that needs to climb the ladder,is you just grasping for air,and trying to make me look bad
Well bad new's MHyour question has not yet been answered correctly,and those that think they have,and it's game,set,and match ,are free to post there real world results using the two components of your question an ideal inductor(no resistance),and an ideal voltage source(no internal resistance)and no means none at all.
How good are you MH,well i now know that (as i suspected) you would not have answered your own question correctly. What i should have done,was to get you to post your answer to Poynt before this thread started,as Poynt (even though we disagree some times)would have posted your answer as you presented it at the end of this discussion.
But there is no need for that,as i have the proof i needsaved and screen shotjust in case you decide to go change things on said post when you realize your mistake.
So take your idiotic remark's,and place them in the waste bin.
Brad

Yes.
It is interesting to consider, which way the energy flows when the pump rotates against that torque.
In the real world,the motor driving the pump would see a higher than startup load,and the energy contained in the flow of water would be converted into heat,by way of the motor now working harder to slow and stop that flow of water,which also is being pushed by the inertia of the water wheel(our inductor).
Brad

Here we go again. Who said that ideal voltages have to be fixed and can't change in time? It's an ideal voltage, I can say it does whatever I want it to do. In my second example I state that the ideal voltage v = 20*t^2. So the ideal voltage increases in value proportional to the square of the time. There is nothing stopping me from saying that.
What about your function generator when you set it to a triangle wave and you connect it to your circuit? Supposing that the triangle wave is somewhat attenuated especially at the peaks because of the load on it from the circuit. Well, what if I said that my "ideal voltage source" had an identical waveform to the somewhat attenuated triangle wave output from your function generator? Under those static load conditions then the ideal voltage source that clones your function generator output and the function generator output are indistinguishable. That means you can look at the output from your function generator as being a timevariable ideal voltage source as long as the load condition remains static and unchanging.
Oh here we go.
The MH paradox is kicking in again.
An ideal voltage source dose not change in timehow many times to you have to be told this?.
You have an ideal voltage source that !you! change in steps over time. Each time period in your question has a set voltage value that dose not change in time. Poynt also is in agreement with this.
Here we once again have MH doing the old !side step!,and adding more confusion to the discussion.
Brad

Oh, C'mon!  these were rhetorical questions. I was not seeking answers to them. I was provoking some thinking.
And dont ever stop doing that verpies,as it would seem that there needs to be more of it.
Of course, but this needs to be stated explicitly, since apparently some people are under the impression, that an ideal voltage source is a constant voltage source :o
An ideal voltage source will deliver the voltage selected by the user,for as long as the user decides to change the voltage value. It will deliver the selected value over any load,and will not dissipate energy,due to the non existence of an internal resistance.
An ideal voltage source dose not,and never will exist.
It is an !oxy moron!.
If an ideal voltage source can deliver an ideal voltage across any load,what happens when it is placed across an ideal short as the load?this would be the same as clipping the two output leads from your ideal voltage source together.
Brad

C'mon yourself. Your question about whether MH knows what an ideal voltage source is, is absolute bunk.
Well he dosnt know what a shorted ideal inductor is,so is verpies remark that far off?
MH adds confusion to this thread ,by instating his later test mixed in with the original question.
After that it seemed you were the one that was lost. I encourage you to avoid the socalled rhetorical questions, and instead try to help Brad understand why his thinking on this affair is a little off the tracks.
And who is it that has tested this very circuit,that can say that !i! am lost ?.
Brad

So, just to be clear here...
I can have an ideal voltage source that puts out 4 volts @ 1 amp but then, after 2 seconds, puts out 10 volts at .5 amps
and later, after 20 seconds puts out 2,000 volts @ 500 amps?
Do we need ideal circuit breakers that do not blow when working with this ideal voltage source?
Honestly, this just completely blows my mind that you (meaning anyone) can just make up some hypothetical device and then assign it ANY operational parameters that you want and then, do an "experiment" with this made up device and then claim that something real was learned?
This sounds like the same crap that Mythbusters might use. I just listened to an audio book that mentioned the Mythbuster's show where they "proved" that Archimedes did not have a death ray. They tried it, it did not work so this was PROOF!
Well, some boys over at MIT did a replication and....boom! It worked as it was supposed to. Those idiots at Mythbusters were ALWAYS changing parameters and then claiming devices to be "impossible". So, they could not find a mirror large enough so, they substituted a bar of soap instead...didn't work...see? Impossible! (Of course I exaggerate here a bit, but not too much)
I apologize...I just can not get my head around making crap up to fit your theory, and when your theory works claiming a victory.
I must be getting old.
Bill
Your not getting old Billyou are spot on,and we see this many times.
The biggest problem with !!some!! of the EE guy's,is they are programmed to make everything fit the way it should,as they have learned.
Myth busters indeed lol,did you see there attempt at the Bedini SSG lol. Now im no bedini fan,but these guys completely bungled that one upand these guy's(the myth busters)get paid a fortune for there idiotic tests.
You will very often(in fact ,im sure you have many times already) see this sort of thing happen,where accurate descriptions are taken,and re arranged to fit in with what should happen,and how something is needed to work.
The fact is,an ideal voltage dose not !!vary!! over time. An ideal voltage will produce a steady voltage value across any load presented to it,until such time as that value is changed by the user. No load will cause a variance in the voltage from the ideal voltage sourceonly the user can change the voltage value.
For MH to state that the voltage can vary in time is incorrect,as we would see that as being like the current value varying in time with a steady slope. An ideal voltage can increase or decrease over time in step's selected by the user,not by the load varying the voltage as it can in a real world voltage source.
Brad

I am not attacking you. In fact the "you" in my posting refers to all those that "don't believe" that an ideal voltage source can vary with respect to time. Like it or not, many people would find just arguing the issue to be strange. It's kind of a mom and apple pie issue and you are of the group saying that you are not allowed to put ice cream on your apple pie. It just makes no sense.
I am not going to address all of the drama in your posting about me nor all of the drama you raise about the question itself. There is no need for all of this, truly. Brad and others are supposed to be trying to answer a simple question and understand all of concepts and the related issues. As far as I am concerned this whole thing should have happened six years ago.
You made reference to a question of yours that was ignored. I don't see a question in post #454. That posting is really hard to digest, BTW. I went back a few more pages and did not see anything so I suppose you are taking about your post #454. If you can tell me what the question is I will try to answer it.
MileHigh
The only reason you are saying !Brad! and others are !!trying!! to answer the question,is because we dont agree with you,and this is something you do all the time.
If you think you are correct beyond being able to be incorrect,then that is nothing more than your own arrogance shining throughand this go's for any of the EE guys that think they have all the answers.
You are dealing with ideal that have not,and cannot be delt with. But you feel that you have all the answers sitting in your hand,and they are correct beyond doubt>bullshit.
Here is one big flaw between the EE guys that think they all agree with one another,and the answer is nailed down solid.
Quote MH>An ideal voltage source does not "contain energy" and likewise an ideal short does not "contain energy.
Im not sure what planet MH is on there?.
Quote Poynt>The energy in the inductor does not "disappear", some is returned to the source during that period.
So we know that the source is the ideal voltage source,and being that Poynt says that the energy in the inductor is returned to the source,that must mean that the ideal voltage source is what contains that returned energybut MH says that the ideal voltage source dose not conatin energy???
So we see here that the two guys that say that each other agree with each others theory ,dont seem to agree on what one of the two components in this circuit can do.
And as far as MHs comment about the ideal short not containing energy,well when it is taken !in! context as it should be,and that being the discussion of the ideal coil having an ideal short across it,then it dose indeed contain energy.
An ideal short is a short with no resistance,and our ideal coil has no resistance,and so when that ideal coil is a continual loop(shorted across it's terminal's),then it is an ideal shortas verpies showed it to be.
So you can see here that there is a discrepancy between the two that think they have the answer nailed down,and also the fact that MH is not the guru that he claims to be here.
How he ever came up withan ideal voltage source dose not contain energy,is beyond me. :o
Brad

@ Poynt
As you believe that MH could answer his question correctly,and following his statementAn ideal voltage source does not "contain energy"
I ask you once againhow is the energy dissipated in the ideal inductor loop during the 0 volt period,so as the negative 3 volt period can start inducing a current flow in the opposite direction to that of the current already flowing?. The energy stored in the inductor loop during the 0 volt period is in opposition to that of what the ideal voltage source wants to induce,and so it must be removed before the next phase can start.
So either you are incorrect,and the energy in the inductor is not returned to the source,or MH is incorrect,and the ideal voltage source can contain this returned energy.
If you are incorrect,then we have not yet derived at a correct conclusion to this question of MHs.
If MH is incorrect,then how is it possible that he could have answered his question correctly?how would he have explained as to where the stored energy in the inductor went before the next part of the cycle started?.
Brad

John.
We must thank the likes of MH. poynt etc. for their patience in inching us along.
wattsup,tinman and me, what I've learned is if you want to do electronics
you've got to go to college and start from DAY ONE and really pay attention.
I'm managing to get a bit but only by constantly referring to very basic
tutorials.
You mean go for your Indoctrination.
You seen a perfect example of how !wrong! you can be taughteven at the best of the best of themMIT.
One of the best physics teachers of all time'sWalter Lewin,got it wrong.
The best of the best teach them selvesNikola Tesla.
Brad

It helped Wattsup to notice that he needs to be more careful with his words.
These are technical posts and every word matters. Making mental shortcuts is sometimes funny but most often it requires "secret decoder rings" from the reader. Those are not good habits for technical communication.
So now we have established that an ideal voltage source merely needs to have:
 zero internal resistance
 zero internal reactance
 an output voltage that is not affected by the load
...but it does not need to output a constant output voltage all the time, and it can be even an ideal alternating voltage source.
We can consider the direction of the current flowing through this ideal voltage source.
So we have 2 variables and each one can have 2 states. This gives us 4 combinations.
Of course this is again the 4quadrant operation, where:
 2 combinations result in energy flowing from the voltage source to the load
 2 other combinations result in energy flowing from the load to the voltage source
Can you enumerate which ones?
Brad will like this:
The current can flow in two directions through the voltage source and the voltage presented by this voltage source can have two polarities, too
.
Well of course it can,but it cant have both at the same time.
Current cant flow in both directions at the same time,and the two voltage polarities cannot exist across the terminals at the same time.
But it would seem that the current can flow out and into this ideal voltage source at the same time???well as far as Poynt is concerned. And MHwell ,he states that the ideal voltage source cannot contain energy at all. So,in MHs case, i have no idea as to where the energy comes from to induce the current flow MH calculated through his ideal circuit ::)
Brad

Brad:
The simple fact is that you have trouble with abstract concepts. An example is an ideal voltage source varying in time, you still refuse to believe that. In the more difficult version of the question the ideal voltage source varies in time. You need to move on with respect to the ideal voltage source business and just accept it.
An inductor can store energy. A capacitor can store energy. A battery can store energy. But an ideal voltage source does not "store energy." "How many Joules are in that ideal voltage source?" is a question that makes no sense. An ideal voltage source is a potentially unlimited source of power. It is simply "power on tap" and it doesn't make sense to think about it storing energy.
If an inductor returns a finite amount of energy back to an ideal voltage source then that energy is not to be "found inside" the ideal voltage source. The concept of an ideal short "containing energy" is simply ridiculous.
It's like a jigsaw puzzle were you want to force pieces of the jigsaw puzzle together that don't fit because you said something and the only way to backup your statement is to force some concepts together that don't fit. It's related to your issue about refusing to accept that you are wrong.
It's time for the discussion to move forward. What is the complete current waveform and more importantly what are the concepts that explain the current waveform? The only part of the question that you have answered so far is the fact that the current doesn't change when the ideal voltage source outputs zero volts. The clues to answering the full question are in this thread.
You have to understand what is taking place in the circuit and why it does what it does if you are going to understand and appreciate the harder version of the question that has already been answered.
MileHigh

This is for those who hold to the misinterpretation of an ideal voltage source as "not being allowed" to vary. First the question, how do we solve a problem involving an ideal AC voltage source? Do we ignore such problems or do we "not allow" them to exist?
In an effort to help resolve this issue, I pose a new problem which is most relative to MH's original.
We have an ideal voltage source that starts at T0 with zero volts and ramps linearly to 4 volts at t1 = 1 second. In parallel with this ideal voltage source is an ideal inductor of 5h. What is the inductor current at T1?
partzman
edit

tinman if you aren't prepared to learn and accept the basics you're a
waste of time.
Ideal voltage source zero internal resistance, able to supply or absorb any amount
of current.
Either independent or dependent as required.
Some things aren't for debate, for example I wouldn't put milk in the fuel tank
of my Honda or I wouldn't try to put petrol in our Tesla p85d.
John.

I'll pick method number #2 of you want me to pick something.
I don't know what you really mean by "falls on its own" in method #1. You seem to be suggesting that it is a passive action by the voltage source. How could that be? How could the voltage source "know" to fall to 3 volts?
I don't why you focus on things like this. If you have a basic understanding of electronics then the question as originally posed is straightforward, selfevident, and not ambiguous. You are reading into something that is not there.
No, there is no problem at all.
You are in "how many angels can dance on the head of a pin" territory. People that understand basic electronics don't see any rhyme or reason in what you are saying.
@MH
Let me just clarify this for the last time.
The question, in the way it is asked implies that with one application of an ideal voltage of 4V to an ideal inductor of 5H will produce in that one action the following effects in a continuous sequence. Consider the effect is a resulting single sine wave from that one first single action of applying that one instance of 4V for a duration of 14 seconds then to infinity.
Method #1
1) the sine wave starts at 4 volts for 3 seconds
2) the same sine wave then drops to 0 volts for 2 seconds
3) the same sine wave then drops to 3 volts for 2 seconds
4) the same sine wave then rises to 0.5 volts for 7 seconds
5) the same sine wave then drops to 0 seconds for infinity.
All this happens with only that first application of 4V at t0 and from there with only that 4V applied to the inductor, the voltage measurements across the inductor would indicate those voltage values from 2 to 6 in those time intervals, but always with that first 4V applied to the inductor. No stopping, manually changing voltage levels. Those voltage levels shift down and up on their own because of the action of that first 4V introduced into that inductor.
That's how your question was asked. That's why I was against your question. But again this is your lack of precision, not mine. You should have seen that something was not understood by the way your question was asked. IT WAS YOUR QUESTION. You should have known that after 23 pages of heated rebuttal that you needed to ask us to first explain what the question means. PRECISELY. Instead you kept pushing this bad question and chastised us for what?
The question the way I understood it was impossible to answer or maybe you would like to take a crack at it as I have explained above. I even tried to make heads or tales of it working out a logic base. Now that I know these are 6 independent events of ideal voltages applied to a 5H inductor, this question just went from 1000% difficulty to 2% difficulty. I was actually impressed that your Method #1 could have been a real scenario and I just could not figure it out. But now I know it was a Method #2 question which is just more rudimentary EE. Basically a false flag event.
Actually I should be commended for trying to at least work some logic into Method #1. hahahahaha
So again, this does not show how a coil works. It only shows how a coil responds. There is nothing in EE that shows how a coil works. That falls into the realm of physics where my Spin Conveyance will shed the light. From there it will migrate to EE and other disciplines.
I have 35 years as a water treatment professional. I have worked with ions and ions don't lie. When you realize how your coil works you will have to start over with your EE. You will have to create new math, new formulas (hehehe), new models. Or, the present math will have new correlations to cause and effect. This is not a bad thing. It will open up a whole new Renaissance of knowledge that Faraday was to chickenshit to realize because his balls were tied tight with his bosses fields. So it was a good stoop.
So here is a question for you. Why does the current never rise above the applied voltage? Why does it always ride at a certain percentage below? Why have I been talking about Half Coil Syndrome?
Now if this has been closed maybe I can go back to the JT thread when I get back from work I can post some new effects patterns that none of you have ever seen before and we can really talk about how a coil responds (not works) instead of this water in a pipe business.
@tinman
OK, here is a close analogy to how a coil responds (not works) with water in a pipe under the DC model.
You have two pumps, one on each end of a length of pipe both pumping into the pipe. Each pump is first connected to a tee connection where the pump is one end, a valve that opens and closes on one end and the pipe length is on the third end. One pump is your positive connection, the other is your negative connection. As your coils are plugged usually one connection is always connected and the other is always pulsed. Each pump has a rheostat. The positive side pump it set at 60%, the negative side pump is set at 40% of available voltage. So if you are pulsing the negative side, only that side pump turns on and off while the other polarity pump is "always on". When the pulsed pump is on, the valve on that same side is closed. When that same pump turns off, the valve on that same side opens. The valve on the always pumping side is always closed. Both pumps draw from the same tank. Each side also has a higher cracking pressure check valve that returns to the tank plus those two valves also return to the tank. By this you will realize that pulsing a coil on the negative will always cost 60% for rebias for 40% change, while if you pulse the positive side of the coil it will cost 40% rebias for 60% change. This analogy is still wrong because we are using water that can only flow. This automatically is not the right way to see it. A close physical way would be if there was a way to introduce a colored die into the water at pulse on and then retract that die at pulse off. The water never really moves. Actually even the die analogy is not right because it is much simpler then that but for physical means they come close enough. Maybe 40% close. hahahahahaha
wattsup

tinman if you aren't prepared to learn and accept the basics you're a
waste of time.
Dont call me a waste of time,you potty mouthed little weasel.
What have you got to show for your self?yes,nothing.
Ideal voltage source zero internal resistance, able to supply or absorb any amount
of current.
MHs response to that isan ideal voltage source dose not contain energy> so go argue with him.
Either independent or dependent as required.
Some things aren't for debate, for example I wouldn't put milk in the fuel tank
of my Honda or I wouldn't try to put petrol in our Tesla p85d.
And you wouldnt try to claim victory when dealing with unknowns and ideals without proof.
You fall under the same spell as MH,and that is not being able to define between a small resistance and no resistance. I have shown you on a number of occasions that the difference is infinite.
Can you accurately calculate the top speed of a Chevy Camaro by measuring the top speed of a mini moke?. No ,i did not think so. But this is what is trying to be done here,using real world applications and devices to conclude how an ideal inductor and an ideal voltage source would react in a circuit,under the conditions that MH has set out.

@ Poynt
The best i can do as far as an inductor i have in the work shop,is a 1 Henry coil with 1.6 ohms of resistance.
This is a very large inductor out of a 300 amp HF mig welder.
Nothing else comes even close to the 5 Henry coil needed,with a low resistance value.
I do have a large MOT,and that is .71 Henry,and a resistance of 1.2 ohms.
I still think we could see the desired effect i am talking about with one of these inductors.
All you need to do,is work out how to switch at the rate needed for either of these two inductors,and maintain an unimpeded current flow through the inductor during the 0 volt portion of the test.You have to design the !ideal! voltage supply ;)
I do have some 6 amp diodes that have only a .3 volt drop across them.
Maybe we could just use one of these as a free wheeling diode across the inductor,to allow for the continuation of the current flowbut how to remove the dead short when the voltage is applied inverted?.
I do have plenty of 20 amp double throw relays,but i think they will be too slow in switching for the job,and large arcs will be formed across the contacts.
I'll leave you i think.
Brad

Well this is how i see it Poynt.
We have a loop of water pipe that represents our ideal inductor loop. In that loop of water pipe we have a pump. Our pump is our voltage source,and the water in that pipe is our current. We start the pump,and this puts pressure(our voltage) on the water(our current) the water starts to flow around our loop. We have a bypass valve in the pump,so as when we switch the pump off(0 volts),the water can still flow in the direction it was(our current is now flowing with the pump off). While the water is flowing,we start the pump up so as it spins in the opposite direction. The pump(our now reversed voltage) wants to now push the water in the opposite direction to that of which it is already flowing. The energy in the moving water is not added to,or stored in the pumpit is working against the pump,and the pump will draw more current to stop the flow of water that is flowing in the wrong direction, before it can start to move the water(current) in the right direction to that of what the pump wants to move it in.
That is how i see it.
Brad
Maybe a fluid coupling would be better?? 4V would be the stall speed as such that the outer "pipe" and the fluid are now not moving relative to each other,, then reverse the input to 4V and it takes twice as long to get to stall again.

@MH
Let me just clarify this for the last time.
The question, in the way it is asked implies that with one application of an ideal voltage of 4V to an ideal inductor of 5H will produce in that one action the following effects in a continuous sequence. Consider the effect is a resulting single sine wave from that one first single action of applying that one instance of 4V for a duration of 14 seconds then to infinity.
All this happens with only that first application of 4V at t0 and from there with only that 4V applied to the inductor, the voltage measurements across the inductor would indicate those voltage values from 2 to 6 in those time intervals, but always with that first 4V applied to the inductor. No stopping, manually changing voltage levels. Those voltage levels shift down and up on their own because of the action of that first 4V introduced into that inductor.
That's how your question was asked. That's why I was against your question. But again this is your lack of precision, not mine. You should have seen that something was not understood by the way your question was asked. IT WAS YOUR QUESTION. You should have known that after 23 pages of heated rebuttal that you needed to ask us to first explain what the question means. PRECISELY. Instead you kept pushing this bad question and chastised us for what?
The question the way I understood it was impossible to answer or maybe you would like to take a crack at it as I have explained above. I even tried to make heads or tales of it working out a logic base. Now that I know these are 6 independent events of ideal voltages applied to a 5H inductor, this question just went from 1000% difficulty to 2% difficulty. I was actually impressed that your Method #1 could have been a real scenario and I just could not figure it out. But now I know it was a Method #2 question which is just more rudimentary EE. Basically a false flag event.
Actually I should be commended for trying to at least work some logic into Method #1. hahahahaha
So again, this does not show how a coil works. It only shows how a coil responds. There is nothing in EE that shows how a coil works. That falls into the realm of physics where my Spin Conveyance will shed the light. From there it will migrate to EE and other disciplines.
I have 35 years as a water treatment professional. I have worked with ions and ions don't lie. When you realize how your coil works you will have to start over with your EE. You will have to create new math, new formulas (hehehe), new models. Or, the present math will have new correlations to cause and effect. This is not a bad thing. It will open up a whole new Renaissance of knowledge that Faraday was to chickenshit to realize because his balls were tied tight with his bosses fields. So it was a good stoop.
So here is a question for you. Why does the current never rise above the applied voltage? Why does it always ride at a certain percentage below? Why have I been talking about Half Coil Syndrome?
Now if this has been closed maybe I can go back to the JT thread when I get back from work I can post some new effects patterns that none of you have ever seen before and we can really talk about how a coil responds (not works) instead of this water in a pipe business.
@tinman
OK, here is a close analogy to how a coil responds (not works) with water in a pipe under the DC model.
You have two pumps, one on each end of a length of pipe both pumping into the pipe. Each pump is first connected to a tee connection where the pump is one end, a valve that opens and closes on one end and the pipe length is on the third end. One pump is your positive connection, the other is your negative connection. As your coils are plugged usually one connection is always connected and the other is always pulsed. Each pump has a rheostat. The positive side pump it set at 60%, the negative side pump is set at 40% of available voltage. So if you are pulsing the negative side, only that side pump turns on and off while the other polarity pump is "always on". When the pulsed pump is on, the valve on that same side is closed. When that same pump turns off, the valve on that same side opens. The valve on the always pumping side is always closed. Both pumps draw from the same tank. Each side also has a higher cracking pressure check valve that returns to the tank plus those two valves also return to the tank. By this you will realize that pulsing a coil on the negative will always cost 60% for rebias for 40% change, while if you pulse the positive side of the coil it will cost 40% rebias for 60% change. This analogy is still wrong because we are using water that can only flow. This automatically is not the right way to see it. A close physical way would be if there was a way to introduce a colored die into the water at pulse on and then retract that die at pulse off. The water never really moves. Actually even the die analogy is not right because it is much simpler then that but for physical means they come close enough. Maybe 40% close. hahahahahaha
wattsup
Method #1
1) the sine wave starts at 4 volts for 3 seconds
2) the same sine wave then drops to 0 volts for 2 seconds
3) the same sine wave then drops to 3 volts for 2 seconds
4) the same sine wave then rises to 0.5 volts for 7 seconds
5) the same sine wave then drops to 0 seconds for infinity.
Wattsup.
I dont see any confusion with the way MH has stated the question.
There are no sine waves in the voltage pattern.
MHs question describes a square wave pattern for the voltagesno sine waves.
I have always pictured the voltage pattern as being exactly what Poynt posted some pages backsee below.
The ideal voltage source,and the ideal coil has never been an issue. The confusion between what myself and MH,was in the words MH chose. When he said the voltage varies in time,i took it as meaning like the current varies in time through the inductorjust MHs bad choice of words that can have many meanings. The correct meaning as far as the question is stated,and the fact that a fixed voltage value is selected for a period of time(dose not vary),would have been to saywhere the voltage value can be changed by the user at any point in time,but where the voltage value is not changed by the loadbeing the inductor.
The only way the voltage can vary from the ideal source at one point in time,is if the user (or timing device) makes that variation. As you can also see,more confusion has been added by talk of ideal AC voltage sourceswhich is not applicable to the question.
Brad

It helped Wattsup to notice that he needs to be more careful with his words.
These are technical posts and every word matters. Making mental shortcuts is sometimes funny but most often it requires "secret decoder rings" from the reader. Those are not good habits for technical communication.
So now we have established that an ideal voltage source merely needs to have:
 zero internal resistance
 zero internal reactance
 an output voltage that is not affected by the load
...but it does not need to output a constant output voltage all the time, and it can be even an ideal alternating voltage source.
We can consider the direction of the current flowing through this ideal voltage source.
Brad will like this:
The current can flow in two directions through the voltage source and the voltage presented by this voltage source can have two polarities, too. So we have 2 variables and each one can have 2 states. This gives us 4 combinations.
Of course this is again the 4quadrant operation, where:
 2 combinations result in energy flowing from the voltage source to the load
 2 other combinations result in energy flowing from the load to the voltage source
Can you enumerate which ones?
Good post verpies. ;)

@ Poynt
The best i can do as far as an inductor i have in the work shop,is a 1 Henry coil with 1.6 ohms of resistance.
This is a very large inductor out of a 300 amp HF mig welder.
Nothing else comes even close to the 5 Henry coil needed,with a low resistance value.
I do have a large MOT,and that is .71 Henry,and a resistance of 1.2 ohms.
I still think we could see the desired effect i am talking about with one of these inductors.
The welder coil has the better L/R ratio of the two (0.625), but it is a far cry from 50, which is what we need to be close to ideal.
From the first plot below (red trace is MH's, violet trace Brad's coil), you can see that the lower value inductance and significantly higher series resistance (dissipating the inductor's energy) produces a result that is far from the ideal. So it seems this scheme as it stands won't work.
However, there is some hope. We can sort of "translate" our L/R ratio of 0.625 to 50 by significantly shortening the timing on the input source. We have little control over the real world L/R ratio, but we do have control over the input timing, which we can manipulate to utilize the "linear" portions of the circuit rise/fall transitions. We will shorten it by a factor of 1.6 x 50 = 80. So now our wave form timing becomes the following:
t0=0s
t1=37.5ms (was 3s)
t2=62.5ms (was 5s)
t3=87.5ms (was 7s)
t4=162.5ms (was 13s)
t5=250ms (was 20s)
From the second plot below you can see we've made a significant improvement in the "idealness" of our experiment, and will produce results good enough to compare to MH's original specs.
So the remaining challenge will be to design a source that can output this wave form.

For comparison sake to see how far off we are from "ideal" using our shortened wave form timing and nonideal inductor, see below for Brad's coil with an R=1m Ohm (close enough for ideal), the same as used for MH's inductor. The wave form timing is still the shortened version.

@MH
Let me just clarify this for the last time.
The question, in the way it is asked implies that with one application of an ideal voltage of 4V to an ideal inductor of 5H will produce in that one action the following effects in a continuous sequence. Consider the effect is a resulting single sine wave from that one first single action of applying that one instance of 4V for a duration of 14 seconds then to infinity.
Method #1
1) the sine wave starts at 4 volts for 3 seconds
2) the same sine wave then drops to 0 volts for 2 seconds
3) the same sine wave then drops to 3 volts for 2 seconds
4) the same sine wave then rises to 0.5 volts for 7 seconds
5) the same sine wave then drops to 0 seconds for infinity.
All this happens with only that first application of 4V at t0 and from there with only that 4V applied to the inductor, the voltage measurements across the inductor would indicate those voltage values from 2 to 6 in those time intervals, but always with that first 4V applied to the inductor. No stopping, manually changing voltage levels. Those voltage levels shift down and up on their own because of the action of that first 4V introduced into that inductor.
That's how your question was asked. That's why I was against your question. But again this is your lack of precision, not mine. You should have seen that something was not understood by the way your question was asked. IT WAS YOUR QUESTION. You should have known that after 23 pages of heated rebuttal that you needed to ask us to first explain what the question means. PRECISELY. Instead you kept pushing this bad question and chastised us for what?
The question the way I understood it was impossible to answer or maybe you would like to take a crack at it as I have explained above. I even tried to make heads or tales of it working out a logic base. Now that I know these are 6 independent events of ideal voltages applied to a 5H inductor, this question just went from 1000% difficulty to 2% difficulty. I was actually impressed that your Method #1 could have been a real scenario and I just could not figure it out. But now I know it was a Method #2 question which is just more rudimentary EE. Basically a false flag event.
I think the question is clear enough. A voltage source is connected across an inductor. The voltage is a step waveform with various voltage levels, what happens?
Some people didn't even know where to begin. That is very telling. What happens? The only unknown is the current, solve for the current.
Honestly, all these words that you are throwing at my question itself are too much. The question is what it is.
MileHigh

An ideal voltage source will deliver the voltage selected by the user,for as long as the user decides to change the voltage value.
Yes, an ideal voltage source will change the voltage across itself only when a user adjusts it.
An ideal voltage source will never alter its voltage when the load tries to change it.
In other words: An ideal voltage source obeys the user  not the load.
If an ideal voltage source can deliver an ideal voltage across any load,what happens when it is placed across an ideal short as the load?
An infinite current flows and the universe explodes ;D

An inductor can store energy.
Yes, unless it is open
A capacitor can store energy.
Yes, unless it is closed.
an ideal voltage source does not "store energy."
but it can absorb energy from a higher voltage source or from a current source.
Where does the absorbed energy go?

Here again are the two versions of Brad's coil (ideal in red, nonideal in violet) current traces on the same plot with the same vertical scaling.
There is obvious expected error with the nonideal inductor (violet trace), but it is sufficiently accurate to allow us to understand how a perfectly ideal inductor would react to the same input.

but it can absorb energy from a higher voltage source or from a current source.
Where does the absorbed energy go?
It gets sucked into the Dark Energy quantum foam vortex. Only to reappear later in the Jim Murray Dynaflux Alternator.
"It produces over 250% more electricity than it takes to run it."

but it can absorb energy from a higher voltage source or from a current source.
Brad, verpies is agreeing that some of the energy stored in the inductor can be absorbed by the voltage source. Or at least I think he is.
What are your thoughts now?

I think the question is clear enough. A voltage source is connected across an inductor. The voltage is a step waveform with various voltage levels, what happens?
Some people didn't even know where to begin. That is very telling. What happens? The only unknown is the current, solve for the current.
Honestly, all these words that you are throwing at my question itself are too much. The question is what it is.
MileHigh
@MH
Your question really involves 6 different experiments, so it should have been questioned as such. Each one would start at their own t0 mark at their respective voltage setting and duration. Each would have produce their individual result. Your question the way it was inferred is that all these events occurred with only one 4V application and nothing else. After the 4V you did not say stop, reset the voltage to 0, wait x seconds, stop reset the voltage to 3, wait x seconds, etc. You said all these events occurred because of the initial first application of 4V. Now do you get it. I don't see why this is so hard to understand.
If you were writing a technical report with such a question, you would have been fired the next day.
And please don't give this "If you were an EEer you would know what the question meant business". You were not asking this question to known EEers you were asking it to @tinman and others here. Wonder why no one else dared answer it? Because they were to smart to get involved in the first place. Dumb me all right.
@tinman
Sine wave, square wave, will not change that the question was just badly presented. Since these are 6 individual experiments with the same 5H, there is no point to answer it anymore because the first 2.4 amps gave out the rest of the answers. hehehe
It helped Wattsup to notice that he needs to be more careful with his words.
These are technical posts and every word matters. Making mental shortcuts is sometimes funny but most often it requires "secret decoder rings" from the reader. Those are not good habits for technical communication.
So now we have established that an ideal voltage source merely needs to have:
 zero internal resistance
 zero internal reactance
 an output voltage that is not affected by the load
...but it does not need to output a constant output voltage all the time, and it can be even an ideal alternating voltage source.
@verpies
I have always appreciated your input regardless of who you respond to it is always a pleasure. That includes @poynt99 and others who know who they are. But you all are not @MH and this is not your question.
So let me ask you this simple simple question.
You have an ideal voltage source of 4V and an ideal inductor of 5H. You apply the voltage at t0 for 5 hours. What will be the voltage read across the terminals at every hour? What will be the current measure at every hour?
You know why I made this easy. hahaha Here is my answer because it has already been said before. The voltage will remain at 4v throughout the 5 hours. The current will rise to 2.4 and remain there for 5 hours. Is this correct or not?
That's all we need to know to ascertain is @MH'S question was well presented of not.
wattsup

Wattsup:
No, the question involves one experiment. A standard way to describe a process is chronologically, which is what I did. Many people have no problem at all with the question. I am attaching Poynt's graph that shows the voltage waveform that corresponds to the question. There is no more need to discuss the question.
What we need is progress towards answering the question.
MileHigh

You have an ideal voltage source of 4V and an ideal inductor of 5H. You apply the voltage at t0 for 5 hours. What will be the voltage read across the terminals at every hour? What will be the current measure at every hour?
You know why I made this easy. hahaha Here is my answer because it has already been said before. The voltage will remain at 4v throughout the 5 hours.
Yes.
No load is incapable of altering the voltage at the output of an ideal voltage source.
The current will rise to 2.4 and remain there for 5 hours. Is this correct or not?
No, at t0 the current will be zero.
at 1h it will be 2.88kA
at 2h it will be 5.76kA
at 3h it will be 8.64kA
at 4h it will be 11.52kA
at 5h it will be 14.4A
...and it will keep increasing like that at the rate of 2880A per hour, ad infinitum.

Brad, verpies is agreeing that some of the energy stored in the inductor can be absorbed by the voltage source. Or at least I think he is.
What are your thoughts now?
My toughts are,because of MHs statement that an ideal voltage source contains no energy,is a clear indication that he would not have been able to answer his question correctly.
You may wish to rethink your stance on MH being able to do so.
Brad

@MH
If you cannot first concede your question was loaded the way it was asked, then you can play with yourself
wattsup

Brad:
My toughts are,because of MHs statement that an ideal voltage source contains no energy,is a clear indication that he would not have been able to answer his question correctly.
You may wish to rethink your stance on MH being able to do so.
This is just useless spinning. You read how I described what an ideal voltage source is. An ideal voltage source can output or absorb an unlimited amount of power, and by extension, an unlimited amount of energy. But it does not "contain energy." It's just a concept, an important tool used all the time in electronics.
Wattsup:
If you cannot first concede your question was loaded the way it was asked, then you can play with yourself
Sorry, but you failed to make an impression. Reading you discussing electronics is difficult and stressful.
Let's focus on moving forward.
MileHigh

My toughts are,because of MHs statement that an ideal voltage source contains no energy,is a clear indication that he would not have been able to answer his question correctly.
You may wish to rethink your stance on MH being able to do so.
Brad
I don't wish to change my stance. I know he could.
But that wasn't what my question was about. I said at one point that some of the inductor's energy would be returned to the source, and you disagreed and said it was not possible. And that had nothing to do with MH's statement, this was your own stance. verpies seems to agree with me on this and I'd like to know if this has caused you to reconsider perhaps?

Brad:
This is just useless spinning. You read how I described what an ideal voltage source is. An ideal voltage source can output or absorb an unlimited amount of power, and by extension, an unlimited amount of energy. But it does not "contain energy." It's just a concept, an important tool used all the time in electronics.
Wattsup:
Sorry, but you failed to make an impression. Reading you discussing electronics is difficult and stressful.
Let's focus on moving forward.
MileHigh
The fact that you dont understand what a !source! Of power is, only means you cannot understand your own question .
Show me any source that can deliver power without that source containing enery.
I await the next MH paradox.
Brad.

Brad, verpies is agreeing that some of the energy stored in the inductor can be absorbed by the voltage source. Or at least I think he is.
What are your thoughts now?
He also stated but what happens to it?.
Then we look at MHs idiotic reply after thatshows how serious he is about the whole thing.
Verpies saidbut what happens to it once it is absorbed?MH says it cannot be contained within the ideal device that verpies and yourself just stated absorbed it.
Im wondering if you are seeing how stupid this sounds.
A source that can provide power, must contain the energy within it, in order to provide that power.
I remain firm on my position that MH could not possibly answer his own question,as he dose not even know what a power source is.
Brad

Brad:
Well, if you are going to push it then we will get to the root of the matter, and it's a use of language issue. It makes sense to say that a capacitor or an inductor can "contain energy." It implies that there is a finite and measurable amount of energy in the device. In contrast, a voltage source has no finite and measurable amount of energy. It doesn't make sense to say that "A voltage source contains some energy." An ideal voltage source is not even a tangible device, it's a theoretical device.
In the world of electronics, nobody uses the language construct of an ideal voltage source "containing energy." That's the way it is, and sometimes you have to go with the flow. Just because you strung those words together does not mean that it is a valid thing to say. In fact it is an invalid thing to say and you should just absorb that fact and move on. When you used the argument that since an inductor can return energy to the ideal voltage source therefore the voltage source "must contain energy," it was simply wrong. You have to think and choose your words better than that. An ideal voltage source supplies energy, it does not contain energy.
Most importantly, we are certainly all in agreement on this fact: The ideal voltage source can supply power to the device it's connected to, and it can even absorb power from the device it's connected to. There is no disputing this fact by anyone in this debate, and that is the critical factor for advancing this debate.
So don't give us your trash talk that I "dont understand what a !source! of power is." You are not fooling anybody by a long shot. It's time for you to move forward in the technical discussion and talk about what we are really here for.
MileHigh

Brad:
The simple fact is that you have trouble with abstract concepts. An example is an ideal voltage source varying in time, you still refuse to believe that. In the more difficult version of the question the ideal voltage source varies in time. You need to move on with respect to the ideal voltage source business and just accept it.
An inductor can store energy. A capacitor can store energy. A battery can store energy. But an ideal voltage source does not "store energy." "How many Joules are in that ideal voltage source?" is a question that makes no sense. An ideal voltage source is a potentially unlimited source of power. It is simply "power on tap" and it doesn't make sense to think about it storing energy.
If an inductor returns a finite amount of energy back to an ideal voltage source then that energy is not to be "found inside" the ideal voltage source. The concept of an ideal short "containing energy" is simply ridiculous.
It's like a jigsaw puzzle were you want to force pieces of the jigsaw puzzle together that don't fit because you said something and the only way to backup your statement is to force some concepts together that don't fit. It's related to your issue about refusing to accept that you are wrong.
It's time for the discussion to move forward. What is the complete current waveform and more importantly what are the concepts that explain the current waveform? The only part of the question that you have answered so far is the fact that the current doesn't change when the ideal voltage source outputs zero volts. The clues to answering the full question are in this thread.
You have to understand what is taking place in the circuit and why it does what it does if you are going to understand and appreciate the harder version of the question that has already been answered.
MileHigh
No need for any more of you babble MH.
My description of an ideal voltage source is correct,and backed up by verpies. So now your just making your self look stupid.
The fact that you think a source that can deliver power ,dose not contain energy,just go's to show how little you know.
That also means that you could not have answered your question correctly.
You would be trying to explain how a car is able to be powered by an engine,without knowing how that engine works.
Good show MH.
Brad

Brad:
Well, if you are going to push it then we will get to the root of the matter, and it's a use of language issue. It makes sense to say that a capacitor or an inductor can "contain energy." It implies that there is a finite and measurable amount of energy in the device. In contrast, a voltage source has no finite and measurable amount of energy. It doesn't make sense to say that "A voltage source contains some energy." An ideal voltage source is not even a tangible device, it's a theoretical device.
In the world of electronics, nobody uses the language construct of an ideal voltage source "containing energy." That's the way it is, and sometimes you have to go with the flow. Just because you strung those words together does not mean that it is a valid thing to say. In fact it is an invalid thing to say and you should just absorb that fact and move on. When you used the argument that since an inductor can return energy to the ideal voltage source therefore the voltage source "must contain energy," it was simply wrong. You have to think and choose your words better than that. An ideal voltage source supplies energy, it does not contain energy.
Most importantly, we are certainly all in agreement on this fact: The ideal voltage source can supply power to the device it's connected to, and it can even absorb power from the device it's connected to. There is no disputing this fact by anyone in this debate, and that is the critical factor for advancing this debate.
So don't give us your trash talk that I "dont understand what a !source! of power is." You are not fooling anybody by a long shot. It's time for you to move forward in the technical discussion and talk about what we are really here for.
MileHigh
We are dealing with facts MH,not some deluded MH version of what an ideal voltage source is.
You are now trying to throw in one of your deluded paradoxes that dose not exist.
In one post you say ideal voltage sources exist,and now you say it's a device that dose not exist in reality.
First you argue with me when i state that an ideal voltage source dosnt exist,and now you agree with mejust to try and bail yourself out of your mistake.
Any voltage source contains energyshow me one that dose not.
If that source could supply an infinite amount of power,then it would contain an infinite amount of energy.
So just provide one example of a voltage source that can supply power,but has no stored energy within that soufce.
Brad

it's a use of language issue. It makes sense to say that a capacitor or an inductor can "contain energy." It implies that there is a finite and measurable amount of energy in the device.
Agreed. The Inductor or Capacitor which is storing
energy is said to be "Charged."
In contrast, a voltage source has no finite and measurable amount of energy. It doesn't make sense to say that "A voltage source contains some energy." An ideal voltage source is not even a tangible device, it's a theoretical device.
Surprisingly, your language and your theory is rather
imprecise Miles. Is there some Wobble here? ???
In the world of electronics, nobody uses the language construct of an ideal voltage source "containing energy." That's the way it is, and sometimes you have to go with the flow.
...
When you used the argument that since an inductor can return energy to the ideal voltage source therefore the voltage source "must contain energy," it was simply wrong. You have to think and choose your words better than that. An ideal voltage source supplies energy, it does not contain energy.
Now that is definitely Wobbly! So you're actually saying that
you believe a Voltage Source contains no Charge? That it is
not a repository of Stored Energy? That it has no Capacity? :o
Miles, that is a Most Unusual Conclusion! ;)

He also stated but what happens to it?.
Then we look at MHs idiotic reply after thatshows how serious he is about the whole thing.
Verpies saidbut what happens to it once it is absorbed?MH says it cannot be contained within the ideal device that verpies and yourself just stated absorbed it.
Im wondering if you are seeing how stupid this sounds.
Brad
Well, in this posting I am going to assume that you are not properly conceptualizing what an ideal voltage source does when you pump power into it.
I cracked a joke because Verpies was just asking a silly question for fun. Any power that is pumped into an ideal voltage source doesn't go anywhere. It's simply gone, you can forget about it. The ideal voltage source is still the same ideal voltage source, nothing has changed.
A source that can provide power, must contain the energy within it, in order to provide that power.
Yeah, it contains an infinite amount of energy. So where does that get you? If a coil pumps more energy into a "container" that stores an infinite amount of energy then voila!, you still have a container filled with an infinite amount of energy.
Or, if you want to be "more sophisticated" you can forget about the container altogether and just call it an ideal voltage source.
So it's time to stop spinning your wheels and stop the trash talk.

So just provide one example of a voltage source that can supply power,but has no stored energy within that soufce.
Brad
This is a ridiculous discussion and you are just wasting time.
Here is the answer to your question: I take a paper napkin and a pen, and I mark a few lines on the napkin, and there is your ideal voltage source and there is no stored energy in sight. The _idea_ that it can supply energy is in your mind only, and that's all that counts.

tinman, forget about concepts and abstractions you obviously can't handle them.
They're no use in the "real world" anyhow.
Forget about 5 Henry inductors they're too big.
Make yourself a Maxwell bridge and have a bit of fun learning what that can do.
John.

They're no use in the "real world" anyhow.
Forget about 5 Henry inductors they're too big.
Make yourself a Maxwell bridge and have a bit of fun learning what that can do.
John.
tinman, forget about concepts and abstractions you obviously can't handle them.
Dont give me that crap John,it's not me here thats talking double dutch.
Concepts and abstractions dont cut it,and if i have to be precise ,then so dose everyone else.
Perhaps take some time,and look up the definition of precision. ;)
Keep on reading,and see who is making no sense in this thread.
You continue on with your wollowing's,and yet know no better yourself.
Brad

Well, in this posting I am going to assume that you are not properly conceptualizing what an ideal voltage source does when you pump power into it.
And here we goMHs paradox that is needed so as he thinks he is not wrong in his statementbut falls apart with another of his own statement's.
So now we have an ideal voltage !source!,that has power pumped into it(a missing component of MHs 2 component circuit),even though it cannot store it ::)
So the ideal voltage source is now just some part that the incoming power passes through(as it cannot store it),and just flows straight into the inductor :o.
A capacitor(a voltage source)that has a voltage across it,contains stored energy.
A battery that has a voltage across it ,contains stored energy.
A sponge that absorb's water,contains that water.
I cracked a joke because Verpies was just asking a silly question for fun. Any power that is pumped into an ideal voltage source doesn't go anywhere. It's simply gone, you can forget about it. The ideal voltage source is still the same ideal voltage source, nothing has changed.
And this is the best yet.Any power that is pumped into an ideal voltage source doesn't go anywhere. It's simply gone
You have excelled your self there MH.You have just defied the laws of physics,and now it has to be rewritten. Energy can be transformed from one form to anotherand also be destroyed by MHs voltage source ;)
A recap on what you said above in the previous paragraph>So now we have an ideal voltage !source!,that has power pumped into it
Then in the next paragraph>Any power that is pumped into an ideal voltage source doesn't go anywhere
So this power that is feeding the voltage source dose not make it to the inductor???,as you have stated it dose not go anywherejust disappears.
I dont even have to go to any other of your post to put all this togetheryou have it all in one post MH :D
Yeah, it contains an infinite amount of energy. So where does that get you? If a coil pumps more energy into a "container" that stores an infinite amount of energy then voila!, you still have a container filled with an infinite amount of energy.
So now your ideal voltage source dose contain energy?,or is this just the supply of energy that go's into the voltage source that just disappears?.
Or, if you want to be "more sophisticated" you can forget about the container altogether and just call it an ideal voltage source.
So it's time to stop spinning your wheels and stop the trash talk.
Have a good look at your own statements above MH,and then ask your selfwho is talking trash.
You are so lost it's not funny anymore,and your buddy minnie is not that far behind you.
Brad

This is a ridiculous discussion and you are just wasting time.
Here is the answer to your question: I take a paper napkin and a pen, and I mark a few lines on the napkin, and there is your ideal voltage source and there is no stored energy in sight. The _idea_ that it can supply energy is in your mind only, and that's all that counts.
Another idiotic statement.
Lets see your paper napkin induce a current into an inductorsee how stupid that sounds MH.
You asked a specific question base around two components.
We know we can get very very close to an ideal inductor in the real world,and !you! have stated many times that we can build an ideal voltage source(which of course we cannot). And now we are going into the land of MHs make believe,by way of lines on a paper napkin to solve MHs question.
And you wonder why i say you are the laughing stock of the forum.
Well we only have to look at the last few of your posts to see why.
Im not interested in following you up the garden path of MH wonder land,where reality give way to paper napkins.
Brad

I read every word. How was that post helpful to anyone?
I encourage you to avoid the socalled rhetorical questions, and instead try to help Brad understand why his thinking on this affair is a little off the tracks
.
My thinking is off track???
It is not me that thinks this ideal voltage source can absorb energy but not contain it.
Did you guys really think about this statement,and how it go's with other agreed upon definitions that have been made about the ideal voltage source?.
Sometime's you EE guys make no sense at all,and it's no wonder people get confused.
We have all agreed that an ideal voltage source dose not dissipate any power,as it's ideal.
If the ideal voltage source can absorb power,but cannot dissipate power>where is this absorbed power if it is not contained within the ideal voltage source?.
Is it just like MH said,and it just vanishes without a traceit's just gonethe energy is destroyed?
Do you see how ridiculous this sound's,and how far away from the fundamentals of science this is.
Here you are Poynt,saying that i should agree with what is being presented by the EE guy's here,but there is no way in hell i am going to settle on anything when statements such as above are being asked to be accepted,as statements such as MHs contradict them selves,and the others make no sense at all.
So my question remainswhat happens to the stored energy (during the 0 volt part of the cycle),when the negative 3 volt part of the cycle start's?
1MHs theory>it just disappearsenergy is destroyed.
2it is stored in the ideal voltage source?
3it is absorbed by the ideal voltage source that cannot dissipate power=stored in the ideal voltage source.
Unlike MHs mockery of this thread,i wish to solve this question correctly. But while there are nonsensical comments likeenergy can be absorbed but not stored in a device that cannot dissipate that absorbed energy,then i will simply not be agreeing to anything.
So maybe you can try and make sense of that ,and let me know what it is,as i know of no other device that can have a voltage across it,but not have stored energy within it. And this rubbish about energy just disappearing by MH,is nothing but a joke.
So to put it simple,how can a component that absorbs energy,but cant dissipate energy,not store that energy?.
Brad

@MH
You really are either willfully ignorant or willfully deceptive.
Give me one reason why for endless pages and pages i would disagree with your question continuously stating to you that ideal voltages do not change in time. @verpies again just confirmed this. You knew all along that we did not "understand" your question because there was no other reason for anyone to object to the question unless they understood the whole process to be one event.
You willfully, and, I have to add with the aid of others who were in the know but decided to just shut up and let the commotion continue endlessly, decided to not clarify the question and leave it as the ambiguous piece of shit it was. You had the choice but you decided to just play along, you fed the fire, you were implicit in this and played us for fools for two weeks.
You knew that one little clarification that would not have given out any of the answers to your question but would have brought more fluency to the question would have solved the whole matter, but you decided to both play your, "I am an EE king and you need to learn boy attitude". So now I know how much of a prick you are and will leave you to play with yourself.
You are an insult to EE teaching and I have to say now that those who knew were the small discrepancy was in this question and did not step forward to provide a correction are no better. But the onus was fully on @MH as he played the teacher fiddle and he knew the question was not in tune but he kept on whipping away enjoying ever minute.
So now, no more trust. You burned that one to hell this time. You played us for fools. You are now off my list. If I ever see you on any of my threads, you will be deleted so don't bother. I have no more trust in your opinions since I know now you are only on this forum to up your own ego. So enjoy it.
@tinman
Listen. I am only going to say this once. If you knew @MH's question involved 6 separate events combined in one sequence, then you had no damn reason to object to his question. So don't pretend you fully understood his question because that would be adding to the pile of bull already on the floor.
@all
All you guys do is play games. Every thread starts holy molly and ends in bullshit. And what did you learn about the coil, F all, but keep on plastering those games all over the placed and wish and wish and wish some more.
wattsup

@MH
@tinman
Listen. I am only going to say this once. If you knew @MH's question involved 6 separate events combined in one sequence, then you had no damn reason to object to his question. So don't pretend you fully understood his question because that would be adding to the pile of bull already on the floor.
wattsup
And when did i object to his question?
I started this thread to answer his question,not object to itso i dont know what page you are on,but it's the wrong one.
I fully understood his question from the day i read it,and did not object to the question at all.
I object to the fact that the EE guys think the answer is simple and straight forward,as when you use ideal components,it is never straight forward.
So i really do not know why you have said what you have said,as i do not object to the question,nor have i ever done soother than the use of ideal components that dont exist being used in the question.
If real world components were used in place of the ideal components,then the answer is pretty straight forward,and that answer would depend on how much information MH required for the answer. Would he want to know things like magnetic field strengths during each part of the cycle?. Would he want to know coil temperatures at the start middle and end of the cycle?.
When you askwhat happens from T=0,leaves a very big list to choose from,and this was unspecified in the question.
Brad

Wattsup:
You are truly baffling, and even Brad says you are baffling. From your comments it's like you are on another planet.
Brad:
The only thing you need to know about the ideal voltage source for this question is that it puts out the waveform as described.
So can you move forward now and try to answer the question?
MileHigh

And when did i object to his question?
I started this thread to answer his question,not object to itso i dont know what page you are on,but it's the wrong one.
I fully understood his question from the day i read it,and did not object to the question at all.
I object to the fact that the EE guys think the answer is simple and straight forward,as when you use ideal components,it is never straight forward.
So i really do not know why you have said what you have said,as i do not object to the question,nor have i ever done soother than the use of ideal components that dont exist being used in the question.
If real world components were used in place of the ideal components,then the answer is pretty straight forward,and that answer would depend on how much information MH required for the answer. Would he want to know things like magnetic field strengths during each part of the cycle?. Would he want to know coil temperatures at the start middle and end of the cycle?.
When you askwhat happens from T=0,leaves a very big list to choose from,and this was unspecified in the question.
Brad
@tinman
Wow, this is playing out to be one hell of drama here.
I cannot believe if you knew what the actual question meant, you would have had a problem answering it then. The only reason I backed you up in this is because I saw this question as to ambiguous and did not openly imply 6 separate MANUAL settings of voltage. In my book there was no other VALID reason to object to it since we all new it was under ideal conditions. We know ideal conditions do not exist but the way the question was asked was in my book, deceptive in the process, not in the substance.
So good luck then.
Added: When did you object to the question? Just look at the JT thread which is the origin.
@MH
When you ask a question, you know, if your life depended on that question, you would be toast by now. Everyone should just go back and read the question from the stand point of an outsider, not your EE brainiacs perspective. Questions are not for those that are in the know. They are for those that want to understand your viewpoint. It's your view you want to bring across and you did a shitty job of it. I don't know where @tinamn is coming from to say he understood it. If he did, then bads on him.
Enough of this.
wattsup

Brad:
If real world components were used in place of the ideal components,then the answer is pretty straight forward,and that answer would depend on how much information MH required for the answer. Would he want to know things like magnetic field strengths during each part of the cycle?. Would he want to know coil temperatures at the start middle and end of the cycle?.
We already discussed a variation on the question. How about you answer it where you add an 0.00001 ohm resistance to the 5 Henry coil? That's the only thing that needs to be added. The answer will be almost identical, you are free to go that route if you want.
There is no need to know the magnetic field strength, and no need to know the coil temperature.
MileHigh

@tinman
Wow, this is playing out to be one hell of drama here.
I cannot believe if you knew what the actual question meant, you would have had a problem answering it then. The only reason I backed you up in this is because I saw this question as to ambiguous and did not openly imply 6 separate MANUAL settings of voltage. In my book there was no other VALID reason to object to it since we all new it was under ideal conditions. We know ideal conditions do not exist but the way the question was asked was in my book, deceptive in the process, not in the substance.
So good luck then.
wattsup
Here is the problem with being able to answer the question using ideal components.
1 if the coil is ideal,then why would not the CEMF be ideal to the EMF that created it?
If the CEMF is ideal,and equal to the EMF,would current ever flow through the coil?.
2 If the voltage source is ideal,it cannot dissipate power. If it absorbs power,and cannot dissipate power,why dose it not store that power as MH says?.
MH says an ideal voltage source can deliver energy,but dose not contain any energy.
He then says that an ideal voltage source can absorb power,and that power just is gonedisappears,no longer exist.
To many holes in the question when ideals are included.
To many holes in MHs explanation.
To many other EE guys agreeing with some of these MH paradoxes that are nonsense.
If we are to come to some sort of correct answer,then these nonsense's need sorting.
To look at the question again
You have an ideal voltage source and an ideal coil of 5 Henrys. At time t=0 seconds the coil connects to the ideal voltage source. For three seconds the voltage is 4 volts. Then for the next two seconds the voltage is zero volts. Then for two seconds the voltage is negative three volts, and then for the next six seconds the voltage is 0.5 volts. Then after that the voltage is zero volts.
What happens from T=0 when the ideal voltage is connected to the ideal coil?.
It is clear that an ideal voltage source delivers a volatge selected by the user,and will maintain that voltage no matter what the load. The voltage value can only be determined and change by the usernot the load.
The user(MH) has selected a voltage of 4 volts to be placed across the ideal coil,for a period of 3 seconds. After the first 3 seconds, he then changes that voltage value to 0 volts,for a period of 2 seconds. After that two seconds is up,he inverts the polarity,and selects a voltage value of 3 volts,for a time period of 2 seconds. After that period of 2 seconds is up,he then selects a voltage value of 500mV for 6 second's. As he dose not mention !negative! at this point,we can assume that the voltage polarity at this point in time,is now back to the original polaritybeing the positive polarity. After this 6 second time period,the voltage value is then selected to be 0 volt's.
I do not see the problem in understanding the voltage wave form that would be associated with this question. What i do question is would that actually be the case,and would this ideal circuit follow the same rules as real world components?.
Brad

He also stated but what happens to it?.
Yes, that is a good question too.
But I'd like your answer on the concept that the source can and will absorb energy from the inductor.
Do you agree or disagree?

What's the problem?
I.V.S. able to supply/absorb any amount of current.
The current through I.V.S. Completely under the control of the external circuit.

I know some here and perhaps many that have followed this thread think that this is some kind of futile exercise that has no bearing on our OU efforts so let's get back to work!
Well, consider this If I should see some kind of odd result in my prospective OU creation, how do I determine if the anomaly I see is the result of some newly discovered function or simply the result of conventional functions that slipped past me because of my lack of understanding. This has happened to me more times than I would like to admit and it is the reason I strive to understand the basics so I will be able to recognize a true anomaly if and when it occurs.
I believe that this was/is the intention of MH when he proposed the original problem as well as others who have contributed along the way. Whether the problem uses ideal components as originally stated or real world components as later stated, the resulting differences are insignificant. The point is, do I understand how to solve this problem or not?
partzman

Yes, that is a good question too.
But I'd like your answer on the concept that the source can and will absorb energy from the inductor.
Do you agree or disagree?
I cannot agree to the ideal voltage source being able to absorb energy if it cannot contain it.
If any energy is absorbed by the ideal voltage source,and that ideal voltage source dose not dissipate energy,then the energy must now be contained in it(the ideal voltage source).
I cannot agree with MHs statement that the energy is just gonedisappears.
MHs question consists of two components onlythe ideal voltage source,and the ideal inductor. He has made this apparent many times when he has stated that EMJ and Wattsup could not answer a simple question that involved on two components.
MH also states that the ideal voltage source can deliver energy,but dose not contain energy.
At T=5s, what would the solution below read?
And then again at T=7s,what would the solution below read?.
Brad

MHs question on the JT thread
What happens to the energy if you pump 100 watts of power into an ideal voltage source for 10 seconds
Poynt
Am i correct in saying>as an ideal voltage source dose not dissipate any energy,the energy is contained in that ideal voltage source.
Brad

My assumption would be that since the Ideal Voltage Source has no means to dissipate nor store energy that the current flow would continue from the inductor through the Source, and at the time of the reversal of the polarity that the source would initiate a stream of charge carriers moving in the opposite direction.
Considering that at the time of a drop in voltage from the source the conductors selfinductance turns the conductor into a source, but not an ideal source, the 5h rating, it has a limit on stored energy and with no input from the source that energy would dissipate.
To hold against a change in current flow the selfinductance would need to be infinite but since it is not then the induced magnetic field would be less and resultant CEMF would be less.
So then,, at the moment the polarity is reversed from the source there would be no impedance to current flow from the source until the prior stored energy is dissipated.

My assumption would be that since the Ideal Voltage Source has no means to dissipate nor store energy that the current flow would continue from the inductor through the Source, and at the time of the reversal of the polarity that the source would initiate a stream of charge carriers moving in the opposite direction.
Considering that at the time of a drop in voltage from the source the conductors selfinductance turns the conductor into a source, but not an ideal source, the 5h rating, it has a limit on stored energy and with no input from the source that energy would dissipate.
To hold against a change in current flow the selfinductance would need to be infinite but since it is not then the induced magnetic field would be less and resultant CEMF would be less.
So then,, at the moment the polarity is reversed from the source there would be no impedance to current flow from the source until the prior stored energy is dissipated.
And the value of that unimpeded current flow at this time would be?
Brad

When T=7s,would the below be correct?.\
Brad

With this thought experiment I am left needing to reconcile a small problem.
If the current flow is an instantaneous acceleration of even a very small mass then that would be an infinite amount of force expended and an infinite amount of energy,, even in this thought experiment that can not happen.
What if the Voltage did not accelerate the charge carriers but rather moved the "cross sectional slice" that is used to view the current flow.
The Voltage then only sets the reference frame of observation and all other parts are already in existence and it becomes a time rate of change of the reference frame,, since that has no mass nor force it has no energy of its own,, therefore instantaneous change is allowed.

Try looking at "abstraction" in Wiki and then it might make more sense.
John.

How John?

tinman,
Seems there's just too many floating variables to nail down any of the 'weird' properties that show up at the extremes of an inductor.
In retrospect, you should have asked about an ideal capacitor, and the paradoxical properties of 'displacement current' ;D

How John?
Perhaps like this:
http://www.oxforddictionaries.com/definition/english/abstraction
The process of considering something independently of its associations or attributes.
"The question cannot be considered in abstraction from the historical context in which it was raised."
"When, for instance, we claim that water can freeze, we consider water simply as such, in abstraction from the conditions in which any given amount of water finds itself."

Perhaps like this:
Thank you MH,,
Now can you suggest how this is any different than using ideals?

Try looking at "abstraction" in Wiki and then it might make more sense.
John.
With all these wonderful pearls of wisdom that you have laid before us John,,
I was wondering if you could put them all together for us to see the pretty necklace,, you know the kind,, with a start, a middle and an end?

Brad:
Your postings #543, 544, 546, 547, 552, 553, 554 and 555 are all just useless trash talk. Poor Brad is having difficulty understanding what an ideal voltage source is and how it works and he can't wrap his mind around a few concepts that are associated with the abstract concept of an ideal voltage source.
So we have to endure yet another ridiculous pushback and a whole litany of what are basically nonsensical statements from you. You are trying to attach real and tangible properties to what is just an abstraction. You are back trying to force pieces of a jigsaw puzzle together that don't fit because apparently that's the only way your brain can process it, or, it's an obstinate refusal to learn.
A discussion about a simple circuit that has an ideal voltage source in it should be about the operation of the circuit, not about how an ideal voltage source works. You are just causing problems where there are no problems and in that sense your behaviour is the same as Wattsup's behaviour where he is freaking out about the very simple and straightforward question.
So that's eight postings that should never have even been made because they are all useless and don't deal with the question, as well as mostly being useless trash talk.
From the very start, there should have been no disagreement at all about what an ideal voltage source is, and how it works. You have to do better than that, and it's time to move on.
MileHigh

I cannot agree to the ideal voltage source being able to absorb energy if it cannot contain it.
If any energy is absorbed by the ideal voltage source,and that ideal voltage source dose not dissipate energy,then the energy must now be contained in it(the ideal voltage source).
I cannot agree with MHs statement that the energy is just gonedisappears.
MHs question consists of two components onlythe ideal voltage source,and the ideal inductor. He has made this apparent many times when he has stated that EMJ and Wattsup could not answer a simple question that involved on two components.
MH also states that the ideal voltage source can deliver energy,but dose not contain energy.
At T=5s, what would the solution below read?
And then again at T=7s,what would the solution below read?.
Brad
No Brad, we are asking you for what would be happening in the circuit at t=5 seconds and t=7 seconds. And if you can't put numbers to it yet, we are asking you to tell us what you think should be happening and why it should be happening. That is the whole point of this exercise, to learn how the circuit works and to understand the concepts and to demonstrate competency.
For example, let's back up for a second. Look at the first three seconds where you have been given the answer. Please explain why there is a linear ramp of increasing current up to 2.4 amps. Why is that? How do you explain it? This is what this thread is all about, and I don't think you have even explained the answer to the first part of the question that has been given to you. Can you explain the ramp up in current to us in your own words? Why is it 2.4 amps at the end? Why isn't it two amps or three amps?
It's too bad your peers have dropped out, I guess they don't know themselves and aren't comfortable trying to brainstorm with you. The only brainstorming I can recall was the usual pieinthesky brainstorming talk about electrons and electron drift and all that jazz that had nothing to do with the question.
It would be great if you met those two goals that I have posted several times.
MileHigh

At T=5s, what would the solution below read?
And then again at T=7s,what would the solution below read?.
I did not follow the discussion and I am not sure what the values for the inductance (L), exciting voltage(V) or resistance (R) are supposed to be in this case. Without these values I cannot make the calculation.
Anyway, whatever these values are, I would use the first formula if R>0 and the second formula if R=0 to calculate the evolution of the current in time i_{(t)} flowing through the inductor. This is because the first formula degenerates into the second formula when R approaches 0 Ohms.
P.S.
These formulas are accurate only when the inductance (L), exciting voltage (V) and resistance (R) do not vary in time. If they do, than much more complicated formulas would need to be used.
But if there is only an initial current I_{0} already flowing at the time t_{0}, then its value should be simply added to the result.

Verpies:
Thanks for the two formulas. Way back earlier in this thread I told Brad you use the appropriate formula for the appropriate circuit. If there is no resistor in the circuit then you don't work with a formula that includes an undefined L/R (or V/R) of infinity. Of course with Brad that was another battle.
The formulas are fine but of course, and just to repeat myself, the purpose of this thread is for Brad and his peers to actually understand the how and the why, and demonstrate full competency with respect to this subject matter. As we know, there is a potential downside in providing formulas because then you rely on the formulas only without thinking.
MileHigh

As we know, there is a potential downside in providing formulas because then you rely on the formulas only without thinking.
Of course. Conceptual understanding should always precede mathematical analysis.
Sadly, often this is not the case.

So my question remainswhat happens to the stored energy (during the 0 volt part of the cycle),when the negative 3 volt part of the cycle start's?
1MHs theory>it just disappearsenergy is destroyed.
2it is stored in the ideal voltage source?
3it is absorbed by the ideal voltage source that cannot dissipate power=stored in the ideal voltage source.
Yes, 3 would be the answer. And you reaffirmed that you can't agree with that.
So to put it simple,how can a component that absorbs energy,but cant dissipate energy,not store that energy?.
You agree that during the period when the voltage source is 0V, it looks essentially like a short across the inductor? Why does the current flatline during this period?
You agree that during the period the voltage source is 4V, the current ramps up? Why does it do that? What would happen if the startng voltage was 4V rather than +4V?
From this you may be able to deduce, at least in concept, what the current does when the voltage source reverses polarity from the initial voltage.

the purpose of this thread is for Brad and his peers to actually understand the how and the why
As far as the selfinductance impeding current flow there is not much to understand.
You set L, V and T,, so run the numbers.
However, there are other things that may also be looked at when the ideal voltage source is in use with a conductor with 0 resistance and 0 capacitance.
This aspect of abstraction is just as valid.
Then also taking the ideal and converting it into real to make sure that there are no violations of known things,, that is that if a view precludes things that we know how to do and use then that view is in error,, is prudent.
When in the course of taking all of these possibly absurd relationships there is something that pops out that is not the same old same old and still allows for what we know and understand then that is worthy of discussion.
Can you have a bidirectional current flow,, I see bidirectional electrical signals all the time while watching my cable TV.

Miles, your conceptualization of an Ideal Voltage Source
reminds one of the demonstrations of Daniel Pomerleau. ???
The Sources of electrical energy in his demonstrations
are simply coils of wire; yet, when Daniel is in communion
with the True Source electrical energy mysteriously appears
and it actually powers whatever devices are connected to
Daniel's coils of wire. ???
In his (Daniel's) case the Source isn't really a Source but more
akin to a Conduit. A conduit which conveys the energy from a
real source somewhere outside our physical dimension. :o
Apparently something has changed in how Ideal Voltage Sources
are taught in today's World. If, that is, your conceptualization is
characteristic of what is taught these days. ::)

oops???
The CW current flow builds a field that tries to make a CCW flow,, the collapsing field tries to make a CW flow,,
The CCW current flow builds a field that tries to make a CW flow,,
So then the selfinductance would add to the collapsing field and would not cancel. ???

Yes, 3 would be the answer.
3it is absorbed by the ideal voltage source that cannot dissipate power=stored in the ideal voltage source.
Thank you.So it is stored in the voltage source,and not just disappears as MH say's. But of course,we knew that was a ridiculous statement anyway.
And you reaffirmed that you can't agree with that.
That is correct,and i draw your attention to verpies comment below.
These formulas are accurate only when the inductance (L), exciting voltage (V) and resistance (R) do not vary in time. If they do, than much more complicated formulas would need to be used.
As at T=5s,the ideal coil loop has a steady DC current flowing through it. At this point in time,there is no induction taking place. At T=5s,a negative 3 volts is placed across the coil. How is the L value of that coil changed when the now negative EMF first has to pull down the existing current flow to a value of 0,before the opposite current flow starts to flow?.
You agree that during the period when the voltage source is 0V, it looks essentially like a short across the inductor? Why does the current flatline during this period?
It flat lines at 2.3 amps because there is no impedance to the current flow during the 0 volt time period,and so it retains that 2.4 amp current flow reached at T=3 seconds. Since both the coil and voltage source are ideal,then no power is dissipated,and so the current flow remains a constant until T=5s.
You agree that during the period the voltage source is 4V, the current ramps up? Why does it do that? What would happen if the startng voltage was 4V rather than +4V?
The current ramps up because the CEMF is reducing,due to the magnetic fields change in time decreasing. If the applied voltage was inverted(negative),then the current would flow in the opposite direction through the current loop.
From this you may be able to deduce, at least in concept, what the current does when the voltage source reverses polarity from the initial voltage.
Yes. It would first have to stop the current flow that already exist in the loop,before it could start to flow in the opposite direction. Being an ideal voltage from an ideal source,it would want to instantly collapse the existing magnetic field around the coil to a non existent state to stop the existing current flow,so as it can start to induce the reversed current flow into that coil.
So what effect dose this have on how the EMF sees the L value of the inductor?.
Brad

If we are to take things as being !normal!,and the inductance value seen by the EMF is not changed due to the existing current flow at T=5 seconds,then i have plotted my current trace below for the entire cycle.
I would first ask that MH say whether it is correct or not,before anyone else says whether or not it is correct.
Brad

Brad:
The request is that you explain how you arrived at that curve and that you demonstrate that you understand the principles at play and you know what you are talking about.
I am getting too much of a Jiffy Pop popcorn vibe due to the seemingly instant magical appearance of that waveform. Please explain the whole waveform.
MileHigh

Brad:
The request is that you explain how you arrived at that curve and that you demonstrate that you understand the principles at play and you know what you are talking about.
I am getting too much of a Jiffy Pop popcorn vibe due to the seemingly instant magical appearance of that waveform. Please explain the whole waveform.
MileHigh
MH:
If they don't have Koolaid in OZ then I doubt they have Jiffy Pop either. Man, what a great product that was. I think you can still buy it in some places. What ever happened to Fizzies? Remember them? Except for a ref. to them in the movie Animal House in 1976, I have never seen nor heard of them. I used to love those things.
Bill

I had no idea that Jiffy Pop was not really available anymore. I am devastated. I don't know what Fizzies are.

I had no idea that Jiffy Pop was not really available anymore. I am devastated. I don't know what Fizzies are.
https://en.wikipedia.org/wiki/Fizzies (https://en.wikipedia.org/wiki/Fizzies)
They were like little tablets in a foil pouch that you dropped into a glass of water and, it carbonated and flavored it to make soda! My favorite was root beer. They sort of worked like an AlkaSeltzer tablet, except with flavoring. Very big in the US in the mid 1960's. The above link says they sold twice as much as KoolAid...I didn't know that.
Bill

http://www.amazon.com/JiffyPopButterFlavoredPopcorn45Ounce/dp/B000CSKKE6

and since I am having an off day,, I keep messing things up,, started a few days ago really,, vacations can do that,,,,
I am going with
2.4
2.8
3.4

Brad:
The request is that you explain how you arrived at that curve and that you demonstrate that you understand the principles at play and you know what you are talking about.
I am getting too much of a Jiffy Pop popcorn vibe due to the seemingly instant magical appearance of that waveform. Please explain the whole waveform.
MileHigh
I have posted that waveform going on what i believe Poynt and your self are asking me to believe will happen.
Now i ask you if you believe that wave form is correct.
Once you have stated a yes or no,i will tell you how i derived at that waveform for the current.
Once that is done,we will discuss the issues i believe may change the outcome of the waveform posted.
So it is a simple yes or no MH.
Brad

The waveform is correct.

I might remind you MH,that your question asks 'what happens from T=0!
I have answered that question in way of the current trace graph,as that is about all that happensrising and falling current flows. You have already stated that i need not worry about the magnetic fields'although there value would just rise and fall with the current. As we are dealing with ideal components,then no heat or power is dissipated from the closed loop.
You question dose not ask as to how i calculated the answer,only what happens through the entire cycle at each time period.
So you have my answer,and im asking you if it is correct.
Yes or no is all thats needed.
Brad

We must thank the likes of MH. poynt etc. for their patience in inching us along.
Yep, you''''re right, present is prepared out of thank, a fansite is now running, 682 members, also including:

The waveform is correct.
I am getting too much of a Jiffy Pop popcorn vibe due to the seemingly instant magical appearance of that waveform. Please explain the whole waveform.
There wasnt any Jiffy Pop (what ever that is) popcorn moment MH. As i stated to you some time back,a generic calculated answer was quite simplerefer to post 569,where i posted the formula used from T=5 seconds to T7 second's,as also posted below.We just subtract or add the calculated value to that of the previous valuedepending on what polarity you stated in your question. Then it's just a matter of playingjoin the dots. But my argument was and still is,that is not the correct answer/current trace when using ideal's.
The 2.4 amps flowing through the loop at T=5 seconds,will be an impedance against the current that is induced at T=5 seconds,that is of the opposite polarity. The calculated current peak value at T=7 seconds,is calculated based around the inductance value,time and a starting current of 0>I=Io. As the polarity is now reversed,that current value is actually I=I2.4amps.
So i am not sure that simply subtracting(due to it being the negative voltage part of the cycle) the calculated peak current value at T=7 seconds from the value at T=5(2.4 amps) seconds is correct,as the impedance value of the coil at T=5 seconds is higher than would normally be seen by the current flow,should the current start with a value of 0.
Brad

Well, you are seemingly making progress, and I am going to ignore the various errors in what you posted because what you are saying is more important in the "error glitches" in your posting.
But I have a question for you: Way back very early in the thread I posted that formula and told you that it was basically the short answer to the question. When I did that you went into a small frenzy and you insulted me like a drunk sailor. So what gives, how do you explain using the formula now when before when I showed you the very same formula you insulted me repeatedly like a drunk sailor?
Don't even bother saying that you don't drink, it's just an expression.

Well, you are seemingly making progress,
But I have a question for you: Way back very early in the thread I posted that formula and told you that it was basically the short answer to the question. When I did that you went into a small frenzy and you insulted me like a drunk sailor. So what gives, how do you explain using the formula now when before when I showed you the very same formula you insulted me repeatedly like a drunk sailor?
Im not sure if you read everything i write or post,but you seem to keep missing the fact that i do not agree with the results of your formula,and believe the values are wrong.
I told you right from the start,that using your generic formula would be an easy way out,but the results would not be that that your formula would show.
and I am going to ignore the various errors in what you posted because what you are saying is more important in the "error glitches" in your posting.
How is it i made errors,when the result is correctas far as your concerned ?.
The method i used seemed to be correct,as the result was what you were looking for.
I do not believe the current value reached at T=7 seconds is correct.
I am not sure that it should be I=I 2.4  1/5 integral V dt.
I believe as verpies said,there is a higher order of math required here.
The 2.4 amps that is flowing opposite to that of the current to be induced by the negative value between T=5 and T=7,will be seen as a high impedance to that current to be induced during the negative portion of the cycle. This impedance value is greater than that which would be encountered by the current flow during the negative voltage part of the cycle to that of if the existing current of 2.4 amps was not there.
Our calculated current for T=7 seconds is 3/5 x 2=1.2 amps. This is calculated on the assumption that I=0 at the time of the calculation. But we have an I value of 2.4 amps of the opposite polarity,not a value of I=0.
As i said,i do not believe that subtracting that 1.2 amps from the negative voltage phase,from the previous 2.4 amp value is the correct way to do this,and gives a wrong answer when done like this.
The inductance value seen by the EMF during the negative phase,must be much lower due to the existing steady state current flow that is apposing the current to be induced during the negative voltage part of the cycle. The other problem is,there is no give in the voltage being applied across that coil loop,in that no matter what the load,the ideal voltage source will apply that full selected voltage across the coil. To me,this would be like trying to stop a spinning flywheel instantly.
Perhaps verpies and/or Poynt might think about that a bit,and see if it is as straight forward as some think it is.
And MH>we need to call a truce here,as all this bickering is doing no one any good.
At the end of the day,we should all be here for the same reason,and that being they very name and nature of the forum.
I am not blaming it all on you ,as i know i have given as good as i have gotten.
Perhaps when we disagree ,we can just agree to disagree.
Don't even bother saying that you don't drink, it's just an expression.
I gathered that. The truth is,i am having a bourbon and coke right nowjust one,for a mate(we call friends mates over herejust in case you take the word !mate! the wrong way) that passed away early this morning.
Brad

Perhaps verpies and/or Poynt might think about that a bit,and see if it is as straight forward as some think it is.
My question to you Brad is what do you think will happen? In other words, please draw out the current wave form according to your theory of how the voltage source and inductor would interact.

Brad:
I am going to ask you the question again:
Way back very early in the thread I posted that formula and told you that it was basically the short answer to the question. When I did that you went into a small frenzy and you insulted me like a drunk sailor. So what gives, how do you explain using the formula now when before when I showed you the very same formula you insulted me repeatedly like a drunk sailor?
MileHigh

My question to you Brad is what do you think will happen? In other words, please draw out the current wave form according to your theory of how the voltage source and inductor would interact.
As i said,it's not so easy to work out when dealing with a voltage source and coil that form a loop that current can flow through unimpeded.
What happens when two 12 volt batteries are looped in series?.
We now have two voltage sources where a voltage cannot be measured at the terminal's,but we have lot's of current flowing through them. How do you measure that current flow if the series loop resistance is 0 ohms?.
i dont think there is much point in posting my thought's,as it seems you and MH have made up your mind's,and anything i say will just be dismissed anyway.
As MH has said that my current values and the trace i have drawn out are correct,and that you have agreed with MH most of the way through this topic,then it is safe to say that you also agree that my provided values,times and current trace are also correct,as far as your concerned.
So is there any real need to go on?.
Perhaps you could show us how you would calculate the peak current at T=7 second's?.
What is the formula and values you would use?.
Brad

i dont think there is much point in posting my thought's,as it seems you and MH have made up your mind's,and anything i say will just be dismissed anyway.
As MH has said that my current values and the trace i have drawn out are correct,and that you have agreed with MH most of the way through this topic,then it is safe to say that you also agree that my provided values,times and current trace are also correct,as far as your concerned.
So is there any real need to go on?.
The point of asking you to draw out the wave form not so we can just dismiss it, it is to see how your thinking has caused you to arrive at your theory, and to see how to steer you toward the correct understanding. Yes of course your wave form trace is correct. It is not MH's way btw, it is just how it is, it is physics and it is reality (even with an ideal or nearideal inductor).
Perhaps you could show us how you would calculate the peak current at T=7 second's?.
What is the formula and values you would use?.
Brad
I would use the formula MH posted. But I prefer to use the sim, which btw, fully supports the formula and the graph you have drawn out.

TinMan extends a Kind hand in a peaceful gesture to move forward!
MileHigh wants more Blood ,and curls The Lip?

Brad:
I am going to ask you the question again:
Way back very early in the thread I posted that formula and told you that it was basically the short answer to the question. When I did that you went into a small frenzy and you insulted me like a drunk sailor. So what gives, how do you explain using the formula now when before when I showed you the very same formula you insulted me repeatedly like a drunk sailor?
MileHigh
OK MH
I will try and explain this one more time.
1stAs i said,i was giving back as good as i was getting,as far as your insulting go's.
I would like to point you toward post two,the very second post on this thread,and your very first post on this thread>quote: You are one strange egg Brad because you think you are "running the show" now but in fact the show is running you.
1. Brad gets up the learning curve and understands the original question and then answers it correctly all by himself and clearly demonstrates that he understands what he is doing.
2. Brad admits that he is wrong when he stated that my response to the harder question is wrong.
You will see my first post saidPlease keep the insults down,and the language clean.
And my last line in that post was>For the record,could you please post your answer to your question above?
As you can see,i asked for insults to be kept down,and also asked you very politely to post your answer. So i think you can see from just the first two posts on this thread,who wanted it to run smooth without insults,and who was the first to cast the first insultand it just escalated from there.
Perhaps you should go and read the first page of the thread,and see who was demeaning to who.
2nd As i have already tried to explain to you,i do not agree that the formula being used is correct from T=5 seconds to T=7 seconds. I only used your formula to show you that i am quite capable of dealing with generic formulas,and calculating generic results. I told you i could do this very early in the threadperhaps you thought i could not. All the information to do so is on the net,and i can learn very quickly.
But once again,i only gave you what you wanted to see,and i gave it correctly.
But as i have said on more than one occasion now,this dose not mean that i believe them to be correct in this situation. In a non ideal situation,this collision of energy would be burnt of as heat,via the internal resistance of the voltage source and the coil it self. But being that we are in an ideal situation,this collision energy cannot be dissipated as heat,as there is no series resistance in the circuit.
Brad

The point of asking you to draw out the wave form not so we can just dismiss it, it is to see how your thinking has caused you to arrive at your theory, and to see how to steer you toward the correct understanding. Yes of course your wave form trace is correct. It is not MH's way btw, it is just how it is, it is physics and it is reality (even with an ideal or nearideal inductor).
I would use the formula MH posted. But I prefer to use the sim, which btw, fully supports the formula and the graph you have drawn out.
As i said,you are not interested in what i have to say,as you have stated that you wish to !!steer!! me in the right direction.
I have already stated my concerns,and whyseveral times now,but maybe you missed them all.
How will the steady state current flow of 2.4 amps effect the seen inductance value by the EMF that wishes to induce a current flow in the opposite direction at T=5 second's?. The reason i ask this,is because that current flow of 2.4 amps,will be seen as a high value impedance to the current trying to be induced during the negative voltage part of the cycle. At the moment,the current value for T=7 seconds, is being calculated on the assumption that the inductance value seen by the now negative EMF is still 5H. I am only asking if this higher impedance value,due to the !already flowing! 2.4 amps of current acting against the current that the negative EMF is trying to induce,will alter the inductance value seen by that negative EMF.
To me it seems that the EMF will see a different inductance value,and there for a different impedance value,when the coil has no current flowing through it at T=0, to that when the current flowing through it is 2.4 amps,but of opposite polarity.
To make this clear,please see diagrams below to see the difference that i am trying to explain.
Brad

Would not the current that is still flowing around the system and the supported magnetic field try and maintain the 4V that initiated it? The voltage becomes imaginary because you would not be able to measure it.
So then would not the change in polarity of 3V then be applied against that stored potential so that the 1.2 change is offset by the 1.6 stored over the same time leaving the 1.6 and an imaginary 1V

@tinman
At least I agree with you the the wave form should not be like that but maybe for another reason.
For me, ideal voltage has zero resistance, ideal inductor has zero resistance so at 4 seconds when the voltage suddenly drops to 0 volts, what is 0  0  at 0 is just like an inductor being disconnected and what happens when an inductor is disconnected especially at 5H, should create a good discharge of the 2.4 * 4 = 8.8 watts. Some will hit back into the ideal voltage but most will just dissipate. At least in real life if you had a 5H coil in your hand changed at 4 volts and 2.4 amps and if you just turned off the power supply, good luck.
So each zero point for me is like an inductive discharge.
Again what is missing is data is the question. We have to suppose to many things. Is it DC ideal voltage. If so then the applied negative is always at zero volts and the positive is the one changing voltage at t0 and at t4 and t6, but then at t6 to t8 both polarities are switched for the applied 3 while the volt meter stays were it is, so you have a different effect because both ends of the ideal voltage are changed.
Anyways, I guess the answer has to be as lazy as the question.
Or, you have to do like 1000 EE students probably have to do when they are confronted with this question is to just play the game, don't break your head, say yes yes yes and just pass. hahahaha
wattsup

As i said,you are not interested in what i have to say,as you have stated that you wish to !!steer!! me in the right direction.
I have already stated my concerns,and whyseveral times now,but maybe you missed them all.
How will the steady state current flow of 2.4 amps effect the seen inductance value by the EMF that wishes to induce a current flow in the opposite direction at T=5 second's?. The reason i ask this,is because that current flow of 2.4 amps,will be seen as a high value impedance to the current trying to be induced during the negative voltage part of the cycle. At the moment,the current value for T=7 seconds, is being calculated on the assumption that the inductance value seen by the now negative EMF is still 5H. I am only asking if this higher impedance value,due to the !already flowing! 2.4 amps of current acting against the current that the negative EMF is trying to induce,will alter the inductance value seen by that negative EMF.
To me it seems that the EMF will see a different inductance value,and there for a different impedance value,when the coil has no current flowing through it at T=0, to that when the current flowing through it is 2.4 amps,but of opposite polarity.
To make this clear,please see diagrams below to see the difference that i am trying to explain.
Brad
Brad, the honest truth is that I don't understand your explanation nor your questions. I can't make sense of them. So rather than spending my time trying to understand your theory, I asked you to plot out your current trace according to your understanding, which gives me instant insight into your thinking. Then I might have a chance of steering you in the right direction. But you state that you can not plot out the current according to your theory, while at the same time, rejecting the present accepted theory which is not only predicted by equation and simulation, but makes sense from the conceptual point of view. I think if you are going to reject a theory, you should have one the you feel is the correct one, and be able to explain it and prove that the existing one is incorrect. So far I have not seen this from you.

Is there a magnetic field around the inductor at t=5
I ask because the ideal voltage source has no impedance, no resistance, no reactance and no capacitance and so it offers nothing to slow down the current flow that is existing.
If I apply a resistance to the current flow what will that magnetic field reaction be with the inductor?
If after t=3 the voltage is set to 4V for another 3 then at t=6 there would be no current flow,, right?
So at t=5 and the 3V is applied then there is no current flow "from" the ideal voltage source but the existing flow continues "through" the ideal voltage source with a resistance of some sort that is slowing down the current flow.
So the brakes are applied by the ideal voltage source but the inductors existing magnetic field mashes the throttle and tries to maintain the 4V???

...like a drunk[en] sailor?
Aye, to you LandLubbers it is just an expression. ;)
To those of us who crewed the "haze gray and underway"
men o' war it was a way of life. Extended periods at sea
develop quite a thirst in the Sailor and when finally in
port to enjoy some "Liberty" sipping the brew was a joyous
relief. ;D
Naturally, too much of the brew leads to some really
odd behaviors which the Sailor is naturally inclined to
indulge in. Hence, the expression of the LandLubbers. 8)
Sailors are not offended by the expression. :)

2nd As i have already tried to explain to you,i do not agree that the formula being used is correct from T=5 seconds to T=7 seconds.
Brad
Please tell us in your own words what the formula means.

I have taken the liberty of changing MH's original problem sightly so there is now a period following T3 to T5 from T5 to T8 with 4 volts applied to the 5H inductor. This is equal to but opposite the initial T0 to T3 so we can now see what happens to the inductor current.
One might find it easier to visualize the magnetic field around the inductor and how it varies in relation to the current. I've included an algebraic proof of this relationship.
𝑬𝒎𝒇 =  𝑵 × 𝚫𝚽/𝚫𝒕 (Faraday’s Law)
𝑬𝒎𝒇 =  𝑳 × 𝚫𝑰∕𝚫𝒕 (Inductance defined in terms of Emf)
∴ 𝑵 × 𝚫𝚽/𝚫𝒕 = 𝑳 × 𝚫𝑰∕𝚫𝒕 (Substitution)
∴ 𝑵 × 𝚫𝚽 = 𝑳 × 𝚫𝑰 (Reduce)
∴ 𝑳 = 𝑵 × 𝚫𝚽∕𝚫𝑰 (Inductance in terms of flux)
∴ 𝚫𝚽 = 𝑳 × 𝚫𝑰 (Flux over time verses inductance and current over time with 1 turn)
partzman
edit

Thank you Partzman,
So the input to spin up the flywheel is the same as is required to stop it.

Thank you Partzman,
So the input to spin up the flywheel is the same as is required to stop it.
Yes, in that analogy you are correct.
partzman

In spirit of the thread lets consider an ideal ferromagnetic as a break from the ideal voltage sources.
What properties would such material have?
For sure it wold not exhibit any hysteresis loss and it would not heat up at all when immersed in a changing magnetic flux.
It would not conduct electric current at all so Eddy currents could not form in it.
Would it have infinite permeability?  I don't think so, anymore than an ideal inductor needs to have an infinite inductance.
What are your thoughts?
What other properties would an ideal ferromagnetic have?

Brad, the honest truth is that I don't understand your explanation nor your questions. I can't make sense of them. So rather than spending my time trying to understand your theory, I asked you to plot out your current trace according to your understanding, which gives me instant insight into your thinking. Then I might have a chance of steering you in the right direction. But you state that you can not plot out the current according to your theory, while at the same time, rejecting the present accepted theory which is not only predicted by equation and simulation, but makes sense from the conceptual point of view. I think if you are going to reject a theory, you should have one the you feel is the correct one, and be able to explain it and prove that the existing one is incorrect. So far I have not seen this from you.
The inductance value is being used to calculate the peak current at T=7 seconds,but the induction process to make this calculation is not correct between the time period T=5 seconds to T=7 seconds. The inductance value being used to make this calculation is not correct. We have started at T=5 seconds,and ended at T=7 seconds to calculate the peak current reached at T=7 seconds. This is on the understanding that the induction process taking place at this time will start with no current flowing through the coil,and no existing magnetic field. We apply our voltage,and current starts to flow,and a magnetic field begins to build(induction). We calculate that using a 5H coil,the peak current value reached at T= 7 seconds will be 1.2 amp's. But as i said,this assumes that there is no current flowing through the coil. This assumes that a current will begin to flow,a magnetic field will begin to form,and the CEMF value will fall as the magnetic fields change in time reduces.
But what we actually have happening here is the opposite.
A current flow and magnetic field already exist .
The current dose the exact opposite to what the calculations we are using say it should.
The current dose not start from 0 and rise to 1.2 amps,it starts from 2.4 amps and falls.
The CEMF produced from this !once again! magnetic field that is changing in time,is now the same as the EMF that created it,and dose not appose it as it should the way we have made the calculation for the time period of T=5 seconds to T=7 seconds.
So we have used a calculation to gain a peak current value at T=7 seconds, that go's on the premise that there will be a Counter EMFthis give us our current rise over time value.
But what we actually have is a current value drop over time,not a rise,and no CEMF,but an EMF working with the induced EMF.
So it would seem to me,that the formula used to calculate the peak current value at T7 seconds,is not the correct formula to use.
If we have no CEMF but an EMF that is not apposing the inducing EMF,then there is no impedance/resistance to to the induced current flow.
So it makes sense that the current flow would sky rocket instantly,as there is no resistance/impedance to slow or stop it.
Brad

Brad:
Let's back up and discuss the inductor impedance over the first three seconds. We all know the drill now, the five Henry inductor makes contact with the ideal voltage source, and for the first three seconds the voltage is four volts.
What is the impedance of the inductor over the fist three seconds?
MileHigh

I have taken the liberty of changing MH's original problem sightly so there is now a period following T3 to T5 from T5 to T8 with 4 volts applied to the 5H inductor. This is equal to but opposite the initial T0 to T3 so we can now see what happens to the inductor current.
One might find it easier to visualize the magnetic field around the inductor and how it varies in relation to the current. I've included an algebraic proof of this relationship.
𝑬𝒎𝒇 =  𝑵 × 𝚫𝚽/𝚫𝒕 (Faraday’s Law)
𝑬𝒎𝒇 =  𝑳 × 𝚫𝑰∕𝚫𝒕 (Inductance defined in terms of Emf)
∴ 𝑵 × 𝚫𝚽/𝚫𝒕 = 𝑳 × 𝚫𝑰∕𝚫𝒕 (Substitution)
∴ 𝑵 × 𝚫𝚽 = 𝑳 × 𝚫𝑰 (Reduce)
∴ 𝑳 = 𝑵 × 𝚫𝚽∕𝚫𝑰 (Inductance in terms of flux)
∴ 𝚫𝚽 = 𝑳 × 𝚫𝑰 (Flux over time verses inductance and current over time with 1 turn)
partzman
Thank you, excellent and very informative and enlightening post, especially the new voltage and current vs. time graph.
MileHigh

In spirit of the thread lets consider an ideal ferromagnetic as a break from the ideal voltage sources.
What properties would such material have?
For sure it wold not exhibit any hysteresis loss and it would not heat up at all when immersed in a changing magnetic flux.
It would not conduct electric current at all so Eddy currents could not form in it.
Would it have infinite permeability?  I don't think so, anymore than an ideal inductor needs to have an infinite inductance.
What are your thoughts?
What other properties would an ideal ferromagnetic have?
I understand fully. But its just pretend world. Anything can be whatever one may want it to be.
Ive been delving back into the resonance projects. Its not pretend. As Ive read over things here, Ive lost interest. Was thinking, as you just did above, of all the hows and whys of it all and it just plain isnt required as the rules are set. So pretty much all the ideal components and equipment are virtually limitless. Pioneer used to make a set of 6x9 speakers back in the early 80s. The magnets were bigger than any other that have been made. The max power handling was labeled 'Unlimited'. It wasnt that they were not telling the truth really. There just wasnt any car amplifiers in those days that exceeded or even came close to the real power handling of the speakers. So in that world back then, they were unlimited in power handling. ;) They were ideal in that world. ;D
Mags

Brad:
Let's back up and discuss the inductor impedance over the first three seconds. We all know the drill now, the five Henry inductor makes contact with the ideal voltage source, and for the first three seconds the voltage is four volts.
What is the impedance of the inductor over the fist three seconds?
MileHigh
Well,as the impedance is just a resistance to the current that wants to flow,is it correct to say that the impedance is a result of the CEMF?,as we know,there is no ohmic resistance.

Methinks a perfect inductor has reactance without resistance. Thus the REAL component
of its impedance would be ZERO.
Has anyone else any trouble with this?
John.

Has anyone else any trouble with this?
Not me.

I understand fully. But its just pretend world. Anything can be whatever one may want it to be.
Not fully because the ideal components that we've been taking about here lately, still have restrictions placed on them even if they do not exist in reality. Thus an ideal voltage source must have zero impedance and an ideal inductor must have zero resistance and capacitance. So it is not anything whatever one may want it to be.
Considering ideal components unburdens the thinker from analyzing their imperfections and allows for clarity of their modeling. Thus such components have great conceptual utility.
I think that an ideal ferromagnetic will still have the S shaped BH curve, albeit without hysteresis and coercivity.
However its permeability would be useradjustable just like the voltage of an ideal voltage source.
What do you think?

Not me.
Are you saying that an ideal inductor has no impedance in reference to MHs question?.
I would think that anything that acts against a current flow,could be seen as an impedance to that current flow. Impedance is just a form of resistance is it not?.
Brad

Are you saying that an ideal inductor has no impedance in reference to MHs question?.
I would think that anything that acts against a current flow,could be seen as an impedance to that current flow. Impedance is just a form of resistance is it not?.
Brad
verpies was responding to minnie's statement about an ideal inductor having reactance, but 0 resistance.
Reactance is the imaginary part of the impedance equation, so no, verpies is not saying that the inductor has no impedance. The impedance however in this case is purely reactive.
MH's question is a good one; what is the inductor's impedance over the first 3 seconds?
Yes, impedance is a form of resistance to current flow. That should give you a hint how to answer the above question.

The inductance value is being used to calculate the peak current at T=7 seconds,but the induction process to make this calculation is not correct between the time period T=5 seconds to T=7 seconds. The inductance value being used to make this calculation is not correct. We have started at T=5 seconds,and ended at T=7 seconds to calculate the peak current reached at T=7 seconds. This is on the understanding that the induction process taking place at this time will start with no current flowing through the coil,and no existing magnetic field. We apply our voltage,and current starts to flow,and a magnetic field begins to build(induction). We calculate that using a 5H coil,the peak current value reached at T= 7 seconds will be 1.2 amp's. But as i said,this assumes that there is no current flowing through the coil. This assumes that a current will begin to flow,a magnetic field will begin to form,and the CEMF value will fall as the magnetic fields change in time reduces.
Where and how did you arrive at the notion that at T=5s or T=7s, that no current is flowing?

Iend=Istart+(1/L)*vt
Iend=2.4+(1/5)*(3*2)
The inductance value is nonpolar,, that is the sign of the Voltage is either used to make it
Iend=2.4(1/5)*(3*2)
or
Iend=2.4+(1/5)*(3*2)
Both leave you with Iend=1.2
If you use
Iend=2.4(1/5)*(3*2)
Then you get Iend=3.6
Which I think is in error due to the double use of the negative sign which turns the negative inductance into a positive one.
However,, I am not sure that this is correct for the ideal condition and is why I am thinking about the mechanism of this interaction.

Where and how did you arrive at the notion that at T=5s or T=7s, that no current is flowing?
Ok,i give up.
I thought i had explained it quite clearly,but apparently not.
Im not saying that there was no current flowing at T=5 second'sim saying the exact opposite.
Because there is a current flowing at T=5 second's,the inductance value that the EMF see's at T=5 seconds is not correct,due to this already flowing current.
See if this help's.
Lets say there is no current flowing through the inductor.
We apply our negative 3 volts for 2 seconds. That gives us a peak current at the end of the 2 seconds of 1.2 amps.
Now we do the same,only this time there is 2.4 amps flowing through the coil loop.
We once again place our negative 3 volts across the coil for 2 second's. We now have the very same answer as far as current value go's ,of 1.2 amps peak at the end of the 2 seconds.
How can it be the same value,when in one case there is no current flowing through the coil,and in the other case there is 2.4 amps flowing through the coil?.
In one case ,we have a rising magnetic field that will produce a counter EMF,and in the other case,we have a falling magnetic field that will produce an EMF that is not !counter! against the applied EMF.
How is it that the same formula can be used for two very different situations?.
Brad

See if this help's.
Lets say there is no current flowing through the inductor.
We apply our negative 3 volts for 2 seconds. That gives us a peak current at the end of the 2 seconds of 1.2 amps.
Now we do the same,only this time there is 2.4 amps flowing through the coil loop.
We once again place our negative 3 volts across the coil for 2 second's. We now have the very same answer as far as current value go's ,of 1.2 amps peak at the end of the 2 seconds.
How can it be the same value,when in one case there is no current flowing through the coil,and in the other case there is 2.4 amps flowing through the coil?.
Brad,
First, understand this; the equation tells us what the final value of current will be, not the "peak" current as you have been calling it. That may be part of your confusion.
Ok, with no current flowing (the beginning of the test) if we apply a 3V supply for 2s, the final current will be 1.2A, not +1.2A as you have stated. The current will ramp down from 0A to 1.2A over 2s.
Now, if we do the same and apply our 3V for 2s but this time there is already +2.4A of current flowing, then the current will ramp down from +2.4A to +1.2A, again a change of current (of 1.2A) in the negative direction.
You will note that the final current in these two cases is not nearly the same; one is 1.2A, and the other is +1.2A.

Are you saying that an ideal inductor has no impedance in reference to MHs question?.
No. I think that an ideal inductor has nonzero reactance and nonzero impedance but it has zero resistance.

Brad,
Ok, with no current flowing (the beginning of the test) if we apply a 3V supply for 2s, the final current will be 1.2A, not +1.2A as you have stated. The current will ramp down from 0A to 1.2A over 2s.
You will note that the final current in these two cases is not nearly the same; one is 1.2A, and the other is +1.2A.
First, understand this; the equation tells us what the final value of current will be, not the "peak" current as you have been calling it. That may be part of your confusion.
No,it's not my confusion at all. As i have stated a time period,then the peak current reached at the end of that time period is the final current value.
Now, if we do the same and apply our 3V for 2s but this time there is already +2.4A of current flowing, then the current will ramp down from +2.4A to +1.2A, again a change of current (of 1.2A) in the negative direction.
And that there is what i think is the problem.
Is it correct to use the same equation when the coil has a positive current already running through it,to make a calculation of the negative current value at the end of the 2 seconds,as to when that same equation is used when the coil has no current flowing through it,to make a calculation of the negative current at the end of the two seconds.
The coil is in a different state when it has current already flowing through it to that if it has no current flowing through it,but yet we use the equation to calculate the negative current value as if there is no current flowing through the coil.
Example 1.
The coil is open.
We apply a voltage of 3 volts(negative to keep in line with MH question only,and keep things clear)across the 5H coil for 2 seconds. At the end of the 2 seconds,the peak(final) current value will be 1.2 amps. That is 3/5 x 2=1.2. At T=0,the voltage is connected across the coil. First a voltage appears across the coil,and then a short time after,a current starts to flow,and a magnetic field starts to build. This increasing magnetic field causes a CEMF to be developed,and this CEMF is what slows the rise time of the induced current. As the magnetic fields change in time is reduced in value,the CEMF is also reduced,and so the current continues to increase over time. If there was no CEMF,then the current would go straight to it's maximum value. So this CEMF can be seen as the impedance to the current flow.
Example 2.
We are now going through the phases of MHs question.
At T=5 seconds,we apply 3 volts across the 5H coil for 2 seconds.
We once again use the same math(equation) 3/5 x 2=1.2 amps.
The difference this time being that there is already a positive current flowing through the coil,with a value of 2.4 amps. This time a magnetic field already exist,that is opposite to that of what the applied EMF wants to create. This time there will be no CEMF as the current value increases,as the current value in the coil is decreasing from T=5 seconds to T=7 seconds,not increasing. As the current is decreasing in value,the EMF across the coil during this decrease is the same as the applied EMF>no counter EMF during this 2 second time period.
If there is no CEMF,then what stops the current going straight to it's maximum value when the 3 volt EMF is applied across that coil?.
Brad

I suggest that you try again Brad because there are several mistakes in each of your examples. Perhaps if you read more about inductors and how they respond to voltage step functions you will be able to put the pieces of the puzzle together and give it another try.

Brad,
You now have everything at hand required to arrive at the full and correct understanding of the circuit current in MH's question. It has been explained, illustrated, simulated, and elaborated upon. It's now up to you.
I know you can do better and that you can "get it", if you give it a fair and persistent go.

Brad,
You now have everything at hand required to arrive at the full and correct understanding of the circuit current in MH's question. It has been explained, illustrated, simulated, and elaborated upon. It's now up to you.
I know you can do better and that you can "get it", if you give it a fair and persistent go.
I have already got it,the way you guys are seeing itas i said to MH,that is not a problem,and the question was never a problem>when dealing with generic equation's.
But i question those equation's,and the situation they are being used in.
In that we are using an ideal voltage source,and that voltage is retained under any load,then i would expect to see a large reverse(negative) current spike at T5 second'sthe instant the 3 volts is placed across the coil.
This slow drop in current from T=5 seconds to T=7 seconds,is not what i think should be the case.
We already know that if the coil became open,then we would see a polarity change in voltage,and this polarity change would then be the same as the EMF applied at T=5 seconds.
The CEMF is no different to that induced by an electric motor,only it is ass about,where the CEMF will increase with motor speed,resulting in a drop in current draw,and with the coil,the CEMF will decrease over time,resulting in a higher current draw. So an increase in CEMF is seen as an impedance to the current flow,by way of reducing the potential voltage difference between the applied EMF,and the CEMF. In the case of the inductor,the CEMF reduces over time,meaning a larger potential difference between applied EMF and CEMF,resulting in a higher current flow value.
We have no CEMF when the 3 volts is applied,due to a collapsing magnetic field that is of the opposite polarity to that of which the applied3v EMF wishes to build.
We now have an EMF+ an EMF>CEMF is gone.
Remove the CEMF from an electric motor while it's running,and what will happen to the current draw?.
Brad

I have already got it,the way you guys are seeing itas i said to MH,that is not a problem,and the question was never a problem>when dealing with generic equation's.
But i question those equation's,and the situation they are being used in.
In that we are using an ideal voltage source,and that voltage is retained under any load,then i would expect to see a large reverse(negative) current spike at T5 second'sthe instant the 3 volts is placed across the coil.
This slow drop in current from T=5 seconds to T=7 seconds,is not what i think should be the case.
We already know that if the coil became open,then we would see a polarity change in voltage,and this polarity change would then be the same as the EMF applied at T=5 seconds.
I can't see how you are "getting it" if you don't see how the equation, and sim works for all cases. So I have to disagree that you get it.
If you are expecting there to be a large reverse current spike, then you're forgetting that the current can't and won't do that in an inductor. By their very nature, inductors don't work that way, and in fact they resist any change in current. As I said, when the 3V step occurs, the inductor current will ramp in a negative direction, no matter what the starting current is. There is no "spike", and the current doesn't instantly go from +2.4A to some negative current value.

Plot the energy exchange,,
0.5*LI^2
For the current at step down points,, I used 0.1 for the step and I ran it down to 0.1
The current graph is fine looking when you do that,, just like the sim even,, but not the energy exchange.
This energy exchange is the same sort of looking thing as the cap to cap dump.
And the charge carriers can not reduce in quantity,,, and the volume of the system for the charge carriers also can not change,, fixed total number even if that is very large.
And no outsource method for heat.
This kind of makes the ideal and the formulas at odds

The difference this time being that there is already a positive current flowing through the coil,with a value of 2.4 amps. This time a magnetic field already exist,that is opposite to that of what the applied EMF wants to create. This time there will be no CEMF as the current value increases...
But the value of the CEMF does not depend on the level of magnetic flux and current flowing through an ideal coil.
The CEMF depends on the rate of change of the flux and current.

Kinetic energy. Flywheel.
If you've got a decent sized flywheel it's not easy to get a spike.
Forget about infinite energy in infinite time, just deal with the
relatively short term.
I have struggled because the whole thing isn't intuitive with the
limited knowledge of inductors which I had.
MH's. question was a hell of a good one, very thought provoking.
John.

Yes John,,
Now picture spinning the flywheel up using compressed air,, now picture slowing it back down using compressed air,, just reverse the direction of the air flow right?

Could well do.
I think the speed of light would put a limit on infinity.

You can do many things that bring things back into equal,, but that is adding in the mechanism that the formulas "as used" with these ideals do not have.
Say if you use pneumatic rams,, then the rams connect the air tank to the flywheel and you have the mechanism,, same thing if you use springs,, or if you use charge separation within the ideal voltage supply <= but it does not have that since it is ideal.
Now what if you are using rams,, what would happen to the pressure inside the ram that is going to stop the flywheel?
Well that depends doesn't it,, an infinite size air tank would not see a pressure change and there you go,, however, there is the adding in of the rams and pipes. <= a mechanism
If you have a limited air tank size then there will be a decrease in pressure as you are spinning up the flywheel and an increase in pressure when slowing it back down.
So using the ideals then we need a mechanism in place that itself does not consume energy but makes the transfer of the potentials equal,, same quantity transferred as stored and then as returned.
Without it the ideal voltage source is only blowing air,, 14.4J worth to spin it up and then another 14.4J worth to stop it,, and like I have said,,, even in this ideal space that can not happen.

picture spinning the flywheel up using compressed air
That would be an accurate analogy of an ideal voltage source and inductor, if the flywheel could also push the compressed air back in.

Exactly.
What is that mechanism?
It has to be there but it is not described by the formulas.
It is like somehow the "SYSTEM" changes
System meaning all of the parts,, the ideal source the ideal inductor the ideal voltage the ideal current the ideal flux,, all of it

Yes John,,
Now picture spinning the flywheel up using compressed air,, now picture slowing it back down using compressed air,, just reverse the direction of the air flow right?
The flywheel analogy makes it pretty damn easy to answer the question. However, I have mentioned the flywheel analogy perhaps 30 times in the past, and I wasn't going to mention it again. This time I switched to a shopping cart analogy, which is just as easy, but it did not hit home and apparently did not register.
Yes reversing the direction of the air flow is perfectly valid. But you did not mention an important thing. As the flywheel spins up in either direction, the speed of the air flow has to keep on increasing proportional to the speed of the flywheel.

Kinetic energy. Flywheel.
If you've got a decent sized flywheel it's not easy to get a spike.
Forget about infinite energy in infinite time, just deal with the
relatively short term.
I have struggled because the whole thing isn't intuitive with the
limited knowledge of inductors which I had.
MH's. question was a hell of a good one, very thought provoking.
John.
Thanks John, but from my perspective this has still not reached a conclusion, and Brad still has not met the two milestones that I stated for him. There is an intermediate question proposed by Partzman to consider which is very educational. Then there is the issue of the second question that I answered right away that was "summarily dismissed" by Brad earlier in the thread as being wrong.
So does the thread die a miserable death, or does Brad try to see it through?
This thread has had the usual "barrel of monkeys" craziness. Just look at the issue of the variable ideal voltage source. We had Brad, Magluvin, Magneticitist, and Wattsup all insisting that it was "not permitted" because they read a definition online or in a book and couldn't think beyond what they read and use their noggins and show some creative thought. Will any of them simply post and admit that they were wrong?
Anybody that cares to read this thread from the beginning will see how truly crazy and nonsensical it could get at times. Does it die, or does Brad see it through?

You can do many things that bring things back into equal,, but that is adding in the mechanism that the formulas "as used" with these ideals do not have.
Say if you use pneumatic rams,, then the rams connect the air tank to the flywheel and you have the mechanism,, same thing if you use springs,, or if you use charge separation within the ideal voltage supply <= but it does not have that since it is ideal.
Now what if you are using rams,, what would happen to the pressure inside the ram that is going to stop the flywheel?
Well that depends doesn't it,, an infinite size air tank would not see a pressure change and there you go,, however, there is the adding in of the rams and pipes. <= a mechanism
If you have a limited air tank size then there will be a decrease in pressure as you are spinning up the flywheel and an increase in pressure when slowing it back down.
So using the ideals then we need a mechanism in place that itself does not consume energy but makes the transfer of the potentials equal,, same quantity transferred as stored and then as returned.
Without it the ideal voltage source is only blowing air,, 14.4J worth to spin it up and then another 14.4J worth to stop it,, and like I have said,,, even in this ideal space that can not happen.
Webby1,
Do I understand you correctly that you are saying 14.4J is consumed from the positive 4v supply during T0 to T3 to produce 2.4 amps in the 5H inductor and then as we apply a negative 4v supply during T5 to T8 to reduce the 2.4amps in the 5h inductor to zero, we consume another 14.4J?
partzman

Sort of,
The system starts at zero stored
We add 14.4J and it is stored
We then use another 14.4J to extract the 14.4J we added
If you will
14.4+14.414.4=14.4
With a resistor you have
+14.414.4=0 dissipated as heat
This should be the outcome for either method but using the formulas and ideals it does not.
If you apply the formula to the negative part of the cycle to get the change in current, that cost is the same as what was stored,, so then you would need to reverse the current value from the source which would be a full stop of current instantly.
If you ramp up the change then it will slowly slow down and stop the current.
The energy differences would be moved into and out of the source, so a sink of ~10.15J from the inductor for a cost of ~4.25J then the last of the extraction of ~4.25J would cost ~10.15J
So it cost me nothing to kill the energy that was stored,, what was taken out equals what was put in to kill it,, but I had to put it in to start with.
This must be wrong,, or incomplete.
Worse is the flywheel analogy using air,, makes an easy think since you are reversing the direction of the input influence to stop the flywheel, so you now have to spend the energy to spin it up,, you also then have to spend the energy to stop it since the air is the brake.

This is crazy maybe,, but to me it seems like you can not have a voltage if you do not have a resistance.
Resistance most likely is the wrong word since it has a definition already,, but there are so many words that describe the same event.
So maybe the ideal voltage source has a virtual variable resistor that automatically changes itself to keep the current and voltage in check?
This maybe as the start to the mechanism that is needed?

High there
Can’t say it better. hope the attached helps.
Anybody that cares to read this thread from the beginning will see how truly crazy and nonsensical it could get at times.

Sort of,
The system starts at zero stored
We add 14.4J and it is stored
We then use another 14.4J to extract the 14.4J we added
If you will
14.4+14.414.4=14.4
With a resistor you have
+14.414.4=0 dissipated as heat
This should be the outcome for either method but using the formulas and ideals it does not.
If you apply the formula to the negative part of the cycle to get the change in current, that cost is the same as what was stored,, so then you would need to reverse the current value from the source which would be a full stop of current instantly.
If you ramp up the change then it will slowly slow down and stop the current.
The energy differences would be moved into and out of the source, so a sink of ~10.15J from the inductor for a cost of ~4.25J then the last of the extraction of ~4.25J would cost ~10.15J
So it cost me nothing to kill the energy that was stored,, what was taken out equals what was put in to kill it,, but I had to put it in to start with.
This must be wrong,, or incomplete.
Worse is the flywheel analogy using air,, makes an easy think since you are reversing the direction of the input influence to stop the flywheel, so you now have to spend the energy to spin it up,, you also then have to spend the energy to stop it since the air is the brake.
OK, let's take a closer look at the symmetrically equal charge and discharge cycles of the 5H inductor with equal and opposite +4/4 voltage sources over equal time periods. I've attached a schematic with Circuit A which is an equivalent to the previous sim, and Circuit B with a repositioned ground connection but otherwise identical in operation to Circuit A.
I will attempt to point out that during the ramp down phase, the energy stored in the inductor is returned to the 4 supply so theoretically no energy is lost with ideal components.
Looking at Circuit B, when switch W1 is closed, L1 begins to receive a positive current flow out of V1 entering the dot end of the coil and exiting the nondot end. When W1 opens and W2 closes, the current stored in L1 still flows in a positive direction out of the nondot end of L1 into the positive terminal of V2 and by the time the current in L1 has reached zero, the energy taken from V1 is returned to V2 due to equal supply voltages and ideal components. The voltage across L1 changes polarity at the switching W1 and W2 but the current begins a gradual decline. These actions are able to be seen in the simulation.
The same energy action occurs in Circuit A but is harder to "see" because of the layout but may become apparent after careful study. The nomenclature below each circuit calls out the voltage polarities across L1 for the various switch positions. Note that they are identical.
partzman

I will attempt to point out that during the ramp down phase, the energy stored in the inductor is returned to the 4 supply so theoretically no energy is lost with ideal components.
Yes but with the compressed air and flywheel analogy the air is not pushed back where it came from when the wheel is braked so the analogy fails.
With this analogy energy has to be actually expended to create the reverse torque and brake the flywheel, because the air leaks out after it hits the flywheel. In ideal electronic circuit there is no such leakage.

Here is the decline in stored energy per 0.1 drop in current,, invert it for the ramp up.
14.400
13.225
12.100
11.025
10.000
9.025
8.100
7.225
6.400
5.625
4.900
4.225
3.600
3.025
2.500
2.025
1.600
1.225
0.900
0.625
0.400
0.225
0.100
0.025
Since there is no resistance within this circuit there is no loss from the stored energy,, so to MAKE it loose the stored energy you need to supply more energy.
How can I say that? well did it take energy to ramp up the inductor? are there any losses that will bleed off that stored energy?
So it will not change by itself.
At t=5,, how much voltage can you measure,, not just have it there because it was put there but measure. The virtual resistor points out that it takes a resistance to have voltage,, how much pressure is in a pipe that has no walls?

Since there is no resistance within this circuit there is no loss from the stored energy,, so to MAKE it loose the stored energy you need to supply more energy.
No, it is enough introduce any series resistance into the L circuit in order to make that circuit perform work on that resistance and convert the stored energy into heat ...or introduce an empty capacitor into the L circuit in order to make that circuit perform work on that capacitance and charge it up.

Yes but with the compressed air and flywheel analogy the air is not pushed back where it came from when the wheel is braked so the analogy fails.
With this analogy energy has to be actually expended to create the reverse torque and brake the flywheel, because the air leaks out after it hits the flywheel. In ideal electronic circuit there is no such leakage.
Good point.
So we need to add a circular chamber around the flywheel as such that it is sealed to the edge of the flywheel so that the face of the flywheel is all that is exposed to the air and chamber,, and the chamber is frictionless.
There are 2 pipes that come out of the chamber on opposite sides tangentially so that one pushes the air and one pulls the air for the same direction of rotation. These pipes are connected tothe air pump
There are 2 magic valves that will connect these pipes to the air chamber as well as they let the existing air within the chamber freely circulate.
The air pump is set so that it will suck air in from one side at 2 psi and it will blow the air out the other side at 2 psi. This pressure then creates a pressure differential of 4 psi.
The valves are opened with the pump turned on, the air tries to move and the drag against the flywheel impedes that motion and exerts a force of acceleration against the flywheel.
After some time of run the valves are closed.
The flywheel and the air within the chamber continue to rotate at the same velocity as each other and at the end velocity the flywheel reached.
At this time we would not be able to measure any pressure differential of the air, only its velocity.
The pump direction is reversed and the valves are opened.

No, it is enough introduce any series resistance into the L circuit in order to make that circuit perform work on that resistance and convert the stored energy into heat ...or introduce an empty capacitor into the L circuit in order to make that circuit perform work on that capacitance and charge it up.
I think I said that if you used a resistor you would be able to extract the stored energy,, dissipated as heat,, and that would equal the input.

Yes and an empty capacitor, too.

How about the Root's Blower (http://www.mekanizmalar.com/roots3.html) with flywheels attached to the impellers and voltage symbolized by the pressure difference between two airtanks?

How about the Root's Blower (http://www.mekanizmalar.com/roots3.html) with flywheels attached to the impellers?
I have an issue with the pulses from that,,
Variable levers are something I play with as well as CVT kind of stuff,, my latest sim is a great CVT :)
My Mini Cooper has a blower :)

I am guessing that if my description of the chamber made sense then there is a part missing.
That would be a bypass tube so that the pressure differential does not exceed 4 psi.
When the air pump is reversed then it will only see an opposing 4 psi pressure differential and stall.
The flywheel and air will still be turning and now the drag will slow down the flywheel.
When it has reached a rate of rotation where the pressure differential the pump sees is less than 4 psi the pump will start pumping and finish stopping the flywheel.

You should say, "Okay, I will discuss it with my peers and go do more research and learn more and improve my skills so that I can answer the question successfully by myself."
After all the ground that has been covered in this thread, and for the innumerable crazy theories and misconceptions that have been addressed and dealt with, don't you feel a bit ridiculous for trolling me like that on the very first page?

I have an issue with the pulses from that,,
Pulses are a minor imperfection that have no bearing on the overall energy balance.
The temperature difference due to air de/compression is a much larger flaw in this analogy.

You could use something other than air,, like nitrogen.
Air is a compressible liquid,, so if this can be made to be close if not the same then the gas could be replaced with a liquid that is not compressible,,
Is the electron cloud itself compressible?
Does Voltage actually compress that cloud into a smaller volume?
I know the electrons can move but I don't think I have thought about them being actually squished.

I am wondering if the ideal voltage source would have the internal voltage as well as the external.
Since it is a difference then is there not the same but opposite internal difference as the external difference?
Then the charged inductor would also have that when it is a source,, as well as when it is being charged up and since there is zero resistance and capacitance the internal and external will be the same but opposite.
When the charge carriers freewheel at t=3,, which is the internal and which is the external,, with no resistance I am thinking that that relationship is a toss of the coin.

Since most of the technical discussion has already occurred, and other sims posted, here is my voltage and current trace in its complete form. R=1m Ohm.
Hopefully Brad wil