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Author Topic: MH's ideal coil and voltage question  (Read 188988 times)

Offline allcanadian

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Re: MH's ideal coil and voltage question
« Reply #15 on: May 08, 2016, 06:19:55 PM »
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

Free Energy | searching for free energy and discussing free energy

Re: MH's ideal coil and voltage question
« Reply #15 on: May 08, 2016, 06:19:55 PM »

Offline Magneticitist

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Re: MH's ideal coil and voltage question
« Reply #16 on: May 08, 2016, 06:30:47 PM »
@AllCanadian   seems a reasonable deduction that follows logic. welcome to the club of having no idea.


Offline allcanadian

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Re: MH's ideal coil and voltage question
« Reply #17 on: May 08, 2016, 08:45:22 PM »
@Magneticists
Quote
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



Offline poynt99

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Re: MH's ideal coil and voltage question
« Reply #18 on: May 08, 2016, 08:53:57 PM »
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).


Offline Magneticitist

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Re: MH's ideal coil and voltage question
« Reply #19 on: May 08, 2016, 09:07:51 PM »
@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

Free Energy | searching for free energy and discussing free energy

Re: MH's ideal coil and voltage question
« Reply #19 on: May 08, 2016, 09:07:51 PM »
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Offline MileHigh

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Re: MH's ideal coil and voltage question
« Reply #20 on: May 08, 2016, 09:10:46 PM »
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

Offline Magneticitist

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Re: MH's ideal coil and voltage question
« Reply #21 on: May 08, 2016, 09:23:36 PM »
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 is--you 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.

Free Energy | searching for free energy and discussing free energy

Re: MH's ideal coil and voltage question
« Reply #21 on: May 08, 2016, 09:23:36 PM »
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Offline MileHigh

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Re: MH's ideal coil and voltage question
« Reply #22 on: May 08, 2016, 09:36:12 PM »

Offline poynt99

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Re: MH's ideal coil and voltage question
« Reply #23 on: May 08, 2016, 10:13:46 PM »
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.

Offline Didymus

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Re: MH's ideal coil and voltage question
« Reply #24 on: May 08, 2016, 11:33:56 PM »
Somebody has to go back to basics, it might as well be me.

The definition of an ideal inductor is a two-terminal 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 two-terminal 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.

Free Energy | searching for free energy and discussing free energy

Re: MH's ideal coil and voltage question
« Reply #24 on: May 08, 2016, 11:33:56 PM »
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Offline Pirate88179

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Re: MH's ideal coil and voltage question
« Reply #25 on: May 08, 2016, 11:55:08 PM »
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

Online Magluvin

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Re: MH's ideal coil and voltage question
« Reply #26 on: May 09, 2016, 12:09:12 AM »
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


Online Magluvin

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Re: MH's ideal coil and voltage question
« Reply #27 on: May 09, 2016, 12:17:49 AM »
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

Offline tinman

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Re: MH's ideal coil and voltage question
« Reply #28 on: May 09, 2016, 01:38:38 AM »
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 said--there 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


Offline poynt99

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Re: MH's ideal coil and voltage question
« Reply #29 on: May 09, 2016, 02:28:13 AM »
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 said--there 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?

Free Energy | searching for free energy and discussing free energy

Re: MH's ideal coil and voltage question
« Reply #29 on: May 09, 2016, 02:28:13 AM »

 

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