Author Topic: MH's ideal coil and voltage question (Read 485033 times)

partzman · « **Reply #165 on:** May 10, 2016, 04:41:21 AM »

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 1e-110. This parameter may have passed a preset limit in LtSpice however.

partzman

tinman · « **Reply #166 on:** May 10, 2016, 04:45:21 AM »

Quote from: MileHigh on May 09, 2016, 10:10:09 PM

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 flat-line 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 infinite--you cannot redefine math at your will.

Quote

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

Magluvin · « **Reply #167 on:** May 10, 2016, 05:08:25 AM »

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

Magneticitist · « **Reply #168 on:** May 10, 2016, 05:11:35 AM »

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.

Magneticitist · « **Reply #169 on:** May 10, 2016, 05:30:38 AM »

Quote from: Magluvin on May 10, 2016, 05:08:25 AM

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?

tinman · « **Reply #170 on:** May 10, 2016, 06:03:46 AM »

Quote from: partzman on May 10, 2016, 04:41:21 AM

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 1e-110. This parameter may have passed a preset limit in LtSpice however.

partzman

Quote

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

tinman · « **Reply #171 on:** May 10, 2016, 07:00:40 AM »

Quote from: Magluvin on May 10, 2016, 05:08:25 AM

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 head-or 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.

Quote

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

.

As i said,and have all along--you 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 answer--you 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 it-the paradox. MH laughed at my two theoretical answers being totally different.
1-being that there is no current flowing through the inductor--that Poynt did agree with--not 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

Magneticitist · « **Reply #172 on:** May 10, 2016, 08:13:45 AM »

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.

wattsup · « **Reply #173 on:** May 10, 2016, 09:12:24 AM »

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

verpies · « **Reply #174 on:** May 10, 2016, 10:12:38 AM »

Quote from: tinman on May 10, 2016, 07:00:40 AM

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 .

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 non-ideal 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)

EMJunkie · « **Reply #175 on:** May 10, 2016, 10:22:30 AM »

@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_Figure-9-Impedance-and-ESR-vs-Frequency-for-T520-vs-T528-equal-parts

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.

verpies · « **Reply #176 on:** May 10, 2016, 10:33:16 AM »

Quote from: EMJunkie on May 10, 2016, 10:22:30 AM

@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.

EMJunkie · « **Reply #177 on:** May 10, 2016, 10:55:18 AM »

Quote from: verpies on May 10, 2016, 10:33:16 AM

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: Ω-jΩ or Ω+jΩ

If you see something like:

Z = 10-j10 (-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/wp-content/uploads/2011/07/ME301-vol-2.pdf

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-4/series-resistor-capacitor-circuits/

Chris Sykes
hyiq.org

tinman · « **Reply #178 on:** May 10, 2016, 01:22:41 PM »

Quote from: verpies on May 10, 2016, 10:12:38 AM

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 non-ideal 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)

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.

Thank you verpies for joining in on this discussion.
You have confirmed my real world answer--an ideal voltage cannot be applied to/placed across an ideal inductor.

Quote

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 non-ideal 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 inductor--as we know.

Brad

tinman · « **Reply #179 on:** May 10, 2016, 01:31:56 PM »

Quote from: verpies on May 10, 2016, 10:33:16 AM

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

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