RL Circuit: Kirchoff's Law is Wrong - Faraday's Law is Correct

[Wannabeagenius]

Hi Guys,

This is going to be a bit controversial but that's ok.

In an RL circuit, both with and without a battery, Kirchoff's law is invoked in virtually every elementary textbook in the United States and perhaps in the world. It is a mistake!

The voltage drop across an inductor is zero. Of course I'm talking about a perfect inductor but I am also talking about an ideal circuit. The usage of Kirchoff's law works because fudging is done to get the correct answer.

The correct law to use is Faraday's law, the closed intergral of Edotdl equals the time rate of change of the magnetic flux. This, when used, does not require any fudging.

I'm hoping that I don't have to justify this but this is from MIT and I can justify it, if necessary.

But that still leaves me with a question. I know that the voltage drop across an inductor is zero, but what does "the induced emf equals minus Ldi/dt" mean physically?

Thank you,
Bob Guercio

 

[Andrew Mason]

Hi Guys,

This is going to be a bit controversial but that's ok.

In an RL circuit, both with and without a battery, Kirchoff's law is invoked in virtually every elementary textbook in the United States and perhaps in the world. It is a mistake!

The voltage drop across an inductor is zero. Of course I'm talking about a perfect inductor but I am also talking about an ideal circuit. The usage of Kirchoff's law works because fudging is done to get the correct answer.

The correct law to use is Faraday's law, the closed intergral of Edotdl equals the time rate of change of the magnetic flux. This, when used, does not require any fudging.

I'm hoping that I don't have to justify this but this is from MIT and I can justify it, if necessary.

But that still leaves me with a question. I know that the voltage drop across an inductor is zero, but what does "the induced emf equals minus Ldi/dt" mean physically?

Thank you,
Bob Guercio

The voltage drop across an inductor connected to a DC source is 0. The voltage drop across an inductor connected to an AC source is not 0.

If dI/dt is not 0, there will be a non-zero induced emf in the coil.

AM

 

[Dale]

The correct law to use is Faraday's law, the closed intergral of Edotdl equals the time rate of change of the magnetic flux.

The laws of circuit theory are a very nice approximation to Maxwell's equations for small circuits relative to the wavelengths of interest. As long as you know it is an approximation and understand the assumptions then you are fine to use Kirchoff's as a simplification of Maxwell's equations. The textbooks you are referring to are fine, although I do wish that mine was a little more explicit on the underlying assumptions.

 

[Wannabeagenius]

The voltage drop across an inductor connected to a DC source is 0. The voltage drop across an inductor connected to an AC source is not 0.

If dI/dt is not 0, there will be a non-zero induced emf in the coil.

AM

Before I justify this, I'm going to wait until others weigh in because it is going to take me a few hours of research. I'll tell you what my source is though.

MIT has taped their elementary Physics course, which by the way ain't too elementary, for the Internet. The professor explains it very well and MIT also has a handout on-line explaining it.

Like I said, if I need to, I'll do the legwork but I really don't feel like it.

By the way, the reason that the voltage drop across an inductor is zero is because an inductor is a conductor. In theory, a voltage drop across a conductor would give infinite current.

Bottom line. Kirchoff's law should not be used, yet it is used in virtually every elementary textbook. Fudging is necessary to get it to work. Faraday's law should be used but isn't. When used, and I've done it, no fudging is necessary.

Let's see what happens.

Bob

 

[Wannabeagenius]

The laws of circuit theory are a very nice approximation to Maxwell's equations for small circuits relative to the wavelengths of interest. As long as you know it is an approximation and understand the assumptions then you are fine to use Kirchoff's as a simplification of Maxwell's equations. The textbooks you are referring to are fine, although I do wish that mine was a little more explicit on the underlying assumptions.

I believe this but in elementary Physics books, that is not explained. And when you can use Faraday's law which is not an approximation, why not use it?

The MIT professor has referred to this as "very embarrassing" and he is correct.

But I'm still trying to understand exactly what "back emf" means physically?

Bob

 

[Dale]

And when you can use Faraday's law which is not an approximation, why not use it?

Because the math is a lot more cumbersome for little or no benefit.

 

[Wannabeagenius]

Because the math is a lot more cumbersome for little or no benefit.

It's not that cumbersome! But if you believe that it is cumbersome, it should at least be explained. And it's not explained in any textbook that I've seen and I'm talking about going back decades!

The induced emf of an inductor is minus the time rate of change of the magnetic flux. Now since the voltage drop across an inductor is zero, what is the physical interpretation of this? This is the question that I am really wrestling with.

At one time I thought that I understood this but now I'm not so sure!

Thank you,
Bob

 

[Upisoft]

The induced emf of an inductor is minus the time rate of change of the magnetic flux. Now since the voltage drop across an inductor is zero, what is the physical interpretation of this? This is the question that I am really wrestling with.

The magnetic flux is produced by the current. The change of the current will result in change of the flux. And you already know what the change of the flux is doing.

 

[Wannabeagenius]

The magnetic flux is produced by the current. The change of the current will result in change of the flux. And you already know what the change of the flux is doing.

Yes. But when I hear induced emf, I think of a voltage drop. There is no voltage drop.

What is the voltage that they are talking about? It's almost as if it's imaginary.

Thanks again,
Bob

 

[Upisoft]

Yes. But when I hear induced emf, I think of a voltage drop. There is no voltage drop.

What is the voltage that they are talking about? It's almost as if it's imaginary.

Thanks again,
Bob

There is voltage on the inductor, and it is (in your perfect case) -emf. The voltage "opposes" to the change of the current. I don't know if this voltage can be called "voltage drop". English terminology is not something I'm best at.

 

[Andrew Mason]

By the way, the reason that the voltage drop across an inductor is zero is because an inductor is a conductor. In theory, a voltage drop across a conductor would give infinite current.

No.

The force that limits current is not resistance. It is inductive reactance (the induced emf = -Ldi/dt). So if you are applying Kirkoff's laws you have to take this into account (ie. the applied and induced emfs in the circuit). This is Lenz' law in operation. A changing current in the inductor creates an electromagnetic effect that opposes the change of current.

The applied AC voltage causes a current to flow in the inductor that lags the voltage by 90 degrees. The current-limiting back emf is greatest (ie voltage drop across the inductor is the greatest) when the rate of change of current is greatest. The rate of change of current is greatest when I = 0.

Conversely, the voltage drop across the inductor is at a minimum (minimum back emf) when the rate of change of current is 0. This occurs at maximum current (ie at a peak or trough of the sinusoidal current graph). So the peaks of the voltage drop across the inductor occur 90 degrees (ie. a 1/4 cycle) ahead of the peaks of the current.

The opposite applies to an AC circuit with a capacitor: current leads voltage by 90 degrees.

AM

 

[Andrew Mason]

Yes. But when I hear induced emf, I think of a voltage drop. There is no voltage drop.

What is the voltage that they are talking about? It's almost as if it's imaginary.

Apply an AC voltage to a circuit containing an inductor and measure the instantaneous voltage across the inductor.

There will be no voltage drop measured across the inductor when di/dt = 0 (ie. maximum |current|, 0 back emf). One will begin to measure a voltage drop as the back emf increases in the inductor. The maximum back emf, and hence maximum voltage drop, is measured when di/dt is maximum - when |I|=0.

AM

 

[Wannabeagenius]

No.

Please go here: http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/lecture-20-inductance-and-rl-circuit/

Go into the video 8 minutes and 15 seconds and see for yourself.

I ain't about to tell this guy he is wrong. MIT also supplies a handout explaining it.

Please also note that the mathematics is not very difficult.

Bob

 

[Dale]

It's not that cumbersome!

Remember that circuit theory is a tool for engineering and design of electrical circuits. Modern CPU's have billions of transistors in them. A computational difference that is "not that cumbersome" suddenly becomes "very cumbersome" when multiplied by billions.

Also, circuit elements all have some variance in their electrical properties. The current and voltage variation caused by these natural manufacturing tolerances is much greater than the errors introduced by using the circuit theory approximations, and in many cases significantly less than the thermal noise.

So, using Maxwell's equations is more cumbersome with absolutely no benefit for most circuits since the errors due to unmodeled effects are greater than the errors due to model inaccuracies.

The induced emf of an inductor is minus the time rate of change of the magnetic flux. Now since the voltage drop across an inductor is zero, what is the physical interpretation of this? This is the question that I am really wrestling with

Andrew Mason already answered this question in the first response. The part in bold is not true in general. In general the voltage drop across an inductor is not zero.

 

[Wannabeagenius]

Remember that circuit theory is a tool for engineering and design of electrical circuits. Modern CPU's have billions of transistors in them. A computational difference that is "not that cumbersome" suddenly becomes "very cumbersome" when multiplied by billions.

Also, circuit elements all have some variance in their electrical properties. The current and voltage variation caused by these natural manufacturing tolerances is much greater than the errors introduced by using the circuit theory approximations, and in many cases significantly less than the thermal noise.

So, using Maxwell's equations is more cumbersome with absolutely no benefit for most circuits since the errors due to unmodeled effects are greater than the errors due to model inaccuracies.

Andrew Mason already answered this question in the first response. The part in bold is not true in general. In general the voltage drop across an inductor is not zero.

You may be correct but please go to http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/lecture-20-inductance-and-rl-circuit/ and go into it 8 minutes and 15 seconds.

This is not a wacko website but MIT.

Does what you're saying agree with this?

Thanks again,
Bob

 

[Upisoft]

You may be correct but please go to http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/lecture-20-inductance-and-rl-circuit/ and go into it 8 minutes and 15 seconds.

This is not a wacko website but MIT.

The voltage on the inductor is E-emf. And since this is superconducting inductor E=0. Still -emf is there to be measured.

 

[Wannabeagenius]

The voltage on the inductor is E-emf. And since this is superconducting inductor E=0. Still -emf is there to be measured.

I don't think that the MIT professor meant that it was a superconducting inductor. I believe the professor mentioned superconductivity only to stress that it was an ideal inductor and, as I understood his lecture, an ideal inductor is like an ideal piece of wire with no voltage drop across it. A real inductor will have a voltage drop across it just as a real wire with real resistance does.

As I recall, the MIT handout on the matter, which can be downloaded from this course site, did not mention anything about superconductors.

I hope that everyone responding to this thread listens to that lecture.

Bob

 

[Dale]

Does what you're saying agree with this?

No, the MIT professor is wrong (at least with the quote taken out of context).

There are three assumptions required for you to use circuit theory instead of Maxwell's equations:
1) the circuit is small relative to the wavelengths involved (lumped-parameter system)
2) no net charge on any component at any time
3) no magnetic coupling between components

Subject to those constraints KVL and KCL are perfectly acceptable means to analyze a circuit with an inductor. Maxwell's equations are not needed unless one of those assumptions is significantly violated.

 

[Wannabeagenius]

No, the MIT professor is wrong.

There are three assumptions required for you to use circuit theory instead of Maxwell's equations:
1) the circuit is small relative to the wavelengths involved (lumped-parameter system)
2) no net charge on any component at any time
3) no magnetic coupling between components

Subject to those constraints KVL and KCL are perfectly acceptable means to analyze a circuit with an inductor. Maxwell's equations are not needed unless one of those assumptions is significantly violated.

I doubt that he is wrong and I'm sure that he would not say that you guys are wrong.

Don't forget that he is looking at it from a Physics viewpoint and is analyzing it ideally with no appoximations made. He is lecturing in a Physics and not an Engineering course.

That does not make him wrong. Not practical perhaps but not wrong.

Please don't forget that MIT has a handout on this issue which, as I recall, was written by another MIT professor. It is very doubtful that both of these gentlemen are wrong!

Bob Guercio

 

[Dale]

I'm sure that he would not say that you guys are wrong

That is exactly what he said, "almost every college book that you read on physics do this wrong". And I would guess that is the reason you linked to that quote.

he is looking at it from a Physics viewpoint and is analyzing it ideally with no appoximations made

Then he is not doing circuit theory, he is doing EM. That is fine if it is what he wants to do, but it doesn't make all of the circuit textbooks wrong.

 

[Wannabeagenius]

That is exactly what he said, "almost every college book that you read on physics do this wrong". And I would guess that is the reason you linked to that quote.

Guys,

You have your way of doing it predicated upon a logical approximation. He would probably agree with that.

In a Physics textbook, you are expected to be a bit more precise. Physics is not engineering and where it is easy enough to be precise, you should be precise. In my opinion, he is saying that it is the wrong approach because it is a Physics and not an Engineering book.

Please also realize, no Physics book even justifies it the way you did. They just do it leaving the reader to believe there is no approximation involved.

I say that is wrong!

However, I'm still wondering about the basic premise that there is not a voltage drop across an inductor versus there is a voltage drop across an inductor.

Tomorrow I'm going to download the MIT writeup on this and post it for you guys.

Then he is not doing circuit theory, he is doing EM. That is fine if it is what he wants to do, but it doesn't make all of the circuit textbooks wrong.

I never said the Engineering textbooks were wrong. I never even thought about it from an Engineering standpoint.

My only concern is Physics and yes, Electricity and Magnetism!

Bob

 

[stevenb]

This is not a wacko website but MIT.

Does what you're saying agree with this?

Prof. Lewin is an awesome professor, and most things he says are right on the mark. But, he is likely a little off base on this issue, at least in the opinion of many. I once got in a discussion with another member here (studiot) about this topic (although we discussed it over at the allaboutcircuits forum). He felt Prof. Lewin was wrong, while I defended Prof. Lewin. In the end I became convinced studiot was right. Prof. Lewin is (perhaps) incorrectly assuming that Kirkoff's Voltage law says that the line integral of electric field around a closed path is zero. Actually, there are books that do say this, but this does not agree with Maxwell's definition of Kirchoff's voltage law. I discussed this point in another thread here. You can find a PDF of the page from his Treatise, in the second post of this thread.

https://www.physicsforums.com/showthread.php?t=405700

The quote from Maxwell's Treatise is as follows, and I can't think of a better horse's mouth to get information like this from.

"In any complete circuit formed by the conductors, the sum of the electromotive forces taken around the circuit is equal to the sum of the products of the current in each conductor multiplied by the resistance of that conductor."

Although the older language is awkward compared to today's terminology, he is basically saying the sum of the potential drops equals the sum of the EMFs. The EMFs can be batteries, or induced EMF as implied in Faraday's Law. Hence, Kirchoff's voltage law should be viewed as a circuit theory equivalent of Faraday's Law, and Kirchoff's current law should be viewed as a circuit theory equivalent of conservation of charge. This makes sense because we don't want circuit theory laws that conflict with fundamental laws in the obvious way you are suggesting if we use Prof. Lewin's definition of KVL.

Prof. Lewin perhaps is not at fault here, and his objections are worthy of note because some recent books don't handle the subject correctly. Still, we can go back to older books and even back to Maxwell himself and see the appropriate definition of KVL.

Aside from this point of confusion. That lecture is great (as are his others) and effectively highlights the nonintuitive aspects of Faraday's Law. Anytime a professor makes us talk about issues like this, he has done his job because an opportunity for learning has been created. Being provocative is more likely to educate than being boring. Still, it's important to identify errors in definition - or redefinition.

 

[Wannabeagenius]

Prof. Lewin is an awesome professor, and most things he says are right on the mark. But, he is likely a little off base on this issue, at least in the opinion of many. I once got in a discussion with another member here (studiot) about this topic (although we discussed it over at the allaboutcircuits forum). He felt Prof. Lewin was wrong, while I defended Prof. Lewin. In the end I became convinced studiot was right. Prof. Lewin is (perhaps) incorrectly assuming that Kirkoff's Voltage law says that the line integral of electric field around a closed path is zero. Actually, there are books that do say this, but this does not agree with Maxwell's definition of Kirchoff's voltage law. I discussed this point in another thread here. You can find a PDF of the page from his Treatise, in the second post of this thread.

https://www.physicsforums.com/showthread.php?t=405700

The quote from Maxwell's Treatise is as follows, and I can't think of a better horse's mouth to get information like this from.

"In any complete circuit formed by the conductors, the sum of the electromotive forces taken around the circuit is equal to the sum of the products of the current in each conductor multiplied by the resistance of that conductor."

Although the older language is awkward compared to today's terminology, he is basically saying the sum of the potential drops equals the sum of the EMFs. The EMFs can be batteries, or induced EMF as implied in Faraday's Law. Hence, Kirchoff's voltage law should be viewed as a circuit theory equivalent of Faraday's Law, and Kirchoff's current law should be viewed as a circuit theory equivalent of conservation of charge. This makes sense because we don't want circuit theory laws that conflict with fundamental laws in the obvious way you are suggesting if we use Prof. Lewin's definition of KVL.

Prof. Lewin perhaps is not at fault here, and his objections are worthy of note because some recent books don't handle the subject correctly. Still, we can go back to older books and even back to Maxwell himself and see the appropriate definition of KVL.

Aside from this point of confusion. That lecture (as well as his others) is great and really highlights the nonintuitive aspects of Faraday's Law. Anytime a professor makes us talk about issues like this, he has done his job because an opportunity for learning has been created. Being provocative is more likely to educate than being boring. Still, it's important to identify errors in definition - or redefinition.

It's a little late and I'll try digesting this tomorrow. But I must say that I'm having a bit of trouble with all of this.

We are talking about a simple ideal inductor and we are not sure if there is a voltage drop across it or not? This is not eleven dimensional space!

Until tomorrow!

Thanks,
Bob

 

[Upisoft]

We are talking about a simple ideal inductor and we are not sure if there is a voltage drop across it or not? This is not eleven dimensional space!

Look at it this way. Is there any voltage drop across the battery(ideal battery)? No. There is only emf. The same is valid for an ideal inductor. The difference is that in the battery the energy source is some kind of electrochemical reaction and in the inductor the energy source is the magnetic field.

The other way to grasp this problem is to look at the capacitor. Basically the capacitor does the same thing but changes places of U and I. So now the argument will be "is there any current going through a charging capacitor"? The correct answer depends on how you look at it. The physical way to look at it is that there is no current going through the dielectric of the capacitor (ideal dielectric). But when you attach an amperеmeter, no matter how close to the capacitor you will measure its charging current.

 

[Dale]

However, I'm still wondering about the basic premise that there is not a voltage drop across an inductor versus there is a voltage drop across an inductor.

We are talking about a simple ideal inductor and we are not sure if there is a voltage drop across it or not?

Whether you call it a "voltage drop" or an "emf", if you place a voltmeter across an inductor you will measure a non-zero value except at DC. This is what Andrew Mason was trying to tell you from the very first response.

The only possible confusion is terminology. stevenb correctly identifies the relevance of Kirchoff's law in its original expression which explicitly uses the term "emf", but since a voltmeter measures a resistive voltage drop and an emf the same way we generally just use the word "voltage" now to refer to electrochemical emf, resistive voltage drop, an inductive or capacitive emf, etc.

 

[Dale]

Please also realize, no Physics book even justifies it the way you did. They just do it leaving the reader to believe there is no approximation involved.

I say that is wrong!

Note that the MIT professor is also using those same approximations without stating them and has led you to believe that there is not approximation involved. So, by your own judgment here the MIT professor is wrong.

 

[Wannabeagenius]

Prof. Lewin is an awesome professor, and most things he says are right on the mark. But, he is likely a little off base on this issue, at least in the opinion of many.

It's a little hard for me to believe this for a couple of reasons.

Professor Lewin is an awesome professor and the lecture that you viewed was from MIT, an awesome University. I don't think that they would allow an error of this nature to continue on the Internet for years.

Secondly, MIT has a writeup that is available on the Internet that explains this in great detail. I have uploaded it as an attachment. This document was originally written by Professor Belcher of MIT and modified slightly by Professor Lewin to tie it in closely with the text used.

Please read this and then see if all of you still disagree.

If I may dare to say and with all due respect, you guys are up against some real heavyweights in the Physics field! If you guys were in my shoes, what would your reaction be to all of this?

Bob

 

[Wannabeagenius]

Note that the MIT professor is also using those same approximations without stating them and has led you to believe that there is not approximation involved. So, by your own judgment here the MIT professor is wrong.

As I understand it, if you consider the ideal case, the approximations are not necessary.

For example, in the real world there is some magnetic coupling between components which, as an approximation, can be ignored. In the ideal world, there is no magnetic coupling! Period! There is no need for an approximation.

Professor Lewin has analyzed the ideal case!

 

[Wannabeagenius]

Note that the MIT professor is also using those same approximations without stating them and has led you to believe that there is not approximation involved. So, by your own judgment here the MIT professor is wrong.

As I understand it, if you consider the ideal case, the approximations are not necessary.

For example, in the real world there is some magnetic coupling between components which can be ignored. In the ideal world, there is no magnetic coupling! Period!

Professor Lewin has analyzed the ideal case!

 

[stevenb]

It's a little hard for me to believe this for a couple of reasons.

Professor Lewin is an awesome professor and the lecture that you viewed was from MIT, an awesome University. I don't think that they would allow an error of this nature to continue on the Internet for years.

Secondly, MIT has a writeup that is available on the Internet that explains this in great detail. I have uploaded it as an attachment. This document was originally written by Professor Belcher of MIT and modified slightly by Professor Lewin to tie it in closely with the text used.

Please read this and then see if all of you still disagree.

If I may dare to say and with all due respect, you guys are up against some real heavyweights in the Physics field! If you guys were in my shoes, what would your reaction be to all of this?

Bob

Correct me if I misinterpreted that document, but doesn't their argument rely on the definition of KVL that says the line integral of electric field around a closed path is zero, or equivalently the sum of potential drops around the circuit is zero?

If this is correct, then this is the source of the disagreement. If we want to appeal to "heavyweights", why not accept the definition from James Clerk Maxwell?

There is one thing I agree with. That is, they are correct that some books are not handling the subject well.

 

[Wannabeagenius]

Correct me if I misinterpreted that document, but doesn't their argument rely on the definition of KVL that says the line integral of electric field around a closed path is zero, or equivalently the sum of potential drops around the circuit is zero?

No! That's Kirchoff's law.

Their argument depends upon the closed line integral of E dot dl equals the negative of the time rate of change of the magnetic flux which is Faraday's law!

Bob

 

[Dale]

If I may dare to say and with all due respect, you guys are up against some real heavyweights in the Physics field! If you guys were in my shoes, what would your reaction be to all of this?

Appeal to authority is a logical fallacy, but by appeal to authority Lewin clearly loses. stevenb has already cited Maxwell, and just looking on my bookshelf I see Serway, Irwin, Nilsson, Riedel, and Rizzoni all of whom disagree with Lewin. Lewin is the one up against the real heavyweights, not us, and he deliberately placed himself in that position.

Additionally there is the inescapable fact that Kirchoff's laws (including inductors) have been used successfully for decades to build devices that function as designed. Every time you use any electronic device you are performing a physical experiment that refutes Lewin's position.

 

[stevenb]

No! That's Kirchoff's law.

Their argument depends upon the closed line integral of E dot dl equals the negative of the time rate of change of the magnetic flux which is Faraday's law!

Bob

You misunderstood me, as I said that's KVL which stands for Kirchoff's voltage law.

I know what Faraday's Law is.

This issue is, what is the correct definition of Kirchoff's Voltage Law (KVL). Did you read my quoted definition for KVL from Maxwell? I'm saying that Prof. Lewin's argument is based on an incorrect definition of Kirchoff's Voltage Law (KVL). He is saying that KVL is defined as "the sum of potential drops equals zero", but Maxwell says the sum of potential drops equals the sum of EMFs. There is a pretty big difference between the two statements. Maxwell deserves some consideration in my view.

Which definition of KVL do you accept?

1. sum of potential drops around a closed circuit equals zero
2. sum of potential drops around a closed circuit equals the sum of EMFs around that circuit

If you accept statement number 1, then Prof. Lewin is justified, but if you accept statement number 2, then he is a little off the mark. I say "a little off the mark" because he is justified to criticize the treatement of KVL in many books. However, he does not criticize them correctly. He needs to focus in on their mis-statement of KVL right from the beginning.

 

[Dale]

As I understand it, if you consider the ideal case, the approximations are not necessary.

For example, in the real world there is some magnetic coupling between components which can be ignored. In the ideal world, there is no magnetic coupling! Period!

Professor Lewin has analyzed the ideal case!

There is no difference between considering the ideal case and using the approximations. That is just two ways of saying the same thing.

The ideal case that you mention is the same as the third assumption
3) no magnetic coupling between components

But note carefully that he also uses the other two assumptions in his work:

He does not consider the finite speed of light so he is assuming
1) the circuit is small relative to the wavelengths involved (lumped-parameter system)

He does not consider the self-capacitance of the wires so he is assuming
2) no net charge on any component at any time

 

[Wannabeagenius]

You misunderstood me, as I said that's KVL which stands for Kirchoff's voltage law.

I know what Faraday's Law is.

This issue is, what is the correct definition of Kirchoff's Voltage Law (KVL). Did you read my quoted definition for KVL from Maxwell? I'm saying that Prof. Lewin's argument is based on an incorrect definition of Kirchoff's Voltage Law (KVL). He is saying that KVL is defined as "the sum of potential drops equals zero", but Maxwell says the sum of potential drops equals the sum of EMFs. There is a pretty big difference between the two statements. Maxwell deserves some consideration in my view.

Which definition of KVL do you accept?

1. sum of potential drops around a closed circuit equals zero
2. sum of potential drops around a closed circuit equals the sum of EMFs around that circuit

If you accept statement number 1, then Prof. Lewin is justified, but if you accept statement number 2, then he is a little off the mark. I say "a little off the mark" because he is justified to criticize the treatement of KVL in many books. However, he does not criticize them correctly. He needs to focus in on their mis-statement of KVL right from the beginning.

I see what you are saying but to be honest, I've never dissected it that thoroughly.

I've always regarded definition 1 as Kirchoff's law which takes into consideration both voltage drops (resistor) and emf's (batteries).

Believe it or not, I once contacted Professor Lewin by email and he responded. I'm tempted to contact him on this matter but I don't know if that would be proper. Perhaps he would regard it as a challenge by amateurs. I just don't know.

Any thoughts on that idea?

Bob