Does Kirchhoff's Law Hold? Disagreeing with a Master

[EEVblog]

He became of #metoo target, due to his own human error. As I already said - I do not agree. Such issues shall be resolved in/by court, not by directorate of MIT - by literally burning all his work, by punishing his students, not actually punishing himself.

That was my first thought, removing his (highly regarded) content is effectively punishing students.

[Sredni]

Please do yourself and anybody else a favor: get hold of a copy of "Fields and Waves in Communication Electronics" by Ramo, Whinnery and VanDuzer and read the first four pages of chapter 4 (The electromagnetics of circuits)...

I guess you really like that book.

It's a nice, no-bullshit, dense and somewhat rigorous textbook. Not perfect by any stretch, but one of my favorites nonetheless. It could use some additional 80-100 pages. All blank. Seriously, that book needs to put some space between paragraphs ;-).
The reason I'm bringing RWvD up so many times is that it has a really nice discussion on the origin of KVL and KCL from Maxwell's equations, and of course the fact that ogden seems impervious to reading it. I'll end up summarizing it here, I know... But pictures, and formulas... they are so time consuming to put in a post...

  It was my textbook in college 37 years ago for "Electromagnetic Fields and Waves"  EECS117A, B, and C.  The teacher was Theodore Van Duzer.

Woah, how was he?

[Sredni]

As an argument I was referring to scientific paper having peer review and corrections. All you have in return is youtube video with introduction "Hello, hello, hello", no written content and no peer review? Whatta crooked mirrors world you are living in?!

The peer-reviewed scientific paper backing Lewin's "Hello hello hello" videos is Romer's paper. I had hoped it was clear.
Lewin himself talks about the paper in one of his recent videos. He said "I may or may not have read it", meaning that since it was published prior to his first superdemo he cannot prove he came up independently with the same experiment. He might as well have, it's not something particularly difficult to conceive. What counts is that both the superdemo and Romer's paper are about the same circuit and the apparent paradox of two voltmeters attached to the very same two points reading two different voltages. (Yes, it's more like oscilloscopes showing waveforms since we're talking about time-varying fields).

What I see is that they both find that the voltage is dependent on the path, and that when placed on the outside of the loop the voltmeters - applied to the very same two points - give different and opposite phase reading.

What?!! Your whole proof is two voltmeters showing different signs?

The whole drama is about two voltmeters showing different readings when attached to the same points (along different paths).
Now you are pretending it isn't?

"what's the summary field (integral E.dl ) of the loop E = E.coloumb + E.induced?" You did not gave clear answer. Is it zero or not?

Here's the answer, assuming that with 'summary' you mean circulation along the circuit's path: the circulation of E_total (integral of (E_coloumb + E_induced) . dl along the circuit's path is equal to minus the time derivative of the flux of B linked by said path.
Yes, it's Faraday's law.

So you refuse to name number because you either do not know it of refuse to acknowledge it being zero?

Oh, for... ****'s sakes!
It's not zero. The value depends on the time-varying flux: it is equal to minus the time derivative of the flux of B linked by said path. Didn't I just tell you that?
You want a value? In the case of Lewin's experiment, IIRC, it's 1V.

left side - keeping track of the signs and considering perfect conducting wires
0.9V on one resistor, 0V in the wire, 0.1V on the other resistor, 0V in the wire ===> circuitation of E.dl = 1V
right side
-dfi(B)/dt = 1V

Please, find a copy of...
--never mind.

EDIT: fixed quoting, and typos and plurals - oh my!

[dayfall]

I have a question.  Why wouldn't a simple loop of wire show that "Kirchhoff is for the birds"?   (Or even a straight wire)

The wire, as in Lewin's more complex circuit, has a resistance of zero.  We can easily create a situation in which we measure a voltage across that loop.  Afterwards a circuit, like Lewin's, can be shown to have different voltages across resistors that are shorted.  Kirchhoff defeated.

Also, in one of his videos does Lewin say something like "The oscillioscope on the left measures the voltage across the left resistor."  How does he know which one it is measuring?

[rfeecs]

  It was my textbook in college 37 years ago for "Electromagnetic Fields and Waves"  EECS117A, B, and C.  The teacher was Theodore Van Duzer.

Woah, how was he?

He was a bit introverted, quiet, maybe the opposite of Lewin's teaching personality.  By the time we got to the last quarter of the course series, there were only about 5 students left.  I was losing interest myself.  Back then there was almost no discussion of computer simulation which has become so dominant in the field now.  We were probably using the first edition of the book.  I remember some of the course was tricks to solve specific problems that wouldn't be needed today.

I agree the book is pretty good.  I have referred back to it myself when I realize that I forgot some basic things and it immediately refreshed my memory with it's basic and direct approach.  It does require vector calculus, so maybe not for everyone.  That's just the nature of the subject.

[bsfeechannel]

The reason I'm bringing RWvD up so many times is that it has a really nice discussion on the origin of KVL and KCL from Maxwell's equations, and of course the fact that ogden seems impervious to reading it.

Well, we have to thank ogden for his "imperviousness". Without him, you, mhz, rfeecs, RoGeorge, radioactive, et al., wouldn't have produced such brilliant expositions. I've learned a lot reading your posts.

I have a question.  Why wouldn't a simple loop of wire show that "Kirchhoff is for the birds"?   (Or even a straight wire)

Very good question. It would. However, Lewin wanted to work the other way around to emphasize the problem a lot of people have when trying to make sense of Faraday's law, having Kirchhoff always holds as a premise. What he did was the classical reductio ad absurdum. By assuming that Kirchhoff always holds, he arrived at a contradiction. Therefore the premise is false.

Also, in one of his videos does Lewin say something like "The oscillioscope on the left measures the voltage across the left resistor."  How does he know which one it is measuring?

You have to imagine the circuit with a battery that he drew on the blackboard with each oscilloscope probe connected across each respective resistor. He didn't demonstrate the circuit with a battery in practice because he believed that the understanding of Kirchhoff's law would be trivial. Then the battery is imaginarily removed. So the left oscilloscope measuring the left resistor is the same oscilloscope which would be measuring the left resistor when we had the battery.

I agree the book is pretty good.  I have referred back to it myself when I realize that I forgot some basic things and it immediately refreshed my memory with it's basic and direct approach.  It does require vector calculus, so maybe not for everyone.  That's just the nature of the subject.

We also used this textbook in college. And the pages recommended by Sredni are really good. However we had vector calculus as a separate course before we were introduced to electromag. That's why I recommended Jordan & Balmain, because it has a simplified review of vector analysis, a.k.a, vector calculus, for use with electromagnetism right in the first chapter. You'll need to study basic Calculus, though, but you can find it easily on the Interwebs.

You see, the biggest mistake of those who struggle with electromagnetism is to think that electricity and magnetism are a property of circuits. They are a property of space. So you need tools to study the properties of space. And that is vector analysis.

[ogden]

The peer-reviewed scientific paper backing Lewin's "Hello hello hello" videos is Romer's paper. I had hoped it was clear.

Fact that you say Romer's paper is backing Lewin's lectures does not magically create new reality. We are talking about Dr.Lewin's claims, not Romer's. It does not matter that Dr.Lewin is doing same experiment and voltmeters are showing similar "phenomena". He did not even made it clear - he know contents of paper or not!!! Math is not the same, claims differ, conclusions are far from equal. You are trying to make impression that you know how science works, but now you produced pure nonsense. There are no Dr.Lewin's scientific papers regarding Kirchoff's circuit rules vs Faradays law. Period. You better just leave it there.

"what's the summary field (integral E.dl ) of the loop E = E.coloumb + E.induced?" You did not gave clear answer. Is it zero or not?

Oh, for... ****'s sakes!
It's not zero. The value depends on the time-varying flux: it is equal to minus the time derivative of the flux of B linked by said path. Didn't I just tell you that?
You want a value? In the case of Lewin's experiment, IIRC, it's 1V.

I asked what is sum of conservative and nonconservative fields (E.coloumb + E.induced), your answer is 1V. I would like to address this *after* simple but important thought experiment:

We have transformer in a box with two twisted-pair wires coming out: primary & secondary. We also have tiny 1 KOhm resistor, it's so small that we ignore it's dimensions. We ignore internal resistance of transformer coils and wires as well. Also it is known that magnetic field of transformer is contained in the box and do not have any effect on wires coming out of it. When transformers primary winding is connected to signal generator, we measure 1V AC peak-peak voltage on it's secondary. Transformer is powerful enough so it is still same 1V AC peak-peak when 1K resistor is connected to secondary.

Three questions: 1) Do you agree that we can treat transformer's secondary winding as AC voltage source and resistor as a load? 2) Do you agree that those are two lumped elements, they form circuit and Kirchoff's circuit law holds here? 3) Do you agree that integral E.dl for this circuit is zero at any given time of observation?

Just tell - you agree or disagree. No evasive BS with pointers to literature this time, please. Part 2 will follow after you answer those questions.

[Sredni]

Disclaimer: Since I wanted to start the new year with the right foot, I spent new year's eve and I am spending new year's day in bed with a fever. It's not high, but I might be omitting letters, signs (especially signs) and symbols. The good(?) news is that I've had the time to find a way to upload images on this site.

Ok, let's get to the questions.

EDIT: on second thought...

This post has been shortened and cleansed to avoid upsetting other children.
Whatever was written here can be found in one or more of the following books (in no particular order, and without mentioning the usual suspects Feynman, Purcell, Griffiths, Ohanian, Jackson):

Panofsky, Phillips
Classical Electricity and Magnetism 2nd ed

John Kraus
Electromagnetism 2nd to 4th ed

Ramo, Whinnery, VanDuzer
Fields and Waves in Communication Electronics 2nd or 3rd ed

Bleaney
Electricity and Magnetism 3rd ed

Nayfeh, Brussel
Electricity and Magnetism

Kip
Fundamentals of Electricity and Magnetism 2nd ed

Lorrain, Courson
Electromagnetic Fields and Waves 2nd ed

"Books" are static paper based documents that can be found in libraries. They are like smartphones, but (usually) bigger, with lots and lots of extremely thin flexible e-ink screens and a very long battery life. Libraries are...
Oh, never mind. Keep on pushing that square peg into that round hole. With a big enough hammer, it will fit.

[ogden]

"3) Do you agree that integral E.dl for this circuit is zero at any given time of observation?"

Absolutely not.

Do you realize that the circuit you are proposing is just Lewin's ring with one resistor?

You wish I don't?

You disagreed that integral E.dl around loop of described circuit is zero, yet for resistor, wires and "black box" you wrote integral E.dl = R*I + 0 + EMF. Do you really claim sum is not zero?

Thing in a box may be piezoelectroc device or just generator - Faraday's law do not apply for such. Please do not stick it everywhere - needed or not. What is coming out of the box - AC 1V peak-peak. Do integral E.dl around the loop of described circuit is zero or not?

Happy and successful NY for everybody! May your every circuit works.

[ogden]

After all, when you compute the circulation of E_total = E_conservative + E_induced you get the sum of the circulation of E_conservative which is zero and all that's left is the circulation of E_induced. Everything checks out.

Wait... E.conservative in the resistor is zero even when on it's terminals is 1V? Are you sure?

[Sredni]

After all, when you compute the circulation of E_total = E_conservative + E_induced you get the sum of the circulation of E_conservative which is zero and all that's left is the circulation of E_induced. Everything checks out.

Wait... E.conservative in the resistor is zero even when on it's terminals is 1V? Are you sure?

I was talking about the circulation. You have to close the loop.
Fever has gone up, I will check again tomorrow, if I survive the night :-)

[ogden]

After all, when you compute the circulation of E_total = E_conservative + E_induced you get the sum of the circulation of E_conservative which is zero and all that's left is the circulation of E_induced. Everything checks out.

Wait... E.conservative in the resistor is zero even when on it's terminals is 1V? Are you sure?

I was talking about the circulation. You have to close the loop.

Indeed to have voltage on resistor terminals, you have to connect it to voltage source. Unconnected, lone resistor for sure will have zero field inside, right?

My way of looking at this box+resistor debate: if black box is able to produce AC 1V alone, I name it EMF source - disregarding it is transformer or just DC-fed AC generator. If I connect resistor, I close the loop, AC 1mA current is flowing and integral E.dl around the loop is zero. There is no magic, just quite simple logic and law of conservation of energy. Do you agree?

[Berni]

This is exactly why the other "less scientific" definition of voltage (The one about how many electrons there are in one spot) is used in circuit analysis and pretty much everywhere else where you need to actually calculate something.

With that definition of voltage the magnetic EMF looks exactly like a battery. There are no longer any paradoxes or ambiguities of what the voltage is inside a changing magnetic field. Put anything you want in that black box and the result is the same, no longer is a magnetic field a special case that must be treated differently. If you have 1V of magnetically induced EMF its the same as a 1V battery, both push the same current trough the resistor load according to Ohms law. Faradays law works perfectly fine with this definition too.

The only thing you have to do is imagine the magnetic field as a electric field that is circulating around the changing magnetic flux lines. Once you do that you have only electric fields acting on electrons and those are always conservative so you never get multiple answers for what voltage is between two points. And this voltage happens to be the exact same voltage voltmeters are showing.

Its just a different way of calculating the same thing, except that this way gives less confusing answers (Particularly in this kind of cirucit)

[bsfeechannel]

Fever has gone up, I will check again tomorrow, if I survive the night :-)

I hope you get better soon.

[radioactive]

Fever has gone up, I will check again tomorrow, if I survive the night :-)

I hope you get better soon.

Yes, get better soon Sredni.  Thank you for sharing your insights and knowledge in this thread.

[ogden]

Yes, get better soon Sredni.  Thank you for sharing your insights and knowledge in this thread.

Yes, get better soon.

Meanwhile I will sum-up our discussion about "1V AC box + resistor":

While measuring 1V AC voltage coming out of the box with voltmeter/scope, we cannot discern - source is transformer, piezo device or just AC generator powered by chemical battery (we agreed that electrons are the same long ago in this thread). Kirchoff's circuit law (KVL) holds for AC voltage source + resistor *only* when AC voltage is generated by anything but transformer or other kind contraption ruled by Faraday's law. When we have transformer in a box, KVL does not hold anymore. To know - KVL holds or not, we have to look inside the box, otherwise we may be mistaken. This is what you are claiming? Anybody else agreeing?

[rfeecs]

The only thing you have to do is imagine the magnetic field as a electric field that is circulating around the changing magnetic flux lines.

Why would you do that?  The changing magnetic field produces an electric field.  The electric field lines are circles around the magnetic field as shown in post #16:
https://www.eevblog.com/forum/chat/does-kirchhoffs-law-hold-disagreeing-with-a-master/msg2003414/#msg2003414

Once you do that you have only electric fields acting on electrons and those are always conservative so you never get multiple answers for what voltage is between two points.

If you have electric field lines that form a loop, as in this case, you have a nonconservative field.

And this voltage happens to be the exact same voltage voltmeters are showing.

So you don't have two voltmeters connected to same point showing different voltages?  This is what happens in the experiment.  Or are we supposed to imagine that it doesn't?

The two voltmeters will only measure the same voltage if the voltmeters and their connecting wires are outside the field region or somehow shielded or arranged so that the field does not affect the measurement.  Fine, but that is not this experiment.  This experiment is set up to demonstrate Faraday's law and non conservative fields.

Its just a different way of calculating the same thing, except that this way gives less confusing answers (Particularly in this kind of cirucit)

You seem to be claiming that a voltmeter only measures an electric field caused by separation of charges.  For example, you have to move charges on the gate of the input FET of the voltmeter to make it read a voltage, I guess is what you mean.

But the voltmeter can't tell you what separated the charges.  It could be a battery, it could be a solar cell, it could be a thermocouple, it could be a hall effect device, it could be a loop of wire surrounding a changing magnetic flux, or it could be a loop of wire rotating in a static magnetic field, to name a few possibilities.  The voltmeter cannot distinguish the charge separation caused by the electrostatic scalar potential from the charge separation caused by all the different other types.

In this case, the wires connecting the voltmeter are passing through a region that has a non-zero electric field.  That electric field, interacting with the charge in the wire, will give you a different reading on the voltmeter depending on the path the wires take through that field.

[ogden]

The two voltmeters will only measure the same voltage if the voltmeters and their connecting wires are outside the field region or somehow shielded or arranged so that the field does not affect the measurement.  Fine, but that is not this experiment.  This experiment is set up to demonstrate Faraday's law and non conservative fields.

This time we are *not* talking about Dr.Lewin's experiment circuit but "transformer in a box + 1K resistor *outside it*".

Most voltmeters (including those Dr.Lewin uses) measure voltage as integral E.dl through their internal resistance. How voltmeters measure is not that important in this case BTW.

[bsfeechannel]

Meanwhile I will sum-up our discussion about "1V AC box + resistor":

While measuring 1V AC voltage coming out of the box with voltmeter/scope, we cannot discern - source is transformer, piezo device or just AC generator powered by chemical battery (we agreed that electrons are the same long ago in this thread). Kirchoff's circuit law (KVL) holds for AC voltage source + resistor *only* when AC voltage is generated by anything but transformer or other kind contraption ruled by Faraday's law. When we have transformer in a box, KVL does not hold anymore. To know - KVL holds or not, we have to look inside the box, otherwise we may be mistaken. This is what you are claiming? Anybody else agreeing?

Let's try to explain. In the "circuit" below V₁ is whatever while V₂ is literally zero. Why? Because the net flux in the area bounded by the path determined by the terminals of V₂ is zero. So this is not a lumped circuit. Simply encasing this loop of wire that is under a varying magnetic field in a box won't lump it.



If you analyze other Maxwell's equations, there is one that shows that you can't have a magnetic monopole. If B is going into the screen, where does it come out? The answer: elsewhere. How could Lewin then create a magnetic monopole when this is not possible? The answer: he showed it a little earlier in the same video where he relegated Kirchhoff to the birds. The answer: solenoids. Solenoids generate a strong magnetic field inside their cores, but very weak outside. In practice we can consider it zero. This happens because outside of a solenoid the lines of magnetic field are very sparse, while in the core they're highly concentrated.

Therefore we need to find a way to lump our loop, so that it can be considered the secondary of a transformer. To do that, we need to gather all the return lines of B and make them cross the area between the loop and V₂, like below. Now, bingo! V₁ = V₂. Now we can box up the loop with the returning lines of B.



Another way of lumping our loop is to make its area perpendicular to the area "seen" by V₂. In that case the scalar product B · dA will be zero, once scalar products can be thought as the "shadow" one vector casts on another. If they are perpendicular, the "shadow" is zero.



You'll recognize these practical arrangements on real inductors. In the picture below, you have a solenoid (a). The area defined by the voltmeter is roughly perpendicular to the area of the solenoid. If you measure like in (b), the voltage will be the real voltage minus one turn. If N is not much grater than one, you can use a toroid as in (c) or other core with a closed magnetic path.

[bsfeechannel]

In short, the loop, the way it is, is unlumpable. Lumpable means that I can measure the same voltage regardless of the position of the voltmeters. What we have is a loop antenna. There is a reason why loop antennas are connected by two-wire transmission lines. So that you guarantee, among other things, that the voltage will always be measured like V₁, and not like V₂. If, instead of this loop antenna, we had, say a lumped generator, that would not be a problem, although we would need the transmission line for other reasons.

[Sredni]

This is exactly why the other "less scientific" definition of voltage (The one about how many electrons there are in one spot) is used in circuit analysis and pretty much everywhere else where you need to actually calculate something.

You keep talking of that as if it were a different physical quantity, measured in different units. It's not. The 'scalar potential' that is part of the (V, A) pair is just the path integral of the conservative part of the total electric field.
It's as if you decomposed a mathematical function into its odd and even parts and than claimed that the odd part is a different, 'less mathematic' definition of function.

You can easily see where that decomposition comes from by writing Faraday's law (use dS for the differential element of area to avoid confusion with the vector potential A, and use E_total instead of E to highlight that it is the resulting field, superposition of coulombian and induced fields). Then express B in terms of the vector potential A,  turn the surface integral on the right into a path integral along the surface contour using Stokes (or "the rotor's") theorem. Now take the integral on the right to the left, changing sign. Bring everything inside the same path integral. You are left with a field whose circulation along a closed path is always zero.



Hey, that's a conservative field! Well of course, you have stripped the induced - non conservative - part from the total field.


For a sanity check, refer to formula 11-3-3 on page 484 of Kraus' "Electromagnetics", 4th edition.

Congratulations, it can be very useful, but it's only half of the story.

I will come back to this post with formulas and drawings when I will be able to scan.
EDIT: yep, I did just that.
EDIT: fixed wrong double integral symbol.
Edit jan 22 - attached pictures

[ogden]

My way of looking at this box+resistor debate: if black box is able to produce AC 1V alone, I name it EMF source - disregarding it is transformer or just DC-fed AC generator. If I connect resistor, I close the loop, AC 1mA current is flowing and integral E.dl around the loop is zero. There is no magic, just quite simple logic and law of conservation of energy. Do you agree?

No. This is the root of your problem in understanding Faraday. You start with the assumption that integral of E.dl around a loop is zero to prove your thesis that integral of E.dl around the loop is zero.

*You* have problem, not me. You say that resistor with voltage drop have zero E field inside. That means that integral E.dl over it is zero meaning that it does not have voltage drop on it! It's paradox, don't you see?

KVL does not work anymore, and that's the point in saying that Kirchhoff is for the birds.

What?! TL;DR. You did not even answer. I will try again. Three Four questions: 1) Does KVL hold when inside box is DC battery? 2) Does KVL hold when inside box is DC-powered AC generator? 3) Does KVL hold when inside box is piezo-based 1V AC voltage generator? 4) Does KVL hold when inside box is transformer? I assume that I know your answer to last question and it is definite "no". I will appreciate your answers to remaining questions.

[ogden]

In short, the loop, the way it is, is unlumpable.

You just disproved yourself. Do you recall drawing lumped model of 1Ohm ring while ago - it was drawn as string of many series batteries + resistors.

Lumpable means that I can measure the same voltage regardless of the position of the voltmeters.

No! It does not mean only that!

Wikipedia is good enough https://en.wikipedia.org/wiki/Lumped_element_model

What we have is a loop antenna. There is a reason why loop antennas are connected by two-wire transmission lines.

Hold your horses! Don't introduce (radiating & leaky) antennas and transmission lines into talk about <= 300 Hz electromagnetism. [edit] Just for fun I tried first loop antenna calculator I can find. Results are: you can't put 300 Hz loop antenna on your table. Zero chances:

The specified conductor length of 1000000 meters is not ideal.
Conductor length should be between 121,250 and 242,500 meters at the specified frequency of 0.0003 MHz.

Other post of yours is unfortunately off-topic because it is not specified that there is Dr.Lewin's experiment in the box. It was defined that box is magnetically shielded and there is no EMF induced in the wires coming out of the box.

[Berni]

Why would you do that?  The changing magnetic field produces an electric field.  The electric field lines are circles around the magnetic field as shown in post #16:
https://www.eevblog.com/forum/chat/does-kirchhoffs-law-hold-disagreeing-with-a-master/msg2003414/#msg2003414

Yes exactly this is the underlying effect due to Einsteins specials relativity, its what makes magnetism and radio waves work.

But notice how you do not need to have a closed loop in order to have a changing magnetic field induce a electric field in a wire. Depending on what definition of voltage you use you get that there is 0V across the ends of the wire (The textbook definition) or that the voltage is equal to the magnetic EMF (The engineering definition).

So if one was to separate the ends of the wire by 5mm and expose it to a magnetic field fast enough to induce 30kV of EMF in it, would the voltage arc over and jump the gap? Or does the loop have to be closed before current could start flowing and then slowly separated to draw an arc?

If you have electric field lines that form a loop, as in this case, you have a nonconservative field.

Yes if you keep it as a loop, but given the wire length has only one path inside of it (Its just a wire afterall) then you can unroll the loop along the length of the wire and you have something that looks exactly like a normal electrostatic field (That im sure we all agree are conservative).

Another way to think about it is that the path was strictly defined by the single wire so we now also have a strictly defined voltage along that path.

So you don't have two voltmeters connected to same point showing different voltages?  This is what happens in the experiment.  Or are we supposed to imagine that it doesn't?

The two voltmeters will only measure the same voltage if the voltmeters and their connecting wires are outside the field region or somehow shielded or arranged so that the field does not affect the measurement.  Fine, but that is not this experiment.  This experiment is set up to demonstrate Faraday's law and non conservative fields.

Well yes and no.

Yes i fully agree that the two voltmeters in Dr. Lewins experiment must show different voltages. I even did the experiment myself and i have shown the circuit working just fine in SPICE. And i fully agree with Dr. Lewins calculated result on the whiteboard where he shows two different voltages across points A and B. There indeed are two diferent voltages there.(In the textbook voltage definition!)

What i don't agree is that they are connected to the same two points. Because the voltmeter probe wires have a significant effect on the result means they can't be ignored in the circuit model. The voltmeters actually being connected to the ends of two different inductors (that represent probe wires), so in my view they are not actually connected to the same two points. A wire no longer acts as a ideal equipotential connection between two points when exposed to certain conditions such as this one (Even if its a superconductor).

You seem to be claiming that a voltmeter only measures an electric field caused by separation of charges.  For example, you have to move charges on the gate of the input FET of the voltmeter to make it read a voltage, I guess is what you mean.

But the voltmeter can't tell you what separated the charges.  It could be a battery, it could be a solar cell, it could be a thermocouple, it could be a hall effect device, it could be a loop of wire surrounding a changing magnetic flux, or it could be a loop of wire rotating in a static magnetic field, to name a few possibilities.  The voltmeter cannot distinguish the charge separation caused by the electrostatic scalar potential from the charge separation caused by all the different other types.

In this case, the wires connecting the voltmeter are passing through a region that has a non-zero electric field.  That electric field, interacting with the charge in the wire, will give you a different reading on the voltmeter depending on the path the wires take through that field.

Exactly that's the point i was making. Voltmeters have no idea what caused the voltage, they just see the effects of the voltage. The other "non textbook" definition is this "effective voltage". If you use that definition then suddenly it does not matter if the source of the voltage is a magnetic field, piezzo effect, thermocuple or a battery. If the black box is pushing 1A trough a 1Ohm resistor across its terminals then the answer is always there is 1V across the terminals.

This is why this "effective voltage" is used in circuit analysis, SPICE and every other case where you would want to actually do math on circuits.

None of this goes against Faraday or Maxwell, its just a slightly different mathematical path trough it that never gives ambiguous multiple results that give rise to this paradox.

[Sredni]

*You* have problem, not me. You say that resistor with voltage drop have zero E field inside. That means that integral E.dl over it is zero meaning that it does not have voltage drop on it! It's paradox, don't you see?

Indeed. But I never said that the resistor has zero E-field inside. I said that the circulation of the conservative part of the E field is zero.
All the resulting field is concentrated at the resistors (one in this case), I've said that many times in my previous posts. The circulation of the resulting field is not zero. The circulation of the conservative part of the resulting field, on the other hand, is zero. But that does not mean either that the path integral of the conservative part of the field is zero inside the resistor, as well.

Try to see it this way: the primary coil is generating a time-varying quasi-static magnetic B field (spatially uniform going up and down). This is turn, links a non conservative time-varying electric field (E_induced) in space. This field is directed along concentric circles (changing direction with time) and goes like r inside the primary coil boundary, and like 1/r outside of it.
Then you place to copper of the secondary coil with the resistor(s) in it.
The free electrons of the copper instantly (with relaxation times of the order or 10^-14 seconds) redistribute themselves in order to satisfy the constitutive equations j = sigma E which means that if sigma is infinite, we will end with a resulting E field in the copper that is zero. This means that the conservative part of the resultant E field is compensating, obliterating it completely, the induced part in the copper. In the resistors sigma is low, so there is a significant resultant E field inside the resistor.

The conservative part of the E field is stronger in the copper as well. It has to be in order to cancel the induced part.

I will come back with some drawings.

What?! TL;DR. You did not even answer.

Well, you did not even read!!!

  I will try again.
1) Does KVL hold when inside box is DC battery? 2) Does KVL hold when inside box is battery-powered AC generator? 3) Does KVL hold when inside box is piezo-based 1V AC voltage generator? 4) Does KVL hold when inside box is transformer?

1) yes, outside and inside
2) it depends. Does the generator have a time-varying B field inside ? Is so and if the flux is neatly tucked inside the box, then 'extended KVL' (which is Faraday under disguise) will appear to work outside, but won't work inside when you cross the flux-varying region.
3) I am not familiar with piezoelectric generators, but if there is no dphi/dt involved we probably can treat them as batteries.
4) 'new KVL' which is Faraday under disguise will appear to work outside and fail miserably inside if you attempt to cross the flux-varying region.