#562 – Electroboom!

[thinkfat]

You do realize that this is not a circuit that is equivalent to the "Lewin Ring", or equivalent to the circuit I gave you to solve, right?

That was my challenge to you, the task was to find the voltages across the resistors and the wires:


And that's what you came up with:


Which of course brings up the question why you found it necessary to add a "transformer secondary" to find the voltage across "2R".

[jesuscf]

Regarding the "tiny voltage sources", the funny thing (which is a bit sad also) is that all the "experimental proof" by Jesse and e.g. Cyriel Mabilde are just cunning demonstrations of how to create paths that enclose a variable amount of magnetic flux. The sad part is that they're unable to see it and will keep claiming it is proof of "KVL holds". But in reality they're not measuring a gradual voltage build-up in the "Lewin Ring", but just the EMF induced in their measurement loop. This under the condition that there is only negligible current flowing through the ring and the measurement loop, of course.

How do you explain the situation where the whole ring is made of resistors, with virtually no wire?  Using the same model it is fairly easy to explain the voltage measured across each resistor:

https://www.eevblog.com/forum/amphour/562-electroboom!/msg3830852/#msg3830852

[thinkfat]

Regarding the "tiny voltage sources", the funny thing (which is a bit sad also) is that all the "experimental proof" by Jesse and e.g. Cyriel Mabilde are just cunning demonstrations of how to create paths that enclose a variable amount of magnetic flux. The sad part is that they're unable to see it and will keep claiming it is proof of "KVL holds". But in reality they're not measuring a gradual voltage build-up in the "Lewin Ring", but just the EMF induced in their measurement loop. This under the condition that there is only negligible current flowing through the ring and the measurement loop, of course.

How do you explain the situation where the whole ring is made of resistors, with virtually no wire?  Using the same model it is fairly easy to explain the voltage measured across each resistor:

https://www.eevblog.com/forum/amphour/562-electroboom!/msg3830852/#msg3830852

I'd say, your calculation of V1 and V2 is 5.555mV off. Also there's no reason why the voltage across any of the resistors should be different from what Ohm's Law says.

[Kalvin]

I would be a little concerned with my probing if I measured V1 something else than 0V across a 0 ohm wire [when |V2 + V3| > 0].

[thinkfat]

I would be a little concerned with my probing if I measured V1 something else than 0V across a 0 ohm wire [when |V2 + V3| > 0].

Really? I wouldn't be surprised at all.

[Kalvin]

I would be a little concerned with my probing if I measured V1 something else than 0V across a 0 ohm wire [when |V2 + V3| > 0].
<removed the image>

Really? I wouldn't be surprised at all.

I would, because that is violating the Ohm's law (U = I*R, and when R is 0 ohms, the voltage should be 0 too), which means that my probing is picking up some interference from some magnetic field affecting my measurements.

[Sredni]

I would be a little concerned with my probing if I measured V1 something else than 0V across a 0 ohm wire [when |V2 + V3| > 0].
<removed the image>

Really? I wouldn't be surprised at all.

I would, because that is violating the Ohm's law (U = I*R, and when R is 0 ohms, the voltage should be 0 too), which means that my probing is picking up some interference from some magnetic field affecting my measurements.

The reason you are surprised is because you gazed into the aby-- into the forbidden zone where the magnetic flux changes.
There, KVL dies.
(To clarify: voltage becomes path dependent and so you can have zero voltage ALONG the conductor in compliance with Ohm's law, and nonzero voltage ACROSS the terminals as you can measure with your voltmeter or oscilloscope)

But if you hide that part of space inside a black box, you will see a nonzero voltage 'across' a generator's terminals. Not surprising at all.

This is the point KVLers seem unable to understand. KVL dies right inside every transformer's secondary. We choose not to see that by not looking inside. But it's all just make-believe.

[Kalvin]

I would be a little concerned with my probing if I measured V1 something else than 0V across a 0 ohm wire [when |V2 + V3| > 0].
<removed the image>

Really? I wouldn't be surprised at all.

I would, because that is violating the Ohm's law (U = I*R, and when R is 0 ohms, the voltage should be 0 too), which means that my probing is picking up some interference from some magnetic field affecting my measurements.

The reason you are surprised is because you gazed into the aby-- into the forbidden zone where the magnetic flux changes.
There, KVL dies.
(To clarify: voltage becomes path dependent and so you can have zero voltage ALONG the conductor in compliance with Ohm's law, and nonzero voltage ACROSS the terminals as you can measure with your voltmeter or oscilloscope)

But if you hide that part of space inside a black box, you will see a nonzero voltage 'across' a generator's terminals. Not surprising at all.

This is the point KVLers seem unable to understand. KVL dies right inside every transformer's secondary. We choose not to see that by not looking inside. But it's all just make-believe.

I just wanted to point this out because in this particular case I know that I am measuring voltage V1 across a 0 ohm wire, but my instrument is showing some other voltage than zero violating the Ohm's law, which is a clear indication that there is something wrong in my probing, and the measurement setup is picking up interference from some magnetic field affecting my measurement.

[Sredni]

I just wanted to point this out because in this particular case I know that I am measuring voltage V1 across a 0 ohm wire, but my instrument is showing some other voltage than zero violating the Ohm's law, which is a clear indication that there is something wrong in my probing, and the measurement setup is picking up interference from some magnetic field affecting my measurement.

What I am saying is that

1. That voltage is NOT interference: it's exactly what that piece of system is supposed to do (hence your probing is correct: you put a voltmeter across the secondary of a transformer and the 12V you measure are not 'interference', despite the fact that if you follow the magnet wires inside the device all you see is... wire)
2. Ohm's law is obeyed (if you use the correct physics, of course)

[Kalvin]

I just wanted to point this out because in this particular case I know that I am measuring voltage V1 across a 0 ohm wire, but my instrument is showing some other voltage than zero violating the Ohm's law, which is a clear indication that there is something wrong in my probing, and the measurement setup is picking up interference from some magnetic field affecting my measurement.

What I am saying is that

1. That voltage is NOT interference: it's exactly what that piece of system is supposed to do (hence your probing is correct: you put a voltmeter across the secondary of a transformer and the 12V you measure are not 'interference', despite the fact that if you follow the magnet wires inside the device all you see is... wire)
2. Ohm's law is obeyed (if you use the correct physics, of course)

Yes, I do believe in transformers, even transformers with a single-turn secondary. You are totally right.

[jesuscf]

Regarding the "tiny voltage sources", the funny thing (which is a bit sad also) is that all the "experimental proof" by Jesse and e.g. Cyriel Mabilde are just cunning demonstrations of how to create paths that enclose a variable amount of magnetic flux. The sad part is that they're unable to see it and will keep claiming it is proof of "KVL holds". But in reality they're not measuring a gradual voltage build-up in the "Lewin Ring", but just the EMF induced in their measurement loop. This under the condition that there is only negligible current flowing through the ring and the measurement loop, of course.

How do you explain the situation where the whole ring is made of resistors, with virtually no wire?  Using the same model it is fairly easy to explain the voltage measured across each resistor:

https://www.eevblog.com/forum/amphour/562-electroboom!/msg3830852/#msg3830852

I'd say, your calculation of V1 and V2 is 5.555mV off. Also there's no reason why the voltage across any of the resistors should be different from what Ohm's Law says.

If that is the case, where is the induced EMF coming from?  Remember, there is no wire in the loop, only resistors, and the only access points to the circuit are the terminals of the resistors.

[thinkfat]

Regarding the "tiny voltage sources", the funny thing (which is a bit sad also) is that all the "experimental proof" by Jesse and e.g. Cyriel Mabilde are just cunning demonstrations of how to create paths that enclose a variable amount of magnetic flux. The sad part is that they're unable to see it and will keep claiming it is proof of "KVL holds". But in reality they're not measuring a gradual voltage build-up in the "Lewin Ring", but just the EMF induced in their measurement loop. This under the condition that there is only negligible current flowing through the ring and the measurement loop, of course.

How do you explain the situation where the whole ring is made of resistors, with virtually no wire?  Using the same model it is fairly easy to explain the voltage measured across each resistor:

https://www.eevblog.com/forum/amphour/562-electroboom!/msg3830852/#msg3830852

I'd say, your calculation of V1 and V2 is 5.555mV off. Also there's no reason why the voltage across any of the resistors should be different from what Ohm's Law says.

If that is the case, where is the induced EMF coming from?  Remember, there is no wire in the loop, only resistors, and the only access points to the circuit are the terminals of the resistors.

I'm not quite sure why the resistance along the path would make a difference. A series of resistors is not fundamentally different from a wire with the same resistance.

[jesuscf]

Regarding the "tiny voltage sources", the funny thing (which is a bit sad also) is that all the "experimental proof" by Jesse and e.g. Cyriel Mabilde are just cunning demonstrations of how to create paths that enclose a variable amount of magnetic flux. The sad part is that they're unable to see it and will keep claiming it is proof of "KVL holds". But in reality they're not measuring a gradual voltage build-up in the "Lewin Ring", but just the EMF induced in their measurement loop. This under the condition that there is only negligible current flowing through the ring and the measurement loop, of course.

How do you explain the situation where the whole ring is made of resistors, with virtually no wire?  Using the same model it is fairly easy to explain the voltage measured across each resistor:

https://www.eevblog.com/forum/amphour/562-electroboom!/msg3830852/#msg3830852

I'd say, your calculation of V1 and V2 is 5.555mV off. Also there's no reason why the voltage across any of the resistors should be different from what Ohm's Law says.

If that is the case, where is the induced EMF coming from?  Remember, there is no wire in the loop, only resistors, and the only access points to the circuit are the terminals of the resistors.

I'm not quite sure why the resistance along the path would make a difference. A series of resistors is not fundamentally different from a wire with the same resistance.

The idea of making the ring out of resistors, no wires, is to show that a resistor will behave as an induced voltage source in series with a resistance.   The net voltage across the terminals of the resistor is then the fraction of the total induced EMF (proportional to the total EMF and ratio of the length of the resistor vs the perimeter of the ring) minus the drop due to Ohms law (proportional to the resistance of the resistor and the current in the loop).

[Sredni]

How do you explain the situation where the whole ring is made of resistors, with virtually no wire?
...
If that is the case, where is the induced EMF coming from?  Remember, there is no wire in the loop, only resistors, and the only access points to the circuit are the terminals of the resistors.

What's the matter, pretty boy? You can't find your EMF?

(Sorry, I could not resist)

[jesuscf]

How do you explain the situation where the whole ring is made of resistors, with virtually no wire?
...
If that is the case, where is the induced EMF coming from?  Remember, there is no wire in the loop, only resistors, and the only access points to the circuit are the terminals of the resistors.

What's the matter, pretty boy? You can't find your EMF?

(Sorry, I could not resist)

I am glad I have this message to reply to!  Why?  Because I just found the solution to Lewin's problem in an Electromagnetics book!  And it doesn't look good at all for team Lewin.  The book is "Electromagnetics" by Branislav M. Notaros, pages 279 and 280 (the edition I have is from 2011).  The example is 6.6.  I have attached the pages to this message.  Sredni, now you have a bibliographical reference that can teach you how to calculate the voltage V_AD.  Don't forget to pay attention to Figure 6.10(b)!

EDIT: fixed page numbers

[bsfeechannel]

I am glad I have this message to reply to!  Why?  Because I just found the solution to Lewin's problem in an Electromagnetics book!  And it doesn't look good at all for team Lewin.  The book is "Electromagnetics" by Branislav M. Notaros, pages 279 and 280 (the edition I have is from 2011).  The example is 6.6.  I have attached the pages to this message.  Sredni, now you have a bibliographical reference that can teach you how to calculate the voltage V_AD.  Don't forget to pay attention to Figure 6.10(b)!

EDIT: fixed page numbers

Notaros says that if σ₁ = σ₂, then v_MN = 0. Incredible! You have two wire resistors with zero volts across their terminals and a current that is i_ind = e_ind / (R₁ + R₂).

Oh yeah, of course. What Notaros didn't tell you, because he obviously presumes you already understand the effect of non-conservative electric fields, is that this voltage is calculated along a straight line between points M and N.

Once defined this path you can model it using mundane circuit theory. No problem. No one is against that.

Build the circuit, immerse it in the varying magnetic field, place your meter right in the middle with the probes stretched along a straight line between M and N and you'll find v_MN he calculated.

[Sredni]

How do you explain the situation where the whole ring is made of resistors, with virtually no wire?
...
If that is the case, where is the induced EMF coming from?  Remember, there is no wire in the loop, only resistors, and the only access points to the circuit are the terminals of the resistors.

What's the matter, pretty boy? You can't find your EMF?

(Sorry, I could not resist)

I am glad I have this message to reply to!  Why?  Because I just found the solution to Lewin's problem in an Electromagnetics book!  And it doesn't look good at all for team Lewin.  The book is "Electromagnetics" by Branislav M. Notaros, pages 279 and 280 (the edition I have is from 2011).  The example is 6.6.  I have attached the pages to this message.  Sredni, now you have a bibliographical reference that can teach you how to calculate the voltage V_AD.  Don't forget to pay attention to Figure 6.10(b)!

Soooo...  the EMF is located on top of the resistors, it seems. Half just above R1, and half just above R2. How many centimeters, exactly? The text does not say. Can you locate with a bit more accuracy? No?
Or maybe...
Maybe that's the "equivalent circuit" that allows you to "solve the problem from the circuit theory point of view" and that is one of the introductory textbooks that do not explain clearly to their easily distracted audience what they intend for V. Oh, wait, but it does explain what V is! Page 269, eq. 6.18

Eq = - grad V

(Eq is what I call Ecoul) and V is... the electric scalar potential. Only half of the potentials required to describe the total electric field. And the text also says so explicitly on page 277, formula 6.43

E(t) = - dA/dt - grad V

"We see that both potentials are needed for E..."
(the same expression I used to express Etot = Eind + Ecoul, even if recently I decided to call the scalar electric potential phi, instead of V - exactly to avoid this kind of confusion you are having)

So...

where is exactly the EMF, again?

(Lewin problem is solved as an exercise on Purcell, Morin: Berkeley Physics vol 2, Electricity and Magnetism 3rd edition)

[jesuscf]

I am glad I have this message to reply to!  Why?  Because I just found the solution to Lewin's problem in an Electromagnetics book!  And it doesn't look good at all for team Lewin.  The book is "Electromagnetics" by Branislav M. Notaros, pages 279 and 280 (the edition I have is from 2011).  The example is 6.6.  I have attached the pages to this message.  Sredni, now you have a bibliographical reference that can teach you how to calculate the voltage V_AD.  Don't forget to pay attention to Figure 6.10(b)!

EDIT: fixed page numbers

Notaros says that if σ₁ = σ₂, then v_MN = 0. Incredible! You have two wire resistors with zero volts across their terminals and a current that is i_ind = e_ind / (R₁ + R₂).

Oh yeah, of course. What Notaros didn't tell you, because he obviously presumes you already understand the effect of non-conservative electric fields, is that this voltage is calculated along a straight line between points M and N.

Once defined this path you can model it using mundane circuit theory. No problem. No one is against that.

Build the circuit, immerse it in the varying magnetic field, place your meter right in the middle with the probes stretched along a straight line between M and N and you'll find v_MN he calculated.

Oh boy bsfeechannel, you are very special, but not in a good way!  In figure 6.10(b) what is the voltage between nodes v_MN if R₁ is equal to R₂?

[jesuscf]

How do you explain the situation where the whole ring is made of resistors, with virtually no wire?
...
If that is the case, where is the induced EMF coming from?  Remember, there is no wire in the loop, only resistors, and the only access points to the circuit are the terminals of the resistors.

What's the matter, pretty boy? You can't find your EMF?

(Sorry, I could not resist)

I am glad I have this message to reply to!  Why?  Because I just found the solution to Lewin's problem in an Electromagnetics book!  And it doesn't look good at all for team Lewin.  The book is "Electromagnetics" by Branislav M. Notaros, pages 279 and 280 (the edition I have is from 2011).  The example is 6.6.  I have attached the pages to this message.  Sredni, now you have a bibliographical reference that can teach you how to calculate the voltage V_AD.  Don't forget to pay attention to Figure 6.10(b)!

Soooo...  the EMF is located on top of the resistors, it seems. Half just above R1, and half just above R2. How many centimeters, exactly? The text does not say. Can you locate with a bit more accuracy? No?
Or maybe...
Maybe that's the "equivalent circuit" that allows you to "solve the problem from the circuit theory point of view" and that is one of the introductory textbooks that do not explain clearly to their easily distracted audience what they intend for V. Oh, wait, but it does explain what V is! Page 269, eq. 6.18

Eq = - grad V

(Eq is what I call Ecoul) and V is... the electric scalar potential. Only half of the potentials required to describe the total electric field. And the text also says so explicitly on page 277, formula 6.43

E(t) = - dA/dt - grad V

"We see that both potentials are needed for E..."
(the same expression I used to express Etot = Eind + Ecoul, even if recently I decided to call the scalar electric potential phi, instead of V - exactly to avoid this kind of confusion you are having)

So...

where is exactly the EMF, again?

(Lewin problem is solved as an exercise on Purcell, Morin: Berkeley Physics vol 2, Electricity and Magnetism 3rd edition)

Can you calculate V_AD yet?  The book is giving you the KVL solution, you just have to put the values and evaluate!  Just be careful, because in Lewin's original problem, the one with one loop, the positions of R₁ and R₂ are swapped.  Let me fix it for you:

\$
V_{AD}  = \frac{{\left( {R_2  - R_1 } \right)EMF}}{{2\left( {R_1  + R_2 } \right)}} = \frac{{\left( {900\Omega  - 100\Omega } \right)1V}}{{2\left( {100\Omega  + 900\Omega } \right)}} = ?
\$

Or you can take a look on how I solved it also using KVL:

https://www.eevblog.com/forum/amphour/562-electroboom!/msg3828206/#msg3828206

[bsfeechannel]

Oh boy bsfeechannel, you are very special, but not in a good way!  In figure 6.10(b) what is the voltage between nodes v_MN if R₁ is equal to R₂?

Who cares about figure 6.10(b)? It's an equivalent circuit. It's an imaginary construct, a math trick. It doesn't exist.

The real circuit is described by figure 6.10(a). There is where you'll come a gutser.

[jesuscf]

Oh boy bsfeechannel, you are very special, but not in a good way!  In figure 6.10(b) what is the voltage between nodes v_MN if R₁ is equal to R₂?

Who cares about figure 6.10(b)? It's an equivalent circuit. It's an imaginary construct, a math trick. It doesn't exist.

The real circuit is described by figure 6.10(a). There is where you'll come a gutser.

I see you just moved to Egypt, living by the shores of the Nile! 

[thinkfat]

PS: "Team KVL", don't get your hopes up too early

[jesuscf]

PS: "Team KVL", don't get your hopes up too early

The video is correct.  He even calculates the correct and unique voltage between V_AD (V_JX at 33:38, something Lewin was unable to do), followed with an explanation of what happens if use  KVL without including the induced EMF (what Lewin did).  So, what is your point exactly?

[Kalvin]

I just wanted to point this out because in this particular case I know that I am measuring voltage V1 across a 0 ohm wire, but my instrument is showing some other voltage than zero violating the Ohm's law, which is a clear indication that there is something wrong in my probing, and the measurement setup is picking up interference from some magnetic field affecting my measurement.

What I am saying is that

1. That voltage is NOT interference: it's exactly what that piece of system is supposed to do (hence your probing is correct: you put a voltmeter across the secondary of a transformer and the 12V you measure are not 'interference', despite the fact that if you follow the magnet wires inside the device all you see is... wire)
2. Ohm's law is obeyed (if you use the correct physics, of course)

This is what I came up with when thinking about this circuit and when KVL holds and doesn't hold.

In figure a) the circuit is inside a magnetic field, the voltage between points A and D depends on path, it is not possible to create a lumped model for the source creating the 1mA current, and KVL doesn't hold.

In figure b) the circuit is only partially inside a magnetic field creating a transformer with a single-turn secondary, thus it is possible to create a lumped model for the source creating the 1mA current like in figure c, and KVL holds.

Figure c is a lumped model for figure b with a (voltage) source creating the 1mA current when the magnetic field is increasing, and KVL holds.

Is my reasoning correct?

Edit: Added some clarifications.

[thinkfat]

The video is correct.  He even calculates the correct and unique voltage between V_AD (V_JX at 33:38, something Lewin was unable to do), followed with an explanation of what happens if use  KVL without including the induced EMF (what Lewin did).  So, what is your point exactly?

Well, it's a Nothing Burger, how our host would be calling it. While you can calculate a voltage for Vad, it is more mathturbation than anything else.

It is not trivial to compute for other than artificial setups with simple paths, perfect symmetry or at least uniformity of the electric field, and actually observing it is also quite complicated, because all measurements would again be taken along paths through a non-conservative electric field, which adds another dimension of error.

So, you'd compute a value for Vad, making various assumptions about the fields involved, then calculate a path based on the same assumptions, then try to make your measurements and calculations match in reality. But at no point you would be sure if any of it is correct. Plus there are geometries where it is not possible to measure it.

The same guy has a good video on the merits of PD to "salvage KVL" on his channel. It is quite thorough IMHO, though a bit long winded, but still worth watching in detail.

https://youtu.be/I1kYKF2x9Ns