Author Topic: Does Kirchhoff's Law Hold? Disagreeing with a Master  (Read 61124 times)

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Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #475 on: December 26, 2018, 11:14:07 pm »
On the contrary, that's exactly where the previous discussion with ogden led, with them concluding that Maxwell's equations are only correct in specific situations (superconducting rings, I believe is what was said). That's another way of saying that sometimes they are wrong.

Typical strategy of forum trolls and "sofa experts" BTW - take just single sentence out of context, twist it as much as possible, draw conclusions out of this new "twisted meaning" to prove own agenda.

For those unaware this is what I ACTUALLY said:

I do not talk about abstract KVL and Maxwell equation "cases". I talk about equations that describes circuit of experiment. Everything seemingly is ok with KVL simple int E.dl = 0, yet I would prefer to split it into EMF source and load, as you already did - thank you for that. Problem arises when Dr.Lewin use plain Maxwell's equation and say that it miraculously tells everything about inner loop of his experiment. I disagree. Maxwell's equation is just EMF source part! Where's physics of load (resistors) in Maxwells equation? If you leave it like that, then it is indeed violation of conservation of energy. Plain Maxwell's equation can be used only to describe superconductive ring (w/o embedded resistors) placed in changing magnetic flux.
« Last Edit: December 26, 2018, 11:17:32 pm by ogden »
 

Offline Psi

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #476 on: December 26, 2018, 11:17:06 pm »
I disagree with the thread title.

Walter Lewin used to be a master. Then he started flinging poo at good people.
So i vote we take Master status from him.  :-DD
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Offline beanflying

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #477 on: December 26, 2018, 11:21:47 pm »
I disagree with the thread title.

Walter Lewin used to be a master. Then he started flinging poo at good people.
So i vote we take Master status from him.  :-DD

So is that why this thread travels toward flinging poo AGAIN emulating the Master?  :palm:

The last few pages and a lot of the thread have been really interesting and in between the poo this thread is more informative than my Uni lecturers way back in the dim dark past ever were.
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Offline mhz

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #478 on: December 26, 2018, 11:31:56 pm »
@berni Ah, perfect. Ok so my point is this, you've modeled the circuit with lumped elements and that gives something that looks correct, but if you set this experiment up in real life, do you really think you'll measure any significant voltage across any of L2, L3, L4 or L5 like spice would tell you is there? The coupled flux isn't confined in those points so you won't. This is where our "modified" KVL breaks down.

We saw Mehdi struggle with this in his first video. This is one of the biggest points of confusion, that there must be "voltage in the wire". I don't remember who, bsfeechannel?, was earlier arguing that this can't be modeled 100% correctly in spice because spice only knows about lumped elements. No matter how many inductors you split the mutual inductance into in spice, it will give an incorrect answer if you use it try to use it to measure a voltage along the wire. The reason is that in the actual experiment, there aren't any lumped inductors, the linked flux in the secondary is not confined to any specific two terminals.

Also, I'd argue that the leads should not be modeled as mutual inductance as they aren't supposed to link any of the flux in the center loop. (This is how Romer defines the experiment).

p.s. I'm not going to be bothering to respond to others that aren't attempting to have a productive conversation

[edit: clarify who I'm talking to since a few posts happened in the interim]
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #479 on: December 26, 2018, 11:58:55 pm »
Ok so my point is this, you've modeled the circuit with lumped elements and that gives something that looks correct, but if you set this experiment up in real life, do you really think you'll measure any significant voltage across any of L2, L3, L4 or L5 like spice would tell you is there? The coupled flux isn't confined in those points so you won't.

So you say that Kirk T. McDonald is wrong in chapter "2.3 Comments" (page 10)? Please tell where and why he is wrong. Prove him wrong.

Excerpt:

« Last Edit: December 27, 2018, 12:09:15 am by ogden »
 

Offline rfeecs

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #480 on: December 27, 2018, 12:48:13 am »
Ok so my point is this, you've modeled the circuit with lumped elements and that gives something that looks correct, but if you set this experiment up in real life, do you really think you'll measure any significant voltage across any of L2, L3, L4 or L5 like spice would tell you is there? The coupled flux isn't confined in those points so you won't.

So you say that Kirk T. McDonald is wrong in chapter "2.3 Comments" (page 10)? Please tell where and why he is wrong. Prove him wrong.

Excerpt:



Did you even read that section?
Quote
Suppose a voltmeter were connected to two points on the upper wire between resistors 1
and 2, as shown in the sketch below. The voltmeter loop is not coupled to the solenoid, so
there is no (or extremely little) EMF induced in this loop, and hence I1 = 0, and the meter
reading would be Vmeter = 0.

This agrees with what @mhz said.

In that section, Dr. McDonald is pointing out that a voltmeter does not read the difference in scalar potential.  He is making up his own definitions of "voltage drop" and "EMF" in the presence of a changing magnetic field.
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #481 on: December 27, 2018, 12:49:54 am »
Do you see the above two equations as equivalent term by term?

I did say "Don't you find it similar". I did not say equivalent, especially term by term. One more illustration of troll & "coach expert" tactics. You just cherrypick out of context whatever you find convenient for you, ignoring what I was ACTUALLY talking about: circuit that has EMF source and load, that both equations describes such and are similar in such sense.
 

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #482 on: December 27, 2018, 12:53:11 am »
Besides, the difference in Fermi levels is the barrier potential (if we agree on how to treat the sign). I guess [NOTE: I should have written 'thought', instead of 'guessed') you were the one saying that you cannot read it with a voltmeter.
With no bias, in equilibrium, the Fermi levels on both terminals are equal, so zero voltage.  Refer to figure 7.3 of Neamen.
I was talking about the Fermi levels of the separated materials, what Neamen call 'intrinsic Fermi levels' when referring to the compound structure. At thermodynamic equilibrium there are no longer Fermi levelS. But yes, out of equilibrium the one Fermi level of the structure splits in separate levels and the voltage corresponding to that difference is what is measured.
Quote
You have to wonder if "potential barrier" is the right phrase for an ohmic contact, which should have little or no barrier.

I guess it is the right denomination, since there are potential barriers nonetheless. In some cases they are not very hard to overcome thanks to the field associated with the charge developed at the contact surface , in others they are very high and steep, but electrons can easily tunnel through them. See Muller and Kamins for an explanation of both versions (section 3.4 in the second edition).

Quote
So what's your point?  Are you saying that there is no net electrostatic potential across the diode terminals?

I am saying that whatever potential is there, it will be canceled by the contact potentials created by the placement of the probes.
But anyway, by putting the limelight on other sources of emf, you raised an interesting point that might benefit the original discussion: could it be that the emf due to changing dB/dt is different from all other emfs? Faraday's law seems to point there.
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Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #483 on: December 27, 2018, 12:54:51 am »
Did you even read that section?

Did you? Carefully enough?

Quote
Quote
Suppose a voltmeter were connected to two points on the upper wire between resistors 1
and 2, as shown in the sketch below. The voltmeter loop is not coupled to the solenoid, so
there is no (or extremely little) EMF induced in this loop, and hence I1 = 0, and the meter
reading would be Vmeter = 0.

This agrees with what @mhz said.

In that section, Dr. McDonald is pointing out that a voltmeter does not read the difference in scalar potential.

From which planet did you just arrive? We are not talking about scalar potentials but EMF. @mhz said that voltmeter will not measure any EMF as Spice do.

« Last Edit: December 27, 2018, 01:00:26 am by ogden »
 

Online Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #484 on: December 27, 2018, 12:56:08 am »
To not repeat a certain arrangement i will also answer this for both definitions of voltage:

A) For definition "Voltage is the integral of all forces pushing on electrons along a given path connecting two points":
The inductors L2 L3 L4 L5 have zero voltage across them at all times (Zero resistance). Any EMF voltage induced in the wire by the magnetic field is instantly countered by the charge separation of electrons.

B) For definition "Voltage is the difference in charge density between two points" (This is what real life voltmeters show)
The inductors L2 L3 L4 L5 have a voltage drop that sums up to the same voltage as the total voltage drop on the resistors.  This voltage in the wire is caused by charge separation pushing electrons towards one end of a wire, resulting in more electrons on one end hence higher voltage on one end.



The proper textbook definition (A) is the one that is used in Dr. Lewins example where he gets two different voltages across the same two points. This is fully correct and there are indeed two voltages there. The reason for the two solutions is that this voltage is including the EMF voltage from the magnetic field, yet the loop is not closed yet as the two points are in different locations in space. Depending on how this path is closed results in a different solution for the EMF voltage and this changes the result. Hence why voltage is path dependent.

So why are we using the other definition (B) if its clearly wrong? Well turns out in real life its rather tricky to measure the voltage according to that definition. Electrical components (such as resistors) represent only a small part of the path around the magnetic fields loop area, because of this all of the electron pushing work is done by the electrical field (caused by charge seperation). Turns out all the voltmeters are actually devices that measure current trough the internal resistance and display the voltage required to push that current. The density of electrons at a point in space can only have 1 single defined value hence why these voltages always have one value rather than multiple. All of this simply makes this definition (B) more useful and as such is used in circuit analysis and spice simulations. Since circuit analysis uses it that forces KVL to use it too.

In the absence of a changing magnetic field around the circuit both definitions of voltage have the same value so it doesn't matter what you use. But in a magnetic field it does matter a lot.

As for the lumped model, it only hides what you want it to hide. Many many tiny inductors in series act the same as one big inductor, so it makes it easier to use a lumped version. Once you lump a segment of a circuit all voltages within the lumped part become meaningless, this is why lumping the inductor as a single one causes problems in this example. The lumping procedure also lumped all our points of interest and messed them up, they no longer show true voltages. However anything outside the area we just lumped is preserved. The rest of the circuit doesn't care how many inductors there are, it just sees a set inductance value across the points. So by only lumping sections of the wire that have no points of interest we preserve all the points we want to measure. Hence why all the points on the ends of the inductors have the correct voltage values.

All wires have some amount of mutual inductance to each other as long as they are not placed at perfect right angles. In this case there is more to it however. The inductors L2 L3 L4 L5 are actually a single inductor (single whole turn of the loop) that has been sliced up in to 4 parts. Because they are part of the same inductors means they share the same magnetic flux and hence are highly coupled inductors. The inductors L6 L7 L8 L9 are another inductor that has been sliced up in to quarters, but since the wire follows the same path as the inner cirucit means that any flux passing trough that loop passes trough this one too. This means all of them are coupled to each other (aka a transformer). The solenoid coil in the middle is also having the same magnetic flux pass trough it hence why its also coupled. In my simulation it has a ideal coupling coefficient of 1, but in reality it would be lower because solenoid is smaller than the loop so some of the flux escapes.

So yeah we are mostly looking at two sides of the same coin here. Its two different ways of explaining the same thing.
 
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Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #485 on: December 27, 2018, 01:17:56 am »
Do you see the above two equations as equivalent term by term?

I did say "Don't you find it similar". I did not say equivalent, especially term by term. One more illustration of troll & "coach expert" tactics. You just cherrypick out of context whatever you find convenient for you, ignoring what I was ACTUALLY talking about: circuit that has EMF source and load, that both equations describes such and are similar in such sense.

I am just trying to pinpoint what I believe to be the origin of your misconception.
Back in one of your exchanges with MHz, you wrote:

Quote
Integral E.dl = 0 of Kirchoff's circuit rule includes *both* EMF source and load. Integral E.dl of Maxwell's equation includes/describes only EMF *source*.

Have you ever tried to compute the integral of E.dl of a RLC circuit with a generator - a lumped circuit, just to see that Kirchhoff and field theory can agree if there is no dB/dt area enclosed by the circuit path? It might clear a lot of things up before trying to attack a non-lumped circuit such as Lewin's.
All instruments lie. Usually on the bench.
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #486 on: December 27, 2018, 01:29:25 am »
Have you ever tried to compute the integral of E.dl of a RLC circuit with a generator

No. Have you?

I use L(di/dt) with real inductors/transformers having inductance and saturation current specs. Works for me well.
 

Offline mhz

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #487 on: December 27, 2018, 01:35:15 am »
Ok so my point is this, you've modeled the circuit with lumped elements and that gives something that looks correct, but if you set this experiment up in real life, do you really think you'll measure any significant voltage across any of L2, L3, L4 or L5 like spice would tell you is there? The coupled flux isn't confined in those points so you won't.

So you say that Kirk T. McDonald is wrong in chapter "2.3 Comments" (page 10)? Please tell where and why he is wrong. Prove him wrong.

Excerpt:




I didn't say that I disagree with McDonald. Unlike many of us here, I'm not generally inclined to disagree with Physics professors. As rfeecs pointed out, McDonald echoes what I said to Berni, that the voltmeter will measure zero (whereas in his spice model, with the loop inductance split up into four lumped inductors, measuring across the inductor will give a nonzero value). 

To be honest, some of McDonald's writing is presented at a level beyond the one I'm at. I'm going to work through Griffiths and hopefully come back to it able to fully digest what he has to say. That said, his footnote number 8 in http://physics.princeton.edu/~mcdonald/examples/voltage.pdf seems to be expressing what I've been trying to get across in my discussion with Berni.
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #488 on: December 27, 2018, 01:44:52 am »
As rfeecs pointed out, McDonald echoes what I said to Berni, that the voltmeter will measure zero

He says exactly opposite: "the result Vmeter = 0 is appealing in that we might naïvely expect the “voltage drop” to be zero between points along a good/perfect conductor.".

You really shall read last paragraph of mentioned chapter carefully.
 

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #489 on: December 27, 2018, 01:45:20 am »
Have you ever tried to compute the integral of E.dl of a RLC circuit with a generator

No. Have you?

Yes. It clears things up. A lot.
You should too, before embarking in discussions such as this.
Luckily for you, you can find a well presented walk-through in Ramo, Whinnery and VanDuzer "Fields and Waves in Communication Electronics". Chapter 4, "The electromagnetics of circuits".
Find a library that has this book, and read the first few pages of chapter 4.

Quote
I use L(di/dt) with real inductors/transformers having inductance and saturation current specs.

You should try to see where that expression comes from.

Quote
Works for me well.

Does not look like that, from this side of the monitor.
All instruments lie. Usually on the bench.
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #490 on: December 27, 2018, 01:54:35 am »
Does not look like that, from this side of the monitor.

You shall visit optometrist then. Thank you for suggestions anyway ;)
 

Offline mhz

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #491 on: December 27, 2018, 02:10:03 am »
To not repeat a certain arrangement i will also answer this for both definitions of voltage:

A) For definition "Voltage is the integral of all forces pushing on electrons along a given path connecting two points":
The inductors L2 L3 L4 L5 have zero voltage across them at all times (Zero resistance). Any EMF voltage induced in the wire by the magnetic field is instantly countered by the charge separation of electrons.

B) For definition "Voltage is the difference in charge density between two points" (This is what real life voltmeters show)
The inductors L2 L3 L4 L5 have a voltage drop that sums up to the same voltage as the total voltage drop on the resistors.  This voltage in the wire is caused by charge separation pushing electrons towards one end of a wire, resulting in more electrons on one end hence higher voltage on one end.

Where does definition B come from? Charge density and Voltage don't even have the same units. [C/m3] vs [J/C]

Quote
The proper textbook definition (A) is the one that is used in Dr. Lewins example where he gets two different voltages across the same two points. This is fully correct and there are indeed two voltages there. The reason for the two solutions is that this voltage is including the EMF voltage from the magnetic field, yet the loop is not closed yet as the two points are in different locations in space. Depending on how this path is closed results in a different solution for the EMF voltage and this changes the result. Hence why voltage is path dependent.

So why are we using the other definition (B) if its clearly wrong? Well turns out in real life its rather tricky to measure the voltage according to that definition. Electrical components (such as resistors) represent only a small part of the path around the magnetic fields loop area, because of this all of the electron pushing work is done by the electrical field (caused by charge seperation). Turns out all the voltmeters are actually devices that measure current trough the internal resistance and display the voltage required to push that current. The density of electrons at a point in space can only have 1 single defined value hence why these voltages always have one value rather than multiple. All of this simply makes this definition (B) more useful and as such is used in circuit analysis and spice simulations. Since circuit analysis uses it that forces KVL to use it too.

In the absence of a changing magnetic field around the circuit both definitions of voltage have the same value so it doesn't matter what you use. But in a magnetic field it does matter a lot.

As for the lumped model, it only hides what you want it to hide. Many many tiny inductors in series act the same as one big inductor, so it makes it easier to use a lumped version. Once you lump a segment of a circuit all voltages within the lumped part become meaningless, this is why lumping the inductor as a single one causes problems in this example. The lumping procedure also lumped all our points of interest and messed them up, they no longer show true voltages. However anything outside the area we just lumped is preserved. The rest of the circuit doesn't care how many inductors there are, it just sees a set inductance value across the points. So by only lumping sections of the wire that have no points of interest we preserve all the points we want to measure. Hence why all the points on the ends of the inductors have the correct voltage values.

I disagree. You can split the total mutual inductance M of the loop into two strings of as many inductors as you want in spice. The value you measure in spice will not be the  actual scalar voltage potential between the ends of the resistors (which is approximately zero as measured by the voltmeter). Lumping can't be done in this kind of circuit  in spice without creating false outcomes.

Quote
All wires have some amount of mutual inductance to each other as long as they are not placed at perfect right angles. In this case there is more to it however. The inductors L2 L3 L4 L5 are actually a single inductor (single whole turn of the loop) that has been sliced up in to 4 parts. Because they are part of the same inductors means they share the same magnetic flux and hence are highly coupled inductors. The inductors L6 L7 L8 L9 are another inductor that has been sliced up in to quarters, but since the wire follows the same path as the inner cirucit means that any flux passing trough that loop passes trough this one too. This means all of them are coupled to each other (aka a transformer). The solenoid coil in the middle is also having the same magnetic flux pass trough it hence why its also coupled. In my simulation it has a ideal coupling coefficient of 1, but in reality it would be lower because solenoid is smaller than the loop so some of the flux escapes.

I see now that you're trying to model the mutual inductance of the "outer loop" i.e. the path formed by the two measurement loops, but not going through R1 and R2. You've arranged the coupling dots in a way that the inner inductors and outer inductors cancel each other out in a way that satisfies there being no flux coupling in the two measurement loops.

Quote
So yeah we are mostly looking at two sides of the same coin here. Its two different ways of explaining the same thing.

Mostly, with perhaps some disagreements on how far you can go with lumped models of inherently non-lumped reality, and the part I'm not following up there about charge density.

[Edit1: fix quoting]
 

Offline mhz

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #492 on: December 27, 2018, 02:26:03 am »
As rfeecs pointed out, McDonald echoes what I said to Berni, that the voltmeter will measure zero

He says exactly opposite: "the result Vmeter = 0 is appealing in that we might naïvely expect the “voltage drop” to be zero between points along a good/perfect conductor.".

You really shall read last paragraph of mentioned chapter carefully.

I have been, and I have to admit I'm quite perplexed with it. In particular equation 35 doesn't seem correct to me. For example, if we assume a point in time where I = 1mA, R1 = 100ohm, R2 = 900ohm, R = 1Mohm and I1 ≈ 0 (as he states) then he seems to be saying 0V = -1V. If someone can help clear this up for me I'd appreciate it.



[Edit 1: fix typos]
« Last Edit: December 27, 2018, 02:29:25 am by mhz »
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #493 on: December 27, 2018, 02:42:58 am »
I have been, and I have to admit I'm quite perplexed with it. In particular equation 35 doesn't seem correct to me. For example, if we assume a point in time where I = 1mA, R1 = 100ohm, R2 = 900ohm, R = 1Mohm and I1 ≈ 0 (as he states) then he seems to be saying 0V = -1V. If someone can help clear this up for me I'd appreciate it.

Equation is correct indeed. Thou it is counterintuitive. Wire segment "a-b" is part of *both* loops - loop of leads and loop containing R1 and R2. Leads loop does not have any EMF induced, source of "voltage drop" in particular wire segment is EMF generated only in the loop containing resistors and it is obviously mirrored in the a-b segment of the leads loop.  All this is not that important. Important part is: voltage will be observed which is contrary to your statement.

[edit] No, you dont'use Ohms law here. You shall use Maxwell's equation to calculate EMF generated in wire segment "a-b"

[edit1] Seems, I know where your frustration comes from. By saying I1=0 he means that current induced by EMF is zero because there is no EMF in the voltmeter leads. On the other hand current will be flowing through voltmeter due to potential difference "voltage drop" between points a & b. This is my explanation. Hope it helps.

[edit2] Obviously I agree that it is kinda incorrect to indicate I1=0, at the same time saying that voltmeter having finite resistance measures something that differs from 0V.
« Last Edit: December 27, 2018, 03:08:38 am by ogden »
 

Offline mhz

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #494 on: December 27, 2018, 03:03:21 am »
I have been, and I have to admit I'm quite perplexed with it. In particular equation 35 doesn't seem correct to me. For example, if we assume a point in time where I = 1mA, R1 = 100ohm, R2 = 900ohm, R = 1Mohm and I1 ≈ 0 (as he states) then he seems to be saying 0V = -1V. If someone can help clear this up for me I'd appreciate it.

Equation is correct indeed. Thou it is counterintuitive. Wire segment "a-b" is part of *both* loops - loop of leads and loop containing R1 and R2. Leads loop does not have any EMF induced, source of "voltage drop" in particular wire segment is EMF generated only in the loop containing resistors.  All this is not that important. Important part is: voltage will be observed which is contrary to your statement.

[edit] No, you dont'use Ohms law here. You shall use Maxwell's equation to calculate EMF generated in wire segment "a-b"

[edit1] Seems, I know where your frustration comes from. By saying I1=0 he means that current induced by EMF is zero because there is no EMF in the voltmeter leads. On the other hand current will be flowing through voltmeter due to potential difference "voltage drop" between points a & b. This is my explanation. Hope it helps.

Thanks for the reply. I annotated his drawing to make my confusion clearer (heh).



The equation in question, repeated is


Green is the path integral on the left side of the equation and red is the path integral on the right side of the equation. Would be great if you can fix it so that the equation balances. By all means use Maxwell's equations to get there.

I suspect I'm not going to fully understand McDonald until I've completely grokked his discussions on vector potential.

 

Offline mhz

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #495 on: December 27, 2018, 03:06:22 am »
The most confusing thing about this is that he seems to be suggesting that there is a scalar potential here that is independent of the path, despite that pesky time varying B field.

I have been, and I have to admit I'm quite perplexed with it. In particular equation 35 doesn't seem correct to me. For example, if we assume a point in time where I = 1mA, R1 = 100ohm, R2 = 900ohm, R = 1Mohm and I1 ≈ 0 (as he states) then he seems to be saying 0V = -1V. If someone can help clear this up for me I'd appreciate it.

Equation is correct indeed. Thou it is counterintuitive. Wire segment "a-b" is part of *both* loops - loop of leads and loop containing R1 and R2. Leads loop does not have any EMF induced, source of "voltage drop" in particular wire segment is EMF generated only in the loop containing resistors.  All this is not that important. Important part is: voltage will be observed which is contrary to your statement.

[edit] No, you dont'use Ohms law here. You shall use Maxwell's equation to calculate EMF generated in wire segment "a-b"

[edit1] Seems, I know where your frustration comes from. By saying I1=0 he means that current induced by EMF is zero because there is no EMF in the voltmeter leads. On the other hand current will be flowing through voltmeter due to potential difference "voltage drop" between points a & b. This is my explanation. Hope it helps.

Thanks for the reply. I annotated his drawing to make my confusion clearer (heh).



The equation in question, repeated is


Green is the path integral on the left side of the equation and red is the path integral on the right side of the equation. Would be great if you can fix it so that the equation balances. By all means use Maxwell's equations to get there.

I suspect I'm not going to fully understand McDonald until I've completely grokked his discussions on vector potential.
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #496 on: December 27, 2018, 03:26:09 am »
I suspect I'm not going to fully understand McDonald until I've completely grokked his discussions on vector potential.




Oh, my... You use Ohms law to claim that voltage between a-b is 1V?  :palm:

I don't even know what to say. How dare you pretend that you mastered Maxwell's equations?!

First, you have to know azimuthal angle between two points, a & b. Let's assume it is PI/4 (45 degrees). According to your data EMF of the resistor loop is 1V. We put 1V and Pi/4 into equation (34): 1V*(Pi/4)/(2*Pi) = 1/8 V. Voltmeter shall show 0.125V in such case (when angle is 45 degrees).

[edit] Forget about "I1=0". It is misleading or even incorrect. All you shall know - there's no EMF induced in the voltmeter leads.
« Last Edit: December 27, 2018, 03:36:40 am by ogden »
 

Offline mhz

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #497 on: December 27, 2018, 03:43:44 am »
I suspect I'm not going to fully understand McDonald until I've completely grokked his discussions on vector potential.




Oh, my... You use Ohms law to claim that voltage between a-b is 1V?  :palm:

I don't even know what to say. How dare you pretend that you mastered Maxwell's equations?!

First, you have to know azimuthal angle between two points, a & b. Let's assume it is PI/4 (45 degrees). According to your data EMF of the resistor loop is 1V. We put 1V and Pi/4 into equation (34): 1V*(Pi/4)/(2*Pi) = 1/8 V. Voltmeter shall show 0.125V in such case (when angle is 45 degrees).

[edit] Forget about "I1=0". It is misleading or even incorrect.

I don't pretend anything, and if you continue using these emoticons or abusive/patronizing language I will just start ignoring you again. Please drop them if you want to continue discussing like adults.

Int E•dl in the wires is 0V (no E fields in perfect conductors, or next to none in real conductors in which case we approximate to 0) and in the resistor is equivalent to I*R (if not then what do you think the contribution of Int E•dl through the resistor is?).

Your response contradicts several of the things McDonald says in his paper, namely that the voltmeter will read 0V (not 0.125V) and that I1 = 0 is misleading/incorrect. So now who should I believe, you or McDonald?

Anybody else wanna take a crack at this?
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #498 on: December 27, 2018, 03:45:49 am »
Have you ever tried to compute the integral of E.dl of a RLC circuit with a generator - a lumped circuit, just to see that Kirchhoff and field theory can agree if there is no dB/dt area enclosed by the circuit path? It might clear a lot of things up before trying to attack a non-lumped circuit such as Lewin's.

See discussion above about McDonald's paper, then think again how non-lumped circuit is Dr.Lewin's experiment.
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #499 on: December 27, 2018, 03:58:50 am »
Int E•dl in the wires is 0V (no E fields in perfect conductors, or next to none in real conductors in which case we approximate to 0) and in the resistor is equivalent to I*R (if not then what do you think the contribution of Int E•dl through the resistor is?).

Faradays law? ... Maybe? From your blog BTW:



Quote
Your response contradicts several of the things McDonald says in his paper, namely that the voltmeter will read 0V (not 0.125V) and that I1 = 0 is misleading/incorrect. So now who should I believe, you or McDonald?

Anybody else wanna take a crack at this?

I do not contradict with McDonald. You do. I will repeat again what he says: "the result Vmeter = 0 is appealing in that we might naïvely expect the “voltage drop” to be zero between points along a good/perfect conductor." He even shows equation (34) how to calculate voltage between a-b points.

I already explained what I think about I1=0:

[edit1] Seems, I know where your frustration comes from. By saying I1=0 he means that current induced by EMF is zero because there is no EMF in the voltmeter leads. On the other hand current will be flowing through voltmeter due to potential difference "voltage drop" between points a & b. This is my explanation. Hope it helps.
« Last Edit: December 27, 2018, 04:21:51 am by ogden »
 


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