Author Topic: Does Kirchhoff's Law Hold? Disagreeing with a Master  (Read 37360 times)

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Offline gildasd

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #75 on: November 11, 2018, 12:11:39 am »
The prof might be right in theory, but in this case Mehdi is right.

Loops can cancel or add to themselves in a electromagnetic situation, and (as I understand it) Mehdi points that this was not considered properly.
It’s not because something appears on a oscilloscope screen and you have Einstein’s hairdresser that it automatically sustains your postulate.

I have worked in a electromotor maintenance factory and I think i would be laughed out if i tried to sense a coil with so sloppily and diagnose the motor with the result.

That said, I cannot fault his other stuff, and he might be right in theory for specific situations.
« Last Edit: November 11, 2018, 12:24:19 am by gildasd »
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Offline rfeecs

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #76 on: November 11, 2018, 04:19:47 am »
He posted a teaser video, sounds familiar:

 

Offline bsfeechannel

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Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #78 on: November 12, 2018, 12:29:52 am »
Let me hazard some thoughts.

The figure below shows a voltage source and two voltmeters. One is connected as closely as possible to the source and another one at a distance very, very far from it. The switch is open, so the second voltmeter reads 0 volts. All conductors are ideal, i.e., resistance is 0 ohms.



Now we close the switch and see the electric field propagating towards the second voltmeter, which still indicates 0V. We can see that, during this transient, Kirchhoff doesn't hold. Because if you add up all the voltages in the loop they will not result 0.



But, and there's always a but, you can REDUCE the long line of conductors to a series of inductors and capacitors, and now Kirchhoff holds. You can explain why the faraway voltmeter is indicating 0V by the fact that, along the line, some capacitor is still uncharged, while some inductor is reacting to the change of its current with an EMF equal to the source voltage.



However, there are no such inductors or capacitors. If we consider their existence, they must be infinitesimal. There are infinite infinitely tiny inductors and capacitors, and for Kirchhoff to hold you will have to apply it to infinite meshes.

This is just a simple example, but we can see that Kirchhoff is not adequate to model situations like that.

For a finite number of lumped components where space and time can be disconsidered, Kirchhoff is fine.

But add space and time and you'd be better off with a theory that takes that into consideration, and that is Faraday-Maxwell.
« Last Edit: November 12, 2018, 12:36:24 am by bsfeechannel »
 

Offline ArthurDent

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #79 on: November 12, 2018, 05:33:19 am »
There may be an infinite number of ways to model this to show what you want to prove and this is my take. The circuit above isn't like the original problem at all partly because the left part of the circuit isn't switched the same way as the right half. The two halves are two different problems.

I'm lumping the induced voltage producer to one area, a transformer secondary instead of a distributed magnetic field, which is one part of the loop that includes a left an a right resistor of equal value. A very astute instructor I had once said: "Things that are in series are in series with each other." This means that the right resistor can be moved anywhere along the entire length of the wire connecting the left resistor and the transformer and it will always read the same voltage.

The video with the leads being moved from left to right to 'prove' there is a difference is the same as placing another induced A.C. voltage (or transformer secondary) in series with the test points either aiding or opposing the induced voltage in the coil.  Phasing is important. 
 

Offline T3sl4co1l

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #80 on: November 12, 2018, 07:04:55 am »
However, there are no such inductors or capacitors. If we consider their existence, they must be infinitesimal. There are infinite infinitely tiny inductors and capacitors, and for Kirchhoff to hold you will have to apply it to infinite meshes.

This is just a simple example, but we can see that Kirchhoff is not adequate to model situations like that.

For a finite number of lumped components where space and time can be disconsidered, Kirchhoff is fine.

But add space and time and you'd be better off with a theory that takes that into consideration, and that is Faraday-Maxwell.

Correct.  As I noted at the start of this thread, you must integrate over space to apply KVL and KCL for the general -- wave mechanical -- case.  For the 1-D case, this effectively gives the differential RLC transmission line element used above, and when solved, gives the Telegrapher's equations.  For the 3D case, you get field solutions of course.

We apply Kirchhoff's laws on coarser elements (e.g., whole circuit loops), when it is justifiable to do so, for example when the signals of interest are slower than the physical scale of the system (so that we need not consider wave mechanics as such).  In that case, a netlist (an abstract schematic drawing) and finitely many L and C can be used.  At lower and lower frequencies, the number of L and C required drops, until at DC, L and C go away completely and we need only consider the network of resistances.

We can likewise consider wave mechanics alone, when it is justifiable to do so.  For example, building active circuitry in as small a form factor as possible, then interconnecting the circuits with transmission lines.  KVL and KCL are broken between the ends of the transmission line, but we can still consider them locally, i.e. at each port of the line.

This is a very powerful design approach, allowing the designer to greatly simplify the circuit: one need not consider every possible coupling between elements in the circuit, but only those close enough together (including self-coupling, i.e., LF-equivalent stray L and C).

Tim
Seven Transistor Labs, LLC
Electronic design, from concept to prototype.
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Offline Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #81 on: November 12, 2018, 07:56:57 am »

However, there are no such inductors or capacitors. If we consider their existence, they must be infinitesimal. There are infinite infinitely tiny inductors and capacitors, and for Kirchhoff to hold you will have to apply it to infinite meshes.

This is just a simple example, but we can see that Kirchhoff is not adequate to model situations like that.

For a finite number of lumped components where space and time can be disconsidered, Kirchhoff is fine.

But add space and time and you'd be better off with a theory that takes that into consideration, and that is Faraday-Maxwell.

And this is exactly why we lump together these stray inductance and capacitance into lumped form that approximates the behavior of an infinitely large mesh as close as possible. Makes math a whole lot easier.

Its not only a problem of a wire. The resistors also have physical dimensions and as such have parasitics. To perfectly accurately model the resistor we would need to have a infinitely large mesh of resistors capacitors indusctors just to model the internal construction of say a metal film type troughhole resistor. Can you simplify this infinite mesh down to a single RLC circuit and still have it perform close enough to give us near perfect results once we do math to it? Yep we sure can so we do it, because doing math to 3 components is easier than to an infinite number of components.

If we tried to calculate the behavior of perfect non simplified models of a circuit we would need a computer that is much much larger than the largest supercomputers we have since we would basically need to simulate every single electron and every single atom of of the circuit. Yet using lumped component models we can get nearly the same result using the computing power of a PC from the 90s.
 

Offline rfeecs

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #82 on: November 12, 2018, 11:25:21 am »
OK, well take the original demo, but instead of two resistors, make the whole loop one big resistor.  That is, a loop made out of resistive material:



Say the EMF induced in the loop is 1V.  Say the total resistance is 1 ohm, so you have 1 amp flowing around the loop.  Take any point in the loop and go around the loop adding up the IR drop.  Go all the way around the loop.  You always end up with 1V, not zero.

So how are you going to model this to make Kirchoff's law work?  An infinite number of resistors dR and an infinite number of voltage sources dV? (I don't think so.)
 

Offline HackedFridgeMagnet

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #83 on: November 12, 2018, 11:32:20 am »
Say the EMF induced in the loop is 1V.  Say the total resistance is 1 ohm, so you have 1 amp flowing around the loop.  Take any point in the loop and go around the loop adding up the IR drop.  Go all the way around the loop.  You always end up with 1V, not zero.

So how are you going to model this to make Kirchoff's law work?  An infinite number of resistors dR and an infinite number of voltage sources dV? (I don't think so.)
No in this case you will end up with 0v if you measure it.
You're misinterpreting Faraday-Maxwell.
 

Offline rfeecs

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #84 on: November 12, 2018, 01:26:20 pm »
Say the EMF induced in the loop is 1V.  Say the total resistance is 1 ohm, so you have 1 amp flowing around the loop.  Take any point in the loop and go around the loop adding up the IR drop.  Go all the way around the loop.  You always end up with 1V, not zero.

So how are you going to model this to make Kirchoff's law work?  An infinite number of resistors dR and an infinite number of voltage sources dV? (I don't think so.)
No in this case you will end up with 0v if you measure it.
You're misinterpreting Faraday-Maxwell.

I am fairly sure Maxwell-Faraday describes a case where the loop is magnetically closed but electrically open circuit.
I'm fairly sure you are misinterpreting it.

But maybe I wasn't clear.  Say you have the resistive loop.  If you could measure say a section 1/10th of the way around.  (I admit it is not easy to accurately measure in the presence of the magnetic field.)  So that section has resistance 1/10th of an ohm and has current 1A flowing.  You should measure 0.1V.  So there are 10 pieces, and if you add them up, you get 1V, not 0V.  You could divide it up in other numbers of sections with the same result. 

How do you get it to add up to 0V?
 

Online KL27x

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #85 on: November 12, 2018, 02:22:36 pm »
Quote
But maybe I wasn't clear.  Say you have the resistive loop.  If you could measure say a section 1/10th of the way around.  (I admit it is not easy to accurately measure in the presence of the magnetic field.)  So that section has resistance 1/10th of an ohm and has current 1A flowing.  You should measure 0.1V.  So there are 10 pieces, and if you add them up, you get 1V, not 0V.  You could divide it up in other numbers of sections with the same result. 
In your example you are measuring just one section of the loop. In the thought experiment, you are measuring the voltage across two points which are connected by two half loops of wire.
 
So in your example, if you measured a 1/10th section, you should measure [0.8V]. 0.9V in one direction, 0.1V in the other direction. It's a closed circuit with two parallel branches. But Lewin very clearly states that is possible for the voltmeter to have two different readings. And he suggests that it would read either [0.1V] or [0.9V] in the other direction, but he doesn't explicitly state this would be due to inductance in the probe wires. In this example you made, I think it would be reasonable to assume a voltmeter that is connected properly would read [0.8V], which is part of what Mehdin seems to have shown, but maybe I am completely wrong. A conundrum in this example is what happens are you move your two points closer together until they approach zero distance apart? Will you essentially measure [1.0V] with the probes essentially touching? That would be cool. Lewin would say the voltage is undefined in the ideal experiment in all these cases. But clearly Medhin can get meaningful voltage measurements in the real world experiment using real copper wire and regular resistors.

In the actual thought experiment, it's not the resistance of the wire that is causing drop in the wire, it's the inductance. The wire has negligible resistance, at least in comparison with the two resistors that were put in the loop. Now the inductance apparently can't even exist if the wire were an actual ideal superconductor, according to Lewin's teaser video.

I'm sure I'm completely wrong and look forward to the corrections. But more or less that's the way I tend to think about it, right now.
« Last Edit: November 12, 2018, 03:16:38 pm by KL27x »
 

Offline Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #86 on: November 12, 2018, 05:35:51 pm »
OK, well take the original demo, but instead of two resistors, make the whole loop one big resistor.  That is, a loop made out of resistive material:



Say the EMF induced in the loop is 1V.  Say the total resistance is 1 ohm, so you have 1 amp flowing around the loop.  Take any point in the loop and go around the loop adding up the IR drop.  Go all the way around the loop.  You always end up with 1V, not zero.

So how are you going to model this to make Kirchoff's law work?  An infinite number of resistors dR and an infinite number of voltage sources dV? (I don't think so.)

Interestingly yes you would get zero volts in this case!



It turns out you can poke any two points in this circle and have the points sit at 0V despite there being 0.8mA peak of current flowing in this simulation. This happens because the voltages on each inductor and resistor pair balance out. Spice follows Kirchhoffs rule so you need the voltages to add up. Since the current is the same means the voltage drop on all resistor is the same, the rest of the drop is in the inductors and since they are the same it distributes equally. This makes the voltage on the resistor chunks the same as on the inductor chunks so each one of these lumped inductive resistor models has 0V across itself all around. But there is voltage within the model across the resistor or inductor alone.

This effect is broken as soon as your ring is not completely even around its radius. If you ware to make the resistive material thicker at some point this upsets the balance and you see a voltage across the thinnest part of the ring.

And yes the inductive probing leads again would mess things up, but again you could fix the probing problem by simply averaging what the left and the right scope sees.

EDIT: Updated the test leads to 2.5uH to reflect them being the same as 1/4 of the loop.
« Last Edit: November 12, 2018, 05:42:44 pm by Berni »
 

Offline Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #87 on: November 12, 2018, 05:56:15 pm »
Oh and i want to add that the same thing would happen in the physical demonstration with 2 resistors if you made the resistors say both 100 Ohm rather than one being 100 and other 900. This makes the demonstration a bit less exiting because then both scopes see the same shape and magnitude except that one scope is seeing the inverted waveform.

As long as the circuit is symmetrical around the two points you measure this is always the result. The same current flows around it so an identical circuit will produce the same voltage drop on both symmetrical sides, but since one of the halves is having the current running the opposite way it also generates an opposite voltage. So once summed up around the loop you basically are subtracting two voltages of the same size so you get 0V.
 

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #88 on: November 12, 2018, 06:31:54 pm »
@RoGeorge, just to let you know that you are not alone.

The only critic I would move to Lewin is in its use of the word "potential" even when the quantity is not defined because it is multivalued. I wonder if his demo would have had a better reception had he used the word "glorp" or "multivalued potential" to describe it.

In his followup video titled "Believing and Science are Very Different" (youtube code watch?v=wz_GqO-Urk4) he also shows a conceptual diagram (I am not calling it a schematic on purpose) demonstrating how the equations give exactly that result: the 'multivalued potential' at the very same pair of points to have two different distinct values at the same time.

It is also possible to push that diagram even further by dragging a voltmeter (and its probes) inside the loop and seeing how the 'multivalued potential' between those two points varies with continuity from +0.9V to -0.1V, as the fraction of the area of the loop containing the voltmeter path goes from 0% to 100% of the area of the loop with both resistors (and the changing magnetic field). And there is also a video made by a detractor (sorry, maybe this is not the right term, opponent might be better) of Lewin that shows just this [note1]... actually confirming Lewin's modeling of the system.

I cannot but wonder what causes this resistance... pun intended... to accept the consequence of the loss of irrotationality (is this even a word?)

I guess one reason is that we place too much faith in our instruments and we tend to believe the numbers they show have a meaning no matter what.
Another might be the difficulty in realizing that the resistor are INSIDE the secondary of the imaginary transformer. So, goodbye lumped component model...
Or probably engineering courses have to cram too much material and leave out too much of the basics, concentrating on the more pragmatic parts.


EDIT: corrected some of my lousy grammar and syntax, but not all.
Also, I would like to add that as long as the probes are outside of the loop (more specifically outside of the region at varying B field) there is no  probing problem at all (you might have to worry about capacitive coupling with the mains, and RFI from RadioBoomBoom, but not about the field that is confined inside the loop. Also self inductance is negligible, as shown by Lewin in the aforementioned video.
 
[note1] Ironically, when we take the measurement path inside the loop, that's where we can interpret the results (changing from +0.9V to -0.1V) as the effect of 'bad probing' (now I have to take into account the intercepted flux).  I guess that's the reason Lewin is staying outside, to avoid confusion.

Finally, maybe getting rid of the voltmeters and their probes, resorting to some other way to measure the 'voltage' across each resistor (placing an ammeter in the loop and using ohm's law?, using calorimetric measurements to infer the dissipated power? substituting the resistors with a voltage dependent light source and measuring the light output?) could help in removing the confusion).

Personally I like to think tiny angels are pushing electrons in the loop, producing a 1mA (oscillating) current. That would cause a 0.1V drop on the 100 ohm resistor and a 0.9V drop on the 900 ohm one. The fact that it's angels and not a generator allows the loop to be closed without anything between the resistors, and that show exactly why KVL is no longer valid. You have to mend it somehow by adding a distributed emf, but that does not remove the fact that KVL is broken.
If you do not believe in tiny angels, well you could use a varying magnetic field that stays all within the loop. A toroidal transformer, maybe?
« Last Edit: November 13, 2018, 04:20:18 am by Sredni »
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Offline rfeecs

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #89 on: November 13, 2018, 05:15:40 am »
OK, well take the original demo, but instead of two resistors, make the whole loop one big resistor.  That is, a loop made out of resistive material:



Say the EMF induced in the loop is 1V.  Say the total resistance is 1 ohm, so you have 1 amp flowing around the loop.  Take any point in the loop and go around the loop adding up the IR drop.  Go all the way around the loop.  You always end up with 1V, not zero.

So how are you going to model this to make Kirchoff's law work?  An infinite number of resistors dR and an infinite number of voltage sources dV? (I don't think so.)

Interestingly yes you would get zero volts in this case!



It turns out you can poke any two points in this circle and have the points sit at 0V...

That is wrong.  Your model is wrong.  It led you to the wrong conclusion.

I realize I went down a dark path when I said the word "voltage".  As has been pointed out, "voltage" and "potential" are words that are not defined well in this situation.  The E field is defined.  The E field in this case forms circular loops.  The integral of E dot dl is also defined here.  As RoGeorge said, charge moving around the loop  experiences a force from E over the distance moved, dl.  That involves energy.  That is defined also.  It is not coming from imaginary coils.  It's coming from the E field which is coming from the changing magnetic field.

Like a lot of people, I have a tendency to say that "the magnetic field induces a voltage between the ends of the wires".  This resistive loop is an example of what happens when there are no wires.

Now I'm not opposed to making this kind of model.  It may be appropriate in some situations.  The model of just an ideal transformer with a single lumped R would be appropriate in some situations.  It depends on what you are trying to model.

As for actually measuring with a voltmeter, if you kept your measurement leads twisted tightly together and arranged the ends tight next to the 0.1 ohm section that you are trying to measure, then you have no magnetic field passing through your measurement loop.  You would measure 0.1V, not zero volts.  That's just Faraday's law. 

 

Online RoGeorge

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #90 on: November 13, 2018, 05:54:22 am »
@Sredni
Indeed, mathematically speaking it doesn't make sense to talk about Potential in a non-conservative field.

Another big confusion is created because in some modern books the definition of Voltage is considered
Code: [Select]
1) Voltage1 = Potential in point A - Potential in point B
2) Voltage2 = The energy required to move a unit of charge between A and B

For conservative fields (e.g. in Electrostatic, or in DC circuits) the two definitions are equivalent, and Voltage1 is the same thing as Voltage2.  The path does not matter.

In non-conservative fields (e.g. Electrodynamics, or in AC circuits) Voltage1 is undefined.  Potential is undefined.  Only Voltage2 makes sense, but beware of the path!  Different paths will give us different Voltage2 values.

NOTE: Our most beloved voltmeters and oscilloscopes measure the Voltage2 type of voltage. They measures the energy to move the unit charge.  They do NOT measure the Voltage1 type of voltage, they do not measure a potential difference.





Mathematically speaking, the notion of Electric Potential doesn't make sense for e.g. AC circuits.
Go figure, Electric Potential doesn't exist for AC!

P.S. Never say that at a job interview.  ;D

Online ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #91 on: November 13, 2018, 06:15:46 am »
It turns out you can poke any two points in this circle and have the points sit at 0V...

That is wrong.  Your model is wrong.  It led you to the wrong conclusion.

Do you find following model as matching your "whole loop one big 1 Ohm resistor" ?



 

Offline rfeecs

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #92 on: November 13, 2018, 07:35:12 am »

Do you find following model as matching your "whole loop one big 1 Ohm resistor" ?



No, that still would have each voltage source + resistor pair cancelling out to 0V.

I'm not sure how to make a model with lumped components that violates KVL.
 

Offline Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #93 on: November 13, 2018, 08:17:18 am »
I do have some resistance wire around here that i could weld into a continuous loop and do the experiment.

Just not quite sure how to keep the probe wires from picking up the magnetic field since we are interested in the actual voltage at the test point. My average out the reading for going left and right only works when probing exact opposite points on the circle.

Using something like a lamp or LED as the 'probe' sounds like a good idea at first but then you realize you still need some wire to connect it to the two points and thus has the same problem. Running it along the ring would give similar results that you get a different voltage depending on the path the wire takes. This essentially makes the magnetically induced voltage disappear and you get to see the voltage drops on sections of resistance again. But what happens if we instead remove the restive ring but leave the probe wires connecting to the same two points in space connected to nothing? Do we measure nothing?
 

Offline rfeecs

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #94 on: November 13, 2018, 08:41:41 am »
But what happens if we instead remove the restive ring but leave the probe wires connecting to the same two points in space connected to nothing? Do we measure nothing?

I'm thinking you have no current path through the meter, so you you measure nothing.
 

Online ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #95 on: November 13, 2018, 08:59:18 am »
No, that still would have each voltage source + resistor pair cancelling out to 0V.

Yes, it cancels to 0V. In every test point labelled "A, B, C, D..." voltage against ground is 0V. That's exactly what I am showing.

Your original "model" contains single 1V EMF source in form of single, rounded 1 Ohm resistor dissipating/cancelling that 1V of EMF - that's OK, but model with 10 sources and 10 resistors in series is not ok anymore? - Apply integral to my 10x0.1V EMF sources and 10x0.1 Ohm resistors to get your single loop back. Do you see what I mean now?

[edit] For those who did not pay attention whole thread, it is advised to read this post, especially to sentence "KVL/KCL reduce to a single point only: they are a differential relation, which must be integrated over the space of interest".

Quote
I'm not sure how to make a model with lumped components that violates KVL.

LOL. You basically said: "you are wrong, but I cannot prove it"  :-DD
« Last Edit: November 13, 2018, 09:33:37 am by ogden »
 

Offline rfeecs

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #96 on: November 13, 2018, 10:03:29 am »

Quote
I'm not sure how to make a model with lumped components that violates KVL.

LOL. You basically said: "you are wrong, but I cannot prove it"  :-DD

KVL holds for lumped circuits.  This isn't a lumped circuit.
I can prove those models are wrong because the voltage around the loop in the model adds up to zero.
Using Faraday's law on the resistive loop, integrating E dot dl around the loop doesn't equal zero.
 

Online ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #97 on: November 13, 2018, 10:58:10 am »
Using Faraday's law on the resistive loop, integrating E dot dl around the loop doesn't equal zero.

Yes. Nonzero current flows in case of electrically short loop. Note that ideal conductor have zero resistance, so in case of short superconductive loop there will be zero volts no matter how, where and using how many scopes you measure. Kirchhoff's Law Holds in this case BTW ;)

Quote
KVL holds for lumped circuits.  This isn't a lumped circuit.

Why not? If you cut 0-resistance loop open and connect 1 Ohm resistor in the opening, then in our 1 Ohm & 1 A example case there will be 1V drop on the resistor and 1V EMF voltage generated in the loop. Loop is voltage source and resistor is load - what's the problem with Kirchhoff's Law in this lumped circuit?

[edit] The same considerations apply to two 0.5 Ohm resistors and two halves of the loop.
« Last Edit: November 13, 2018, 11:55:14 am by ogden »
 

Offline HackedFridgeMagnet

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #98 on: November 13, 2018, 11:38:15 am »
Pls keep the thread polite and leave out the emoticons as they can distract.

Back to the thread title.

Does Kirchhoff's Law Hold?
My belief is if you can model something and if you can verify this with measurements then your model holds.
Every model has limitations but what are the limitations of KVL?

So AFAICT it will hold in Time varying magnetic fields.
People use it in steady state analysis all the time.
But not so easy in Transmission lines.
Teslacoil suggested you can use it at either end.

So to dispprove KVL within Time varying magnetic fields I suggest you need to prove it in the lab.
IMO I don't think Dr Lewin did this.
 

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #99 on: November 13, 2018, 11:48:55 am »
So to dispprove KVL within Time varying magnetic fields I suggest you need to prove it in the lab.
IMO I don't think Dr Lewin did this.

Agreed. Rfeecs already mentioned paper explaining "Super Demo" experiment and it's results:

We've discussed this many times in previous threads.  Dr. Lewin gave his world famous SUPER DEMO as he refers to it in 2002, I guess.  But he didn't invent it.  It is an exact recreation of the experiment in this 1982 paper:
http://www.phy.pmf.unizg.hr/~npoljak/files/clanci/guias.pdf

In the paper, everything is explained very simply without drama.  It's no mystery.  The meter wires are part of the circuit and the orientation of the wires determines the results.
 


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