### Author Topic: Does Kirchhoff's Law Hold? Disagreeing with a Master  (Read 88839 times)

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#### bsfeechannel

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #800 on: February 06, 2019, 12:58:22 pm »
The phenomenon is exactly the same.

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#### ogden

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #801 on: February 06, 2019, 01:59:54 pm »
The phenomenon is exactly the same.

You say that circuit of Dr.Lewin's experiment at RF frequencies like 3GHz acts same way as 300Hz?

#### bsfeechannel

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #802 on: February 06, 2019, 02:34:49 pm »
The only spot i disagree with Dr. Lewin is the use of KVL on a circuit that has not been modeled to include the magnetic properties of wires.

The wires do not have "magnetic properties". The only way for a wire to "react" to magnetic field is if it is moving in relation to a frame of reference.

This is stated clearly in Feynman's chapter 22:

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So what is always true is that the sum of the electric field E and the cross product of the velocity of the conductor and the magnetic field B—which is the total force on a unit charge—must have the value zero inside the conductor:

F/unit charge = E+v×B = 0 (in a perfect conductor), (22.12)

where v represents the velocity of the conductor. Our earlier statement that there is no electric field inside a perfect conductor is all right if the velocity v of the conductor is zero;

So since our conductor is static, the electric field inside it is "immune" to the magnetic field.

When current flows, however, it will generate a magnetic field that will oppose the magnetizing field of the solenoid. But Lewin deliberately and carefully chose resistors with values sufficiently high so that the current will be small and generate a negligible opposing field.

If my calculations are not wrong, just to give you an idea, you need 100mT @150Hz for a 0.01m² (10x10cm) loop to generate a 1V EMF. 1mA flowing through the loop will generate less 70µT near the surface of a 0.4mm² (21AWG) wire (J ~ 3A/m²), and, since its intensity is inversely proportional to the distance of the center of the wire, it will be much less elsewhere.

So we can pretty much consider our conductors "unaware" of the magnetic field.

But they are real conductors, and what are in fact real conductors? Just resistors with a very low resistance. And resistors do allow the existence of an electric field inside them.

Let's suppose that in Lewin's experiment, each piece of conducting wire had a 0.1Ω resistance. So now the EMF will be incident all along the circuit, since there will be no ideal wire to zero the electric field along any stretch of the path of the circuit.

So, the total series resistance will be 1000.4Ω. Subjected to a 1V EMF, it will generate a 999.60015µA current. Multiplied by 0.1Ω that will be precisely 99.960015µV. It is not zero, but it is closer to zero than to 250mV.

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Once they are modeled everything works fine and gives identical results without any sort of paradox.

Once you understand that Faraday's law invalidates Kirchhoff's law for varying magnetic fields, the paradox goes away. Believe me.

EDIT: Replaced "the charges inside it are immune" with "the electric field inside it is "immune""
« Last Edit: February 06, 2019, 05:10:34 pm by bsfeechannel »

#### bsfeechannel

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #803 on: February 06, 2019, 02:40:01 pm »
The phenomenon is exactly the same.

You say that circuit of Dr.Lewin's experiment at RF frequencies like 3GHz acts same way as 300Hz?

Let me see. To explain Lewin's circuit, we need Maxwell's equations. To explain how a loop antenna works, we need Maxwell's equations. I'm starting to see a coincidence there.

#### ogden

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #804 on: February 06, 2019, 04:45:11 pm »
You say that circuit of Dr.Lewin's experiment at RF frequencies like 3GHz acts same way as 300Hz?

Let me see. To explain Lewin's circuit, we need Maxwell's equations. To explain how a loop antenna works, we need Maxwell's equations. I'm starting to see a coincidence there.

#### bsfeechannel

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #805 on: February 06, 2019, 05:03:22 pm »

Does that make you happy?

#### Berni

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #806 on: February 06, 2019, 05:42:29 pm »
The only spot i disagree with Dr. Lewin is the use of KVL on a circuit that has not been modeled to include the magnetic properties of wires.
The wires do not have "magnetic properties". The only way for a wire to "react" to magnetic field is if it is moving in relation to a frame of reference.

This is stated clearly in Feynman's chapter 22:
Yes we can argue about what a wire is, ask a chemist and he will see a large number of copper atoms arranged in to a rod shape with some on the outside being bound to oxygen. From the point of view of circuit analysis the wire is a magnetic component because the current inside it interacts with the magnetic field, much like a capacitor is an electrostatic component (Even if its just parasitic capacitance between two wires rather than an actual component with parallel plates inside)

To prove my point that you can measure 0V or 250mV depending on what you want to see i have put together some examples:

Here is the case of an ideal transformer mesh model where we see those dreadful 250mV across one of the wire sections inside it. Below are multiple ways of constructing the physical cirucit to match that. Notice that the field is taken care of in 3 different ways, yet the result is the same. As long as the magnetic effects are contained to its wingdings it doesn't matter what the field does. Ideal cirucit analysis wires are not capable of interacting with fields, wherever it needs to interact it needs a circuit component to represent the mathematical transfer function for the field its interacting with(Inductors for magnetic fields, Capacitors for electrostatic fields).

Now lets see how you can get the result of 0V across the wire section. Now that the fields effects are not contained inside the transformer means that it is affecting the wires traveling around it, this makes those wires a transformer secondary too. When you model that in you now have a mesh model that behaves correctly even when the field is not contained. We didn't ignore the field, we just 'expanded' the transformer to cover it.

So there you go, that's how you measure 250mV in there, if you don't believe me go on and try building one of these circuits to see for yourself. Circuit analysis stays perfectly consistent when you model things correctly.

#### bsfeechannel

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #807 on: February 06, 2019, 09:11:42 pm »
From the point of view of circuit analysis the wire is a magnetic component because the current inside it interacts with the magnetic field, much like a capacitor is an electrostatic component (Even if its just parasitic capacitance between two wires rather than an actual component with parallel plates inside)

From the point of view of circuit analysis a wire is a wire. It has zero ohm resistance and does not interact with any magnetic field, because there can't be any mangnetic fields in a lumped circuit. Period.

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To prove my point that you can measure 0V or 250mV depending on what you want to see i have put together some examples:

You want to see 250mV. The thing is that they are not there. None of those circuits are equivalent to Lewin's circuit, so let's forget them.

The only approximations are 2b and 2c even though they also contain errors.

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So there you go, that's how you measure 250mV in there, if you don't believe me go on and try building one of these circuits to see for yourself.

I can imagine a bunch of other circuits that will give me 250mV. Only that they are not Lewin's circuit.

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Circuit analysis stays perfectly consistent when you model things correctly.

If you arrange a circuit and decree in your head that it is a model of another, it may appear consistent to you, but don't expect others, especially those with a minimum understanding of the physical phenomenon, to believe you.

#### Berni

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #808 on: February 07, 2019, 06:16:51 am »
Okay i do admit that figure 1b is diferent then Dr. Lewins circuit because it has a shielding box that actually intersects the circuit, but the point of that one was to show that the box does NOT alter the behavior of the circuit as the voltmeters still show the exact same values.

But all of other circuits are the same circuit on the blackboard or represent his experimental setup. Only difference being of the 2 extra voltmeters added around it and since voltmeters have infinite impedance they don't affect the operation of the circuit in any way, they just show voltages.

Its all about connecting the voltmeter in the right way. If its 0V that you want to see just look at circuits in figures 2a 2b 2c and see it is indeed 0V. Happy now?

There is no paradox with this circuit as long as its correctly modeled.

#### rfeecs

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #809 on: February 08, 2019, 12:50:42 am »
With regard to the issue of "He's not probing it right":

A reasonable way to probe this, given that you want the voltage difference of two separated points and realizing that there are stray fields around, might be to use coaxial probes, and do a differential measurement between the two points.

For example, use typical scope probes and pull the clips off, leaving just the probe points with coax shielding almost all the way to the tips.  Use channel 1 and channel 2 for the two probes and do a CH1-CH2 measurement:

So everything is in one plane, 2 dimensional.

Now what do you measure?  Does it matter if the scope is on the left side or the right side?

What if you measure a quarter turn of the wire?  Do you get 0V, or something else?

#### bsfeechannel

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #810 on: February 08, 2019, 02:51:14 am »
If its 0V that you want to see just look at circuits in figures 2a 2b 2c and see it is indeed 0V. Happy now?

I don't want to see 0V. Nature shows it is 0V.

Mehdi and Mabilde reconstructed Lewin's circuit. Mabilde got even to the point of recreating Lewin's solenoid.

They thought like you: there must be a voltage in the wires. Lewin goofed it up. He bad probed the whole thing. He doesn't know how to lump model his circuit.

When they measured exactly zero volts they got puzzled, and invented each their own completely different convoluted pseudo-scientific theory to explain what the Maxwell's equations predict with simplicity and elegance: there are no voltages in the wires, nor in the probes, Kirchhoff doesn't hold for varying magnetic fields in a circuit, Faraday's law is what accounts for the voltages across the resistors, nothing else.

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There is no paradox with this circuit as long as its correctly modeled.

The only thing your circuit is modeling is the lack of understanding of electromagnetism.

Your version contradicts Mehdi's version which contradicts Mabilde's version.  Which shows that the assumption that Kirchhoff always holds for varying magnetic fields is self-contradictory, i.e. is a paradox. Which is what Lewin brilliantly showed.

Since it is impossible to lump model Lewin's circuit nor any other circuit with varying magnetic fields in them, the paradox ceases to exist when you realize that the only way to solve them is to apply Faraday's law and give Kirchhoff and his doggone "law" to what it deserves: the birds.

#### seagreh

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #811 on: February 08, 2019, 06:43:27 am »
Does Kirchhoff's Law Hold?

Is it about Mehdi and/or Mabilde against Prof. Lewin and/or Robert H. Romer or is it about Gustav Robert Kirchhoff against Michael Faraday and/or James Clerk Maxwell ?

1845 Gustav Robert Kirchhoff - still being a student at this time - wrote an article in a scientific physics journal “Annals of Physics”
https://gallica.bnf.fr/ark:/12148/bpt6k151490/f509.item
About electric current passing through a plane, in particular through a circular one; from the student Kirchhoff.

And at https://gallica.bnf.fr/ark:/12148/bpt6k151490/f525.item last 3 paragraphs at this page you find his two rules.
As well as next page https://gallica.bnf.fr/ark:/12148/bpt6k151490/f526.item first paragraph.
Rule 2) being, what everybody calls the KVL today.

As it is in German, a very good translation you find in ‘Elementary Treatise on Electricity’ from James Clerk Maxwell (Oxford 1881) - page 127.
https://archive.org/details/elementarytreati00maxwrich/page/126

…Kirchhoff has stated the conditions of a linear system in the following manner, in which the consideration of the potential is avoided.
(1)   (Condition of ‘continuity.’) At any point of the system the sum of all the current which flow towards that point is zero.
(2)   In any complete circuit formed by the conductors the sum of electromotive forces  taken round the circuit is equal to the sum of products of the currents in each conductor multiplied by the resistance of that conductor….

Kirchhoff did not say

Σ Vk = 0

But he basically said, the sum of voltage drops (he expressed as product of resistance and current) equals to the sum of EMFs.
He did not say if emf needs to be based on:

•   electrochemical effects (galvanic cells)
•   electromagnetic induction (motional emf or transformer emf)
•   solar cell or photodiode
•   fuel cell based
•   Peltier effect
•   Seebeck effect
•   Hall effect

Hence, applying Kirchhoff’s second rule would lead to:

I * R1 + I * R2 = Σ EMFs
1 mA * 100 Ω + 1 mA * 900 Ω = Σ EMFs
1 V = Σ EMFs

and not to   ‚1V = 0‘

Although the Σ EMFs within a mesh can be as well zero, in case the mash contains no emf, only power consuming elements. But then the sum of voltage drops is zero as well.
How, to know if you measure an emf or a voltage drop? Somehow funny some people argue, ‘you cannot measure an emf’! Partly true, as you measure a small current and conclude to a voltage drop (via a known high resistor) - hence you measure a voltage drop on your internal resistor. But, still you can measure the voltage (even of an emf)! You can measure the voltage of your car battery, the voltage of a transformer, a generator, a solar cell…

To be sure if acertain element is an emf or a voltage drop you need only to watch the current flow direction. For emf current flow is from minus to plus!

Although, Kirchhoff is not wrong and thus his rule holds, he won’t tell us where the emf is hidden exactly !
And he won't help you too much with induction in you probe wires.
Now we need to consult Mr. Faraday.

#### Berni

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #812 on: February 08, 2019, 09:22:35 am »

So everything is in one plane, 2 dimensional.

Now what do you measure?  Does it matter if the scope is on the left side or the right side?

What if you measure a quarter turn of the wire?  Do you get 0V, or something else?

Typical coax cable doesn't really give any shielding to magnetic fields so the scope would show different values depending on what side it is on. If the shielding material on the coax cable is made of a infinite permeability material then you should get the same value no matter where you put the scope. Same difference as Figures 1 and 2 in my post, its a different magnetic path.

I don't want to see 0V. Nature shows it is 0V.

Mehdi and Mabilde reconstructed Lewin's circuit. Mabilde got even to the point of recreating Lewin's solenoid.

They thought like you: there must be a voltage in the wires. Lewin goofed it up. He bad probed the whole thing. He doesn't know how to lump model his circuit.

When they measured exactly zero volts they got puzzled, and invented each their own completely different convoluted pseudo-scientific theory to explain what the Maxwell's equations predict with simplicity and elegance: there are no voltages in the wires, nor in the probes, Kirchhoff doesn't hold for varying magnetic fields in a circuit, Faraday's law is what accounts for the voltages across the resistors, nothing else.

I KNOW the voltage appears as a net field at the ends of the wire.

If you know enough about circuit meshes you also know that you can't have any voltage jumps within the same net, hence why a component is introduced into the mesh to represent the voltage jump, this component is what represents the potential across the ends of the wire. There is Faradays law itself inside that component.

How is it surprising that it shows 0V ? I don't remember any of them finding it particularly surprising, all they did is gave some extra explanation apart from just religiously saying "Its because Faradays law" to help people understand it from different perspectives.

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There is no paradox with this circuit as long as its correctly modeled.

The only thing your circuit is modeling is the lack of understanding of electromagnetism.

Your version contradicts Mehdi's version which contradicts Mabilde's version.  Which shows that the assumption that Kirchhoff always holds for varying magnetic fields is self-contradictory, i.e. is a paradox. Which is what Lewin brilliantly showed.

Since it is impossible to lump model Lewin's circuit nor any other circuit with varying magnetic fields in them, the paradox ceases to exist when you realize that the only way to solve them is to apply Faraday's law and give Kirchhoff and his doggone "law" to what it deserves: the birds.

Can you please show me a quote where i say that Kirchhoff always holds for varying magnetic fields? (Cause i really don't remember saying that, but i have not deleted any of my posts so you should have no problem finding where i said it)

Mesh modeling is supposed to abstract away electromagnetism because there are no fields present in circuit meshes, its one of the tasks that modeling has!

What lack of understanding do i have of electromagnetism? If you say i'm lacking it then you also surely have something to quote me for where i said something wrong about it.

All i did was show that Dr. Lewins circuit can be lump modeled just fine and that the resulting circuit mesh behaves identically while causing no paradoxes. If you are going to keep accusing me of saying some made up things i never said in the whole tread, then we might as well end the discussion here.

#### bsfeechannel

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #813 on: February 08, 2019, 04:44:52 pm »
If you know enough about circuit meshes you also know that you can't have any voltage jumps within the same net, hence why a component is introduced into the mesh to represent the voltage jump, this component is what represents the potential across the ends of the wire. There is Faradays law itself inside that component.

You should get rid of this dogma.

The real and profound meaning of Faraday's law is exactly that there can be "voltage jumps" in a circuit mesh.

A voltage can appear in your mesh without any generator associated with it. Maxwell showed that electric and magnetic fields are a property of space, not a property of circuits. In our previous discussions we showed you that there can be voltages in an empty space under a varying magnetic field without any circuit. This is the proof that there can be voltages without any generators associated with them.

The source of your confusion lies in the fact that to apply circuit analysis we have to take care of any varying magnetic field first and see if it is confined to a specific region of the circuit. If it is, we declare that a forbidden zone and treat the effects of that field at the interface (the terminals) of that part of the circuit. If we already know the relation between voltages and currents at the interface, good. If we don't, we have to calculate it using Maxwell. Then the rest of the circuit can be solved using Kirchhoff, because there won't be no varying fields left. All the "voltage jumps" will be known and treated properly.

In short, we change the path of the circuit to avoid those fields. That's the meaning of this picture in Feynman's lectures.

No one pays the due attention to the dashed line that he calls gamma (Γ). See how it avoids the path of the unlumped circuit. So the schematics that you see on paper or on your screen are following the line gamma. Sometimes this line coincide with some wire (DC circuits, for instance), most of the times it doesn't.

If it is not possible to consider the effects of the varying magnetic field confined to a specific region of your circuit, or to specific regions (plural), in other words, if it is physically impossible to draw that dashed line away from any fields, you're doomed. Your circuit analysis is cactus.

And that's precisely what happens to Lewin's circuit.

So you're confounding the technique with the physical principle, which is something that I've already pointed out before.

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All i did was show that Dr. Lewins circuit can be lump modeled just fine and that the resulting circuit mesh behaves identically while causing no paradoxes.

Unfortunately you didn't. Your, if we can say, "model" reveals gross errors in the understanding of electromagnetism AND circuit modeling and analysis, too. Sorry.
« Last Edit: February 08, 2019, 04:48:12 pm by bsfeechannel »

#### Berni

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #814 on: February 08, 2019, 05:10:11 pm »
Yes so you take the dotted Γ line in a path that encloses no net field, everything else is solved using Maxwell and everything is fine.

If this is the wrong way to apply circuit analysis can you then explain why all the voltmeters still show correct values regardless if the type of analysis applied is Faradays loop equation or just circuit analysis as a transformer? If the voltmeters are supposed to show something else feel free to point it out.
(Regarding to this post: https://www.eevblog.com/forum/chat/does-kirchhoffs-law-hold-disagreeing-with-a-master/msg2182160/#msg2182160 )

If its wrong you should be able to make it spit out wrong results too (at least using some special case circuit if not in general).
« Last Edit: February 08, 2019, 05:12:06 pm by Berni »

#### rfeecs

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #815 on: February 08, 2019, 05:48:49 pm »

So everything is in one plane, 2 dimensional.

Now what do you measure?  Does it matter if the scope is on the left side or the right side?

What if you measure a quarter turn of the wire?  Do you get 0V, or something else?

Typical coax cable doesn't really give any shielding to magnetic fields so the scope would show different values depending on what side it is on. If the shielding material on the coax cable is made of a infinite permeability material then you should get the same value no matter where you put the scope. Same difference as Figures 1 and 2 in my post, its a different magnetic path.

There are no magnetic fields where the coax is.  You yourself simulated the field pattern for a solenoid.  There is only an electric field.  The coax should have some effect with an electric field present.

#### Berni

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #816 on: February 08, 2019, 06:14:27 pm »

So everything is in one plane, 2 dimensional.

Now what do you measure?  Does it matter if the scope is on the left side or the right side?

What if you measure a quarter turn of the wire?  Do you get 0V, or something else?

Typical coax cable doesn't really give any shielding to magnetic fields so the scope would show different values depending on what side it is on. If the shielding material on the coax cable is made of a infinite permeability material then you should get the same value no matter where you put the scope. Same difference as Figures 1 and 2 in my post, its a different magnetic path.

There are no magnetic fields where the coax is.  You yourself simulated the field pattern for a solenoid.  There is only an electric field.  The coax should have some effect with an electric field present.

Yes the conductive shield around a coax does make it immune to electrostatic fields as the charges in the shield will redistribute to perfectly oppose it. However the non conservative field caused by the vicinity of a magnetic field is not the same thing. The charges need to be able to flow along the length of the cables shield in order to counter the effects and the shield would need to be superconductive to create a opposing field that is just as strong. Since the coax abruptly ends this is not possible.

The reason Romer uses coax cable is to make sure none of the effect comes from capacitive crosstalk since parts of the circuit are undergoing rapid voltage changes during the experiment. If coax cable was capable of hiding its inner conductor from magnetic effects then Romer would have seen no voltage.

However coax cables are good at maintaining a low leakage inductance between the inner conductor and the shield, so applying a voltage on one end between the conductor and shield will make it to the other end of the cable unaffected (between the conductor and shield) even if the cable passes trough strong magnetic fields. In other words the conductor and shield appear to occupy the same physical space so there is no loop area and so Faradays law says the voltage is zero no matter how strong of a field there is.

EDIT: Sorry i incorrectly remembered the coax part. Romer didn't use any coax, just said that the wires have to run close to each other.
« Last Edit: February 08, 2019, 08:23:35 pm by Berni »

#### rfeecs

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #817 on: February 08, 2019, 07:26:45 pm »
Yes the conductive shield around a coax does make it immune to electrostatic fields as the charges in the shield will redistribute to perfectly oppose it. However the non conservative field caused by the vicinity of a magnetic field is not the same thing.
This is a quasi-electrostatic situation, at least the way Romer set up the experiment.  The electric field is constant during the measurement period.

Once again you are saying that charge can tell the difference between "different" electric fields?

I don't think so.  But I agree that the charge on the outer surface of the coax shield will redistribute itself to create an opposing field resulting in no tangential field along the outer surface of the coax.  The same thing happens on the inner surface.  So the center conductor sees no E field in the direction along it's path.

Consequently the charge on the center conductor does not rearrange itself in response to the outside field.  It is shielded by the outer conductor.

Quote
The reason Romer uses coax cable...
I don't see any mention of coax cable in Romer's paper.

#### Berni

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #818 on: February 08, 2019, 08:36:41 pm »
Sorry i remembered Romers paper a bit incorrectly. He just says about wires having to be run close to each other, not actually using coax.

The difference between a purely electrostatic field and one induced by a magnetic field is the non conservative part. This gives the field ability to push electrons around closed loops of conductor while a electrostatic field can only redistribute them but not sustain a current apart from the very brief transient as they redistribute.

Since the inner conductor forms a continuous loop trough the voltmeter, it means that the the field can push the electrons around it and create a current that the voltmeter detects as voltage across its internal resistance. The shield however does not form a continuous loop and as such can't experience any current trough it.

If you are to connect the ends of the two shields together with a wire you would get currents flowing inside of it and there would be effects on the voltmeters readings. How exactly they would be affected depends on how exactly the ends are connected. If the conductivity of the shield is sufficiently high then you can probably get the result of 0.1V or 0.9V or anything in between depending on what path you run that shield connecting wire. But this is going a bit off topic, its not really relevant. My point is that the coax shield in this configuration has no effect.
« Last Edit: February 08, 2019, 08:38:24 pm by Berni »

#### bsfeechannel

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #819 on: February 08, 2019, 10:03:52 pm »
Yes so you take the dotted Γ line in a path that encloses no net field, everything else is solved using Maxwell and everything is fine.

If this is the wrong way to apply circuit analysis can you then explain why all the voltmeters still show correct values regardless if the type of analysis applied is Faradays loop equation or just circuit analysis as a transformer? If the voltmeters are supposed to show something else feel free to point it out.
(Regarding to this post: https://www.eevblog.com/forum/chat/does-kirchhoffs-law-hold-disagreeing-with-a-master/msg2182160/#msg2182160 )

If its wrong you should be able to make it spit out wrong results too (at least using some special case circuit if not in general).

It is wrong because you added "compensation" to your probes to account for a non existing voltage in the wires.

You know, this is the problem with using Spice as a learning tool. Spice is a stupid software with no critical thinking. If your premises are wrong it will accept them as truth and will sheepishly confirm whatever wrong conclusions you have drawn.

Ditch this devilish program at once and learn electromagnetism as it should: studying the classics, like Feynman's lectures and such.

#### Berni

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #820 on: February 08, 2019, 10:25:04 pm »
It is wrong because you added "compensation" to your probes to account for a non existing voltage in the wires.

You know, this is the problem with using Spice as a learning tool. Spice is a stupid software with no critical thinking. If your premises are wrong it will accept them as truth and will sheepishly confirm whatever wrong conclusions you have drawn.

Ditch this devilish program at once and learn electromagnetism as it should: studying the classics, like Feynman's lectures and such.

Where is the compensation? I'm just treating the probe wires the same way as the circuit wires since they follow the same path. Should wires that connect voltmeters be treated in some special way?

And what is wrong with SPICE? Its just a set of tools for analyzing circuit meshes because doing it manually by pen and paper becomes very time consuming for large circuits. It doesn't invent any new way of modeling circuits, just automates the existing circuit analysis math that people had to do by hand before computers came around Can you recommend a better alternative for predicting the behavior of a circuit that includes active nonlinear components? A good circuit to show as a example on might be the Royer oscillator as it involves magnetic components.

#### bsfeechannel

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #821 on: February 08, 2019, 11:37:00 pm »
Where is the compensation? I'm just treating the probe wires the same way as the circuit wires since they follow the same path. Should wires that connect voltmeters be treated in some special way?

The "transformers" you introduced there. They don't exist. Transformers are lumped components. They must have terminals where there be no varying magnetic fields. You already know how to lump part of a mesh under varying magnetic fields. If you look at Lewin's circuit again you'll see no way to draw the gamma line.

What your circuit is modeling is something else, not Lewin's circuit.

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And what is wrong with SPICE?

Spice is a simulation software, not a learning tool. As I said, it has no mechanism to check whether your assumptions are sound or not. And that's your problem. Your assumptions are wrong. Hence you are drawing the wrong conclusions, and since Spice is just confirming the conclusions you draw from wrong assumptions, you think your reasoning is right.

Spice is already inadequate as a tool to learn about circuits. It is an insanity to try to use it to understand electromagnetism.

#### seagreh

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #822 on: February 09, 2019, 01:42:12 am »

The inner circle in Lewin’s circuit underlies varying magnetic flux and all other nasty things as well.
Kirchhoff doesn’t hold, because it won’t add up to zero! The sky is FALLING. We solve this with Maxwell’s equations.

The outside ‘circle’ (consisting of the probe leads) are NOT underlying a varying magnetic field. is this what you are stressing ?
Hence, now Kirchhoff holds in this domain, because it adds up to zero! The sky is BLUE!

0.9V + 0.1V = 0 V

#### Berni

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #823 on: February 09, 2019, 12:01:03 pm »
Yes that is another way to draw the physical representation of my lumped circuit, it looks slightly different but behaves identical. Much like Dr. Lewins physical experiment, the fields in there are not the same (Field closes around on his table and not at the edge of the universe), but the end result is identical so does it matter?

I was never suggesting SPICE is a good tool for learning about electromagnetism. Its just a automated implementation of manual pen and paper mesh circuit analysis that is thought in every electronics engineering school on the planet. Mesh circuits have no fields or physical dimensions, just nodes, volts and amps. It doesn't ignore fields in a physical circuit, just hides them away into a lumped component. This abstraction makes it great instead for learning about the circuits behavior under given conditions, especially when circuits get large and contain 100s of components.

So then on the topic of the correct way to analyze a circuit.

This is a slightly more complex circuit than Dr. Lewins experiment, increasing the component count to 3 parts. However one of the parts now exhibits non-linear behavior due to being a semiconductor. But it is still simpler than 99% of circuits you can find on the internet, so determining its behavior should not be difficult right?

How would you analyze this circuit in the correct way?

What happens to to the current in the inner and outer loops upon connecting the battery power source?

#### rfeecs

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##### Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #824 on: February 10, 2019, 12:36:59 am »
The difference between a purely electrostatic field and one induced by a magnetic field is the non conservative part. This gives the field ability to push electrons around closed loops of conductor while a electrostatic field can only redistribute them but not sustain a current apart from the very brief transient as they redistribute.

Since the inner conductor forms a continuous loop trough the voltmeter, it means that the the field can push the electrons around it and create a current that the voltmeter detects as voltage across its internal resistance. The shield however does not form a continuous loop and as such can't experience any current trough it.

I don't see why you would need current to shield a static electric field.

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...My point is that the coax shield in this configuration has no effect.

I'm not convinced.  I expect the coax will shield the center conductor from the induced electric field.  So you would measure the same voltage whether the scope was on the left or the right side.

Smf