Author Topic: Does Kirchhoff's Law Hold? Disagreeing with a Master  (Read 37206 times)

0 Members and 1 Guest are viewing this topic.

Offline bsfeechannel

  • Frequent Contributor
  • **
  • Posts: 492
  • Country: 00
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #100 on: November 13, 2018, 01:54:51 pm »
Using Faraday's law on the resistive loop, integrating E dot dl around the loop doesn't equal zero.

It doesn't. However, every infinitesimal gain of energy represented by E · dl is lost as heat in the infinitesimal resistor dR.



« Last Edit: November 13, 2018, 01:57:36 pm by bsfeechannel »
 

Offline HackedFridgeMagnet

  • Super Contributor
  • ***
  • Posts: 1914
  • Country: au
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #101 on: November 13, 2018, 02:24:01 pm »
[edit]Apparently this is the setup Faraday used.

Not sure, but wouldn't this be a better setup to what ElectroBoom's/Lewins experiment should be using? [/edit]




The torroid would capture most of the stray flux.
So instead of worry about bad probing, you know what flux is linked.



« Last Edit: November 13, 2018, 07:01:30 pm by HackedFridgeMagnet »
 

Offline HackedFridgeMagnet

  • Super Contributor
  • ***
  • Posts: 1914
  • Country: au
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #102 on: November 13, 2018, 02:26:51 pm »
couldn't link image
 

Online Berni

  • Super Contributor
  • ***
  • Posts: 1864
  • Country: si
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #103 on: November 13, 2018, 05:35:20 pm »
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.

Well if you have both meters connected simultaneously just like in Dr. Lewins experiment you actually get the same voltage pulse on both meters no matter if the resistive ring is connected or not (At least inside LtSpice). This clearly points to a problem with the experimental setup since we shouldn't be measuring voltage in a circuit with 0 components. Yes if you leave only one meter connected you will indeed read 0V because the probe wires are floating. Both meters connected provides a return path trough each meter so they can measure something. This shows that the meters are not correctly showing the voltage at points A and B but a voltage somewhere else (Like across the meters terminals)

And that makes sense since the ring is supposed to have 0V across any two points on it. Connecting to nothing also produces 0V so the readings on the meters are the same in both cases.

I do want to do this continuous resistive ring as a physical experiment, but i first need to figure out the correct way to probe this.
 

Offline ogden

  • Super Contributor
  • ***
  • Posts: 1576
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #104 on: November 14, 2018, 12:42:40 am »
I do want to do this continuous resistive ring as a physical experiment, but i first need to figure out the correct way to probe this.

Toroidal iron core AC mains transformer will greatly help to contain magnetic flux. AC mains seems to be good enough "test signal" source as well. All you need to do - add test winding and have a fun. Beauty of using AC voltage here - you don't even need scope. Any multimeter is good enough.
 

Offline Sredni

  • Regular Contributor
  • *
  • Posts: 177
  • Country: aq
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #105 on: November 14, 2018, 03:11:10 am »
"Well if you have both meters connected simultaneously just like in Dr. Lewins experiment you actually get the same voltage pulse on both meters no matter if the resistive ring is connected or not"

This is a good point. You will have a loop with the meters' internal resistances taking the place of the resistors. With 10 meg input impedance and a 1V emf, the current in the loop would be 1/20 of a microamp. If the internal impedances are the same, the voltmeters will measure the same voltage (edit: the problem with a sinusoidal stimulus is that we lose the sign, but this adds drama). But what happens if the internal resistance are different, say 10 meg, 1 meg? Would they not measure different voltages, despite being connected to the same points?


EDIT
To get the same voltage ratio as in Lewin's experiment the internal impedances (at the test signal frequency) should be 9meg, 1meg.

Probably this is where we reach the language barrier between physics and engineering. An engineer that has different readings from instruments connected to the very same poiints will try to interpret this as a measurement error (load effects) or as wrong probing (in this case using the emf to make the trusted KVL to balance again). A physicist, on the other hand, knows that there is no problem at all, it's just that the so called 'voltage' is no longer positional: it does not depend only on the points where you place the probes, it also depends on the path followed by the probes.
The engineer will say: "see? KVL checks out: one instrument measures 0.1V, the other 0.9V, and then there's the emf of 1V, when I take all signs into account, the result is zero. KVL rulez!". A physicist will say: "see? the voltage is path dependent! If I sum the voltages around this loop I get 1V. That's the emf. KVL is for the birds!"

You might think both are right, but... the engineer is getting his balance check still assuming voltage is positional - and that's not true. The physicist gets his balance check and has the satisfaction to add his impression of Walter Kronkite saying "...and that's the way it is".
« Last Edit: November 14, 2018, 04:15:38 am by Sredni »
All instruments lie. Usually on the bench.
 

Offline bsfeechannel

  • Frequent Contributor
  • **
  • Posts: 492
  • Country: 00
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #106 on: November 14, 2018, 04:32:39 am »
I do want to do this continuous resistive ring as a physical experiment, but i first need to figure out the correct way to probe this.

Lemme give it a burl using what I think I know about Maxwell.

Let's suppose we have a changing magnetic field B confined to the area Σ, and we connect a voltmeter like in the picture below. This can be achieved for example when you have a ferromagnetic core that manages to concentrate most of the lines of B in its transversal area. So the very few lines of magnetic field in the area Σ' can be considered negligible.



Now let's remove the right hand side of the loop between the voltmeter probes. We end up with a new loop with an area Σ+Σ'.



Faraday-Maxwell says that the emf generated by this loop will be the derivative relative to time of the integral of the scalar product of B and dA on the surface Σ+Σ`. dA is the infinitesimal vector element of surface perpendicular to the surface Σ+Σ`. However B is zero over Σ'. So the scalar product over Σ' will be zero. This means that the voltage measured by the voltmeter will be that as if the loop covered only the area Σ.



Now let's remove the left arc and replace the right arc of the loop.



We can see that B is totally outside of this new loop and, according to Faraday-Maxwell, will not induce a voltage. So the voltmeter is going to show 0V.

Now that we know what is going on, we can REDUCE the problem to Kirchhoff and put everything together.



In relation to the voltmeter, the left arc is a generator in series with half of the total resistance of the loop. The right arc is the other half and is just a load. It doesn't contribute to the emf the voltmeter sees.

If you move the voltmeter to the left side of the loop, the left arc is now the load and the right side is voltage source in series with an internal resistance. Notice that source is inverted.




But, of course, we are smart people. We place a voltmeter immune to a magnetic field in the center of the loop. And we have the equivalent Kirchhoff reduction.



And bingo! We have our 0V.

However, if we change the position of the voltmeter and of the leads we are going to have whatever voltages we want between 0 and 1V, in absolute terms.

So what do we take from that? Kirchhoff doesn't always hold. Kirchhoff does not expect that the readings of your voltmeter varies depending on the position of your test gear in relation to your device under test. You need to inspect what you have under Faraday-Maxwell and then reduce it to Kirchhoff, i.e., you have to define things that are not predicted by Kirchhoff before you apply it.

We are accustomed to measuring the open-circuit voltage of secondary of transformers and we do not care about Faraday-Maxwell, but what we are really doing is an implicit reduction to Kirchhoff because we always have the configuration shown in the second picture of this post.
« Last Edit: November 14, 2018, 04:48:35 am by bsfeechannel »
 

Offline Arznei

  • Contributor
  • Posts: 19
  • Country: de
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #107 on: November 14, 2018, 07:15:21 am »
To be honest, I don't understand what all the fuss is about. What Dr. Lewin is pointing out with his experiment is, as he has repeated often by now, basic physics. The most basic definition of a voltage between two points is expressed as the line integral of the electric field from point A to point B. It is obvious that this only works properly as long as the line integral is _independent_ of the path chosen. If it becomes, for whatever reason, dependend on the path chosen you can no longer define a potential function for that electric field. You don't even need physics for that, thats just basic math at this point. With no potential function the term "voltage" becomes imprecise, because it does not take into account the path of measurement.

Now, if you design an experiment to include a changing magnetic field you get curl E != 0 at some area of the experiment making the very definition of voltages and potentials meaningless. There is this very nice paper [1], that has been shared on this forum multiple times, explaining to the last detail how it is that even with perfect measuring tools you will get two different voltage readings between two points. Note that in this paper bad probing is not a problem. The area where the probes are have no magnetic field, meaning there is _no_ voltage induced into the probes. I think this very fact of voltage not beeing defined is what Dr. Lewin wants to bring across, and I can understand him getting frustrated when people continue to disagree with him there. I was scared of this paper at first, but believe me there is no complicated math involved.

Now to the part of KVL not holding any more: it seems people have different understandings of what KVL means. If you know it as "summed up voltages across a loop in a circuit are zero" then _yes_ it does not hold in this case. Because in the experiment shown by Dr. Lewing the voltages do not sum up to zero. And I will trust him when he says that at MIT they did the experiment with superconductors _which cannot have an electric field inside them_ so there is no voltage across the "ring" that magically makes KVL hold.

If you interpret KVL to say "summed up voltages across a loop in a circuit sum up to the magnetically induced voltage" then this version of KVL holds. It is not nearly as practical as the first one though. You can, for example, not use it in a schematic. Because a schematic does not define the physical location of a component - and therefore, you don't know where the wires are and you don't know what the induced EMF of the loop will be.

Now if you are an engineer, most often you will not deal with "real physical devices" but you will _model_ them to make a equivalent circuit. That is where the voltage source, that many people want to add to the circuit, comes into play. Yes, as long as you are careful with your circuit you can model the effects of EMF into a voltage source. But the very fact people start to talk about "bad probing" all the time shows why this is now only a model of the real experiment. If this were a voltage source it wouldn't matter how I place my probes. They will always read the same voltage. With an EMF they do not. The concept of "bad probing" is that the measurements depend on the way you place your probes. Hence, the voltage you get is *path dependent*

It is of course a common problem, that a model of reality will fail at some point. So to still work with your model, you need to _know how and why_ it can fail. So, as long as you know what you do you can use KVL, but you need to keep in mind that it will only work as long as your actual physical device is carefully designed to match your model (or the other way around).

[1] http://www.phy.pmf.unizg.hr/~npoljak/files/clanci/guias.pdf
« Last Edit: November 14, 2018, 07:24:05 am by Arznei »
 

Offline ogden

  • Super Contributor
  • ***
  • Posts: 1576
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #108 on: November 14, 2018, 07:17:16 am »
However, if we change the position of the voltmeter and of the leads we are going to have whatever voltages we want between 0 and 1V, in absolute terms.

No. We can't measure whatever voltage we want between 0 and 1V. You shall check model of ring split ring into 10 parts or 4 parts each having 1/4V and 1/4R accordingly. BTW we talked about such case in this thread already.

Quote
So what do we take from that?

That you shall double-check your theory.

Quote
Kirchhoff doesn't always hold.

I'm afraid that you did not prove that.
 

Offline Sredni

  • Regular Contributor
  • *
  • Posts: 177
  • Country: aq
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #109 on: November 14, 2018, 07:32:44 am »
Quote
No. We can't measure whatever voltage we want between 0 and 1V.

Almost right. If the emf is 1V, with equal resistances in the loop you can measure any voltage you want between -0.5V and +0.5V.
This goes from intercepting the whole flux in one sense to intercepting the whole flux in the opposite sense.

Quote
You shall check model of ring split ring into 10 parts or 4 parts each having 1/4V and 1/4R accordingly. BTW we talked about such case in this thread already.

Trying to localize in lumped components an inherently distributed phenomena will lead you to contradictions. Trust me, I know because I too did it, before seeing the light. :-)
« Last Edit: November 14, 2018, 07:34:37 am by Sredni »
All instruments lie. Usually on the bench.
 

Offline ogden

  • Super Contributor
  • ***
  • Posts: 1576
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #110 on: November 14, 2018, 07:33:23 am »
Now to the part of KVL not holding any more: it seems people have different understandings of what KVL means. If you know it as "summed up voltages across a loop in a circuit are zero" then _yes_ it does not hold in this case. Because in the experiment shown by Dr. Lewing the voltages do not sum up to zero.

Experiment of the Dr.Lewin does not show it (voltages do not sum up to zero). It is explained in the paper you BTW refer to: http://www.phy.pmf.unizg.hr/~npoljak/files/clanci/guias.pdf.

Quote
And I will trust him when he says that at MIT they did the experiment with superconductors _which cannot have an electric field inside them_ so there is no voltage across the "ring" that magically makes KVL hold.

We are doing experiment with superconductor loop and single, lets' say 100 Ohm resistor. What will be EMF voltage on particular resistor during experiment?
 

Offline ogden

  • Super Contributor
  • ***
  • Posts: 1576
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #111 on: November 14, 2018, 07:40:38 am »
Almost right. If the emf is 1V, with equal resistances in the loop you can measure any voltage you want between -0.5V and +0.5V.

:palm:

Prove it. Download spice (LTspice) and try to model it with more than few lumped elements.

Quote
Trying to localize in lumped components an inherently distributed phenomena will lead you to contradictions. Trust me, I know because I too did it, before seeing the light. :-)

Try me. Please explain  why your model of two batteries and two resistors is OK to show that you read 0V, but mine with 10x0.1V batteries and 10x0.1R resistors is not?

« Last Edit: November 14, 2018, 07:47:00 am by ogden »
 

Offline Arznei

  • Contributor
  • Posts: 19
  • Country: de
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #112 on: November 14, 2018, 07:46:31 am »
Yes, he did show exactly that. Open up a physics book of your choice and it should tell you the KVL in presence of magnetic fields is that the sum of all voltages equals the induced EMF. At least from what I can tell that is also what any professor at university taught me.

If you take a ring of superconducting material and connect it to a resistor at one side, the voltage across the resistor will read whatever is appropriate according to faraday. If, however, you measure across the superconducting ring you will read 0 Volts.
 

Offline ogden

  • Super Contributor
  • ***
  • Posts: 1576
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #113 on: November 14, 2018, 07:50:14 am »
If you take a ring of superconducting material and connect it to a resistor at one side, the voltage across the resistor will read whatever is appropriate according to faraday. If, however, you measure across the superconducting ring you will read 0 Volts.

Wait... Please explain. Im afraid that I do not follow you here. Are you saying that if I route my test leads on resistor side, I will read voltage whatever is appropriate according to Faraday, but when I place my test leads on superconducting ring side, I read 0V?
 

Offline Arznei

  • Contributor
  • Posts: 19
  • Country: de
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #114 on: November 14, 2018, 07:51:37 am »
Yes.

Edit: Now I'm sorry, this is not a chat but a forum so i should probably explain myself a little. What you get from faraday is the voltage *across the closed loop*. So thats Superconducting Ring + Resistor. From the resistance of the total loop (which is 0R for super conductor + resistance of resistor) you calculate the the current. Now take the resistance of the super condcutor, which is 0, and multiply it with the current. You will get 0V across the super conductor.
« Last Edit: November 14, 2018, 07:57:23 am by Arznei »
 

Offline Sredni

  • Regular Contributor
  • *
  • Posts: 177
  • Country: aq
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #115 on: November 14, 2018, 08:04:33 am »
"Prove it"

That's what Lewin did. See his video "Science and believing are different things"  and follow his mesh analysis.
He is doing it with two voltmeters outside the main loop where B is changing. As long as you are outside you can always see two meshes with one of the voltmeter inside: one that do not contain the B field, and another one that contains it. As far as the B-encircling mesh go, in one case you have a positive orientation passing trought the small resistor, in the other one a negative orientation passing trough the big resistor. And this gives rise to the different voltages shown by the two external voltmeters.

But you can expand on that and bring the voltmeter inside the B-encircling loop. In this case for the inner voltmeter you will see two meshes that are both inside the B field: one positively oriented passing through the small resistor and the other negatively oriented passing through the big resistor. Depending on what fraction of the area is intercepted in the small or big R mesh, you end up with a different voltage reading.

You should interpret the square mesh as conceptual diagram, the actual area intercepted depends on the path of the probes in the real wold, but still goes from intercepting all the B-varying region in one sense, to intercepting all the B-varying region in the other. In between you get all the values of voltage comprised between the reading of the voltmeter on the left and the reading of the voltmeter on the right. All of this from the same two points!

Do the math, solve the circuit and you will see the light.

Now, I understand the urge to choose, among all these values the one that makes you feel better, namely the one from a path the equally splits the area (I should say the normal flux to be precise) between the two meshes, but that's just an arbitrary decision.

EDIT to answer the question posted below: removed another extra space at the bottom.
Also to clarify: I have zero batteries in my circuit, just the emf. And the two resistors are lumped component part of the physical circuit, not a tentative to model a distributed component.
As for your attempt to model a distributed circuit (where voltage is non single-valued) with a lumped circuit (where voltage is single-valued) to show that voltage is not multi-valued, I guess it's logically flawed.

Solve the mesh circuit with the emf, do the math. You will see the light.
« Last Edit: November 14, 2018, 08:20:25 am by Sredni »
All instruments lie. Usually on the bench.
 

Offline ogden

  • Super Contributor
  • ***
  • Posts: 1576
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #116 on: November 14, 2018, 08:10:26 am »
Yes.

Edit: Now I'm sorry, this is not a chat but a forum so i should probably explain myself a little. What you get from faraday is the voltage *across the closed loop*. So thats Superconducting Ring + Resistor. From the resistance of the total loop (which is 0R for super conductor + resistance of resistor) you calculate the the current. Now take the resistance of the super condcutor, which is 0, and multiply it with the current. You will get 0V across the super conductor.

Oh, really? When it is convenient, you just forget about EMF.

We take 1.0V battery with 0.001 Ohm  internal resistance, connect it to 1 Ohm resistor. 1A current will flow. Now take the resistance of the battery  which is 0.001, and multiply it with the current. Result is 0.001V. So battery miraculously is not 1V anymore but 0.001V?  :palm:
 

Offline ogden

  • Super Contributor
  • ***
  • Posts: 1576
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #117 on: November 14, 2018, 08:12:48 am »
"Prove it"

That's what Lewin did.

TL;DR. I did not find answer to my very simple question I asked:

Please explain  why your model of two batteries and two resistors is OK to show that you read 0V, but mine with 10x0.1V batteries and 10x0.1R resistors is not?

[edit] Dr. Lewin just re-created experiment http://www.phy.pmf.unizg.hr/~npoljak/files/clanci/guias.pdf. Unformtunately for fanboys of Dr. Lewin, that experiment does not disprove KVL.
« Last Edit: November 14, 2018, 08:20:18 am by ogden »
 

Offline Arznei

  • Contributor
  • Posts: 19
  • Country: de
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #118 on: November 14, 2018, 08:18:53 am »
Yes.

Edit: Now I'm sorry, this is not a chat but a forum so i should probably explain myself a little. What you get from faraday is the voltage *across the closed loop*. So thats Superconducting Ring + Resistor. From the resistance of the total loop (which is 0R for super conductor + resistance of resistor) you calculate the the current. Now take the resistance of the super condcutor, which is 0, and multiply it with the current. You will get 0V across the super conductor.

Oh, really? When it is convenient, you just forget about EMF.

We take 1.0V battery with 0.001 Ohm  internal resistance, connect it to 1 Ohm resistor. 1A current will flow. Now take the resistance of the battery  which is 0.001, and multiply it with the current. Result is 0.001V. So battery miraculously is not 1V anymore but 0.001V?  :palm:

No, the result of 0.001V is the voltage across the internal resistance of the battery, not the internal voltage source. Which is why the voltage across the terminals of the battery will have dropped by 0.001V, what are you trying to get at now?

I did not forget about EMF, how do you come to that conclusion? Faraday gives us the voltage across the closed loop. You have a loop consisting of 2 components, one is superconducting and will not have any voltage across it. The other is a resistor, which can have a voltage across it if there is current flowing through it. So, naturally, all voltage in the circuit will be across the resistor.

Think of the way this voltage is created: the change of magnetic field puts a force on the charges in the ring+resistor. Now inside a superconductor it doesn't take energy to move a charge (hence resistance of 0R) so your electric field is 0. Inside the resistor it does tage energy to move the charges across, so you will have an electric field of nonzero, therefore a voltage across the resistor.
 

Offline ogden

  • Super Contributor
  • ***
  • Posts: 1576
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #119 on: November 14, 2018, 08:33:44 am »
I did not forget about EMF, how do you come to that conclusion? Faraday gives us the voltage across the closed loop. You have a loop consisting of 2 components, one is superconducting and will not have any voltage across it. The other is a resistor, which can have a voltage across it if there is current flowing through it. So, naturally, all voltage in the circuit will be across the resistor.

"all voltage in the circuit will be across the resistor" .. where superconductor loop terminals are connected meaning same voltage will be present on the both ends of the loop as well. So you do not measure different voltages by manipulating with test leads, this is dumb. Actual effect of test lead placement is explained in the document linked here so many times already.
« Last Edit: November 14, 2018, 08:42:10 am by ogden »
 

Offline Arznei

  • Contributor
  • Posts: 19
  • Country: de
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #120 on: November 14, 2018, 08:48:41 am »
I did not forget about EMF, how do you come to that conclusion? Faraday gives us the voltage across the closed loop. You have a loop consisting of 2 components, one is superconducting and will not have any voltage across it. The other is a resistor, which can have a voltage across it if there is current flowing through it. So, naturally, all voltage in the circuit will be across the resistor.

"all voltage in the circuit will be across the resistor" .. where superconductor loop terminals are connected meaning same voltage will be present on the both ends of the loop as well. So you do not measure different voltages by manipulating with test leads, this is dumb. Actual effect of test lead placement is explained in the document linked here so many times already.

No, this is precisely what this is all about. If you measure at the same two points of a circuit you can get different readings on a voltmeter in the presence of an EMF. The reason this is the case and is *not dumb* is that in the presence of EMF the electric field is not a conservative vector field. Your voltage will vary depending on which path you take.

You want to know the voltage across the superinductor, you integrate along a line through the superconductor. Your voltage will be zero, because inside an ideal conductor the electric field is always zero.

You want to know the voltage across the resistor, you integrate along a line through the resistor. Your voltage will not be zero.

I hope we agree, that the voltage induced by the changing magnetic field will be across the resistor. Take any textbook, look up the introdcution to magnetic indcution. You will find an "almost closed loop" split up at some point and a resistor attached to that "split". Every textbook will tell you the voltage across the resistor is the closed-loop-voltage as dictated by faraday.

Now assuming for a second that the same voltage is also present across the superinductor. In that case the "closed-loop" voltage will be zero. That directly contradicts faradays law.
 

Offline ejeffrey

  • Super Contributor
  • ***
  • Posts: 1683
  • Country: us
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #121 on: November 14, 2018, 10:14:00 am »
You can't do that with changing magnetic fields.  Curl-E = -dB/dt, so there is no scalar potential that describes electron motion.  That is what Walter Lewin was showing.  Everything else is just noise.

Yeah, but that "noise" is a very big claim that two points on the same circuit measure differently, he states that as a fact and uses a flawed demonstration to try and prove it. This is why many people have a big problem with this.

There is no way to fix the demonstration that keeps the character because the whole point is that "voltage" is undefined in the circuit in question.  So there is no consistent way to define a measurement of it.  I don't get why this is hard to understand.

Yes, there are always good and bad measurement techniques and you always have to worry about measurement technique causing errors.  But you always assume there is some underlying "true" value you are trying to get at.  Here, the "flawed" measurement is actually intrinsic to the problem.
 

Offline RoGeorge

  • Super Contributor
  • ***
  • Posts: 1194
  • Country: ro
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #122 on: November 14, 2018, 10:40:55 am »
First I want to point out that SPICE is for simulating, not for proving.

Second, SPICE is not aware of the electric and magnetic fields.  SPICE is for solving circuits using Kirchhoff, and it can work only after you, the SPICE user, converts the geometry of a real circuit into a lumped circuit, with equivalent coils and capacitors.  You, the user, have the responsibility to take the real circuit where Kirchhoff might fail, and convert it into an equivalent lumped circuit with all kinds of coils and capacitors and transformers, and transmission lines, so SPICE can crunch the numbers using only Kirchhoff and no Maxwell.  It is the SPICE user who must take into considerations not only the geometric size and shapes, but also the permeability and the dielectric properties of the surrounding materials. That is a lot left outside SPICE, and even more, the topology of the lumped circuit needs to be changed when the external conditions change.

A lumped circuit is an equivalent circuit, where you, the user, are responsible for finding the correct topology of coils and capacitors and their values in such a way that Kirchhoff's law can be applied.
https://en.wikipedia.org/wiki/Lumped_element_model

Spice is solving its I/V matrix equations based on Kirchhoff and lumped circuits.  That is why in SPICE you won't find any circuit that does not obey Kirchhoff.

In real circuits, any piece of wire is a capacitor, a coil and a resistor in the same time.  This is crazy, and makes it almost impossible to calculate a lumped equivalent that holds with the real world.  We have some dedicated shapes for capacitors and coils where the formulas for L and C are well known, but that's all.  For a random piece of wire there is no clear formula without fields and material constants, and Maxwell.  If we go into transmission lines or antennas, things goes even more complicated.  There are no antennas in SPICE.  There is no causality in SPICE.  Propagation in lumped circuits is considered instantaneous, which in the real world is not true.

If we go further, let's say to calculate the beam of a particles accelerator, SPICE can not help.  Spice does not know Maxwell, and does not know relative speeds and Special Relativity, SPICE knows only Kirchhoff.

If we go even further, for satellites working in different gravitational fields, SPICE is again unaware of General Relativity.  It is us who need to take Special or General Relativity into consideration, and invent a lumped circuit in such a way that we capture all these effects into our imaginary lumped circuit.

Maxwell's equations stand no matter what.  That is why they are so praised.  Maxwell's equations does not need inductors and capacitors.  Kirchhoff does.  Again, Kirchhoff is great for static fields (AKA conservative), but can not be applied for non-conservative fields.  It was not meant for such non-conservative fields.

In order to use Kirchhoff for non-conservative (loosely speaking "variable" fields), we found a workaround, a trick.  We stuff the real circuit with all kinds of imaginary parts, and not only we add those imaginary parts, but the values and the topology of the lumped circuit keeps changing based on external fields and external surrounding of the real circuit, yet our real circuit never changed during this, only the surroundings changed.  We call this stuffed imaginary circuit a lumped equivalent circuit of the real thing.

This fake stuffing we do in order to obtain a lumped circuit is very, very fake and unnatural thing to do, yet we considered it NORMAL, because we are used to it, and because it makes our calculations easier.

In fact, we did it so often that the lumped circuit became to us a second nature, while the real circuit is looked rather as a fussy oddity that needs a lot of care (like electric and magnetic shielding) for it in order to work as expected.  We got so used with our imaginary fake lumped circuits that we now consider reality an anomaly.  We call reality parasitic hum, parasitic inductance coupling, parasitic capacitive coupling, parasitic ground loops, parasitic antennas, parasitic transmission lines, parasitic everything.

Suddenly, reality become for us just a parasitic effect, and we like to consider it that way just to accommodate our distorted lumped circuits with the cruel reality.  That has a lot to do with how human mind works, and from here we are in the realm of human psychology and philosophy.

"A fake repeated over and over becomes reality" <- just an old saying.

Peace!
« Last Edit: November 14, 2018, 08:52:40 pm by RoGeorge »
 
The following users thanked this post: orin, ogden, mhz

Offline ejeffrey

  • Super Contributor
  • ***
  • Posts: 1683
  • Country: us
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #123 on: November 14, 2018, 11:27:27 am »
To confirm what RoGeorge said:

Once you get to the point you can't make a sufficiently accurate lumped element circuit model, you have to resort to field solvers such as Sonnet, ADS/Momentum, HFSS, or COMSOL.  Those model what is actually going on in a geometrical model of your circuit.  You can't extract a voltage from those.  I actually tried this once indirectly -- I was trying to make a nice heat map of a crosstalk simulation result.  When it had weird discontinuities I looked at what I was doing and realized it made no sense.
 

Offline ogden

  • Super Contributor
  • ***
  • Posts: 1576
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #124 on: November 14, 2018, 01:22:39 pm »
I did not forget about EMF, how do you come to that conclusion? Faraday gives us the voltage across the closed loop. You have a loop consisting of 2 components, one is superconducting and will not have any voltage across it. The other is a resistor, which can have a voltage across it if there is current flowing through it. So, naturally, all voltage in the circuit will be across the resistor.

"all voltage in the circuit will be across the resistor" .. where superconductor loop terminals are connected meaning same voltage will be present on the both ends of the loop as well. So you do not measure different voltages by manipulating with test leads, this is dumb. Actual effect of test lead placement is explained in the document linked here so many times already.

No, this is precisely what this is all about. If you measure at the same two points of a circuit you can get different readings on a voltmeter in the presence of an EMF. The reason this is the case and is *not dumb* is that in the presence of EMF the electric field is not a conservative vector field. Your voltage will vary depending on which path you take.

You want to know the voltage across the superinductor, you integrate along a line through the superconductor. Your voltage will be zero, because inside an ideal conductor the electric field is always zero.

You want to know the voltage across the resistor, you integrate along a line through the resistor. Your voltage will not be zero.

Ok. I am still trying to connect theory to my real-world knowledge of transformers. Dr. Lewin do not use superconductor in his transformer. So we can take transformer w/o superconductor as well - AC mains transformer. So you say that my AC mains transformer with 0.1 Ohms secondary winding does not actually output 12VAC that I measure with my (10MOhm input) RMS multimeter?  :-//
« Last Edit: November 14, 2018, 03:39:07 pm by ogden »
 


Share me

Digg  Facebook  SlashDot  Delicious  Technorati  Twitter  Google  Yahoo
Smf