Author Topic: Does Kirchhoff's Law Hold? Disagreeing with a Master  (Read 61142 times)

0 Members and 1 Guest are viewing this topic.

Online ogden

  • Super Contributor
  • ***
  • Posts: 2566
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #125 on: November 14, 2018, 02:57:56 am »
ogden, 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.

I did not ask to prove Maxwell's equations using SPICE  :palm: Before you even consider to spread your wizdom, you really shall read what we actually were talking about.
 

Offline Berni

  • Super Contributor
  • ***
  • Posts: 2422
  • Country: si
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #126 on: November 14, 2018, 06:50:17 am »
There is another way of looking at it. Circuits analysis calls things like this a over-defined circuit. In the real world these circuits don't happen but can happen in diagrams with ideal components.

You don't even need magnetic fields to create such a circuit. Lets go break Kirchhoffs law again! :box:

For example this is an over-defined circuit:


What is the voltage across A and B? Well looking from one side its 1V, looking from the other side is 2V. And look Kirchhoffs voltage law is wrong again! The voltages don't add up anymore.

What happens if you do this in real life? You just get lots of current and the 1V that Kirchhoffs is missing to make it work is in the internal resistances of the voltage sources and wires. This is essentially a battery charger circuit.

Okay okay everyone knows you can't just parallel together voltage sources or series together current sources. That's like dividing by zero in math. So lets take a different circuit then:


When the switch is open its pretty easy to solve the voltages everywhere. A to B is 1V and C to D is 0V (Assuming the capacitor was not given any energy upon its creation in the universe). But then lets close the switch and ask what are the voltages at that moment.

So lets look at it from the left side first:
Voltage source is putting 1V there so A to B must be 1V. And switch is closed so A=C and B=D. So in that case the voltage between C and D is also 1V. Solved

Now lets try solving it from the right side:
If we know the voltage across the capacitor then we can know the voltage between C and D. We do have a formula for the voltage on a capacitor:


So the 1/C part is simple, we know its 1uF so that works out to 1e6. We also know V(t0) is 0V. All we need is the integral of the current. Since the switch has been closed for 0 seconds means the integral is also 0 so from this we conclude the voltage on C and D is 0V....wait didn't we say it was 1V before?Aha! Dr. Lewin is on to something, it does matter from what direction you look at it!

Alright fine the switch basically didn't do anything because no time has passed. Lets fix out question a bit then "What is the average voltage on the nodes in the span of 1ms of the switch closing". Okay now we are integrating from t=0 to t=0.001 .This time we do need to calculate the current, since this is one loop this is easy. We just sum up all the voltages and resistances and use Ohms law. So we get I = 1V / 0 Ohm ... yeah we can't divide by zero so the current is undefined. So the integral is undefined. So the voltage on C and D is also undefined. Okay fine we could say the current is infinite since approaching division by zero limits towards infinity. Well then the integral is also infinite and we find C and D have infinitely large voltage. But as soon as we introduce a resistance of not 0 in there we can calculate it just fine and it becomes a mathematically valid circuit.

This is the same as trying to solve:
x2 = -3
x = ?
Yes i know this has a complex number answer, but when doing DC circuit analysis what does a current of 2+j3 Amps look like in DC?

Circuit analysis methods break when you introduce these over-defined circuits, its not just SPICE that will throw a angry error message at you for trying to simulate one. Analyzing it by hand simply gives you weird results like 0, undefined or infinity in places where there should be a sensible number, or you get multiple results for the same voltage or current depending on what path you calculate.

Dr. Lewins circuit is also such a circuit because he is forcing a current around while the resistors try to define there own current. Its not only Kirchhoffs law that breaks in such circuits, you can break many other laws since the math simply does not come together.
« Last Edit: November 14, 2018, 06:54:45 am by Berni »
 

Offline RoGeorge

  • Super Contributor
  • ***
  • Posts: 1657
  • Country: ro
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #127 on: November 14, 2018, 09:51:54 am »
ogden, 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.

I did not ask to prove Maxwell's equations using SPICE  :palm: Before you even consider to spread your wizdom, you really shall read what we actually were talking about.

I think SPICE was used in the previous pages to prove Kirchhoff holds, you used it in page 4, but maybe I misunderstood why you used it there.  My apologies if it was so.

Anyway, I shouldn't have started a 2AM rambling speech with a name, in the first place.  Name removed.

Offline Sredni

  • Regular Contributor
  • *
  • Posts: 185
  • Country: aq
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #128 on: November 14, 2018, 04:30:56 pm »
@ogden

you probably missed my answer because I added it to my previous post to keep the number of posts down.
Here it is again:

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.

So to avoid wasting this post space with just a repetition, let me add this

Forget the circuit with two resistors and the voltmeter. That appears to be too complex.
Think of just two voltmeters hooked up to form a loop. Place the coil inside the loop. The instruments will measure voltage in opposition of phase with amplitudes partitioned according to the ratio of their internal impedance (suppose one with 9 Mohm impedance, the other with 1 Mohm impedance at the frequency of the test signal - ok to have decent reading with homelab created magnetic field we'd probably need far lower impedences, but consider this to be a thought experiment).
So, which one is probing 'the right way'? (please don't say "the right one" :-) )
Are they both wrong? And yet it checks out: the sum of the voltages around the loop is not zero, it's the emf.
Engineers like to think that to be a 'generalized' KVL. The sum of voltages and emfs around a loop gives zero, but somehow some of them freak out when they see two voltmeters, attached to the very same two endpoints, reading two different values (that add up exactly to what they expect!!!) and call that 'bad probing'. Go figure.
All instruments lie. Usually on the bench.
 

Offline Vtile

  • Frequent Contributor
  • **
  • Posts: 986
  • Country: fi
  • Ingineer
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #129 on: November 14, 2018, 05:28:05 pm »
This is a good thread to ask. Does anyone know the original publication of these two Kirchoff laws 1st and 2nd, not voltage and current laws as usually wrongly tittled.

There is probably also pointed out that this our typical "ideal" circuit analysis way of drawing circuits is also ideal and centralized model that brakes down miserably when you go smaller in time domain, or go further in component size (ie. cross continent powerlines).
« Last Edit: November 14, 2018, 05:33:43 pm by Vtile »
 

Online ogden

  • Super Contributor
  • ***
  • Posts: 2566
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #130 on: November 14, 2018, 05:44:53 pm »
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.

As you disagree with my model where between any two points of 1Ohm ring where 1A current is running, there's always is 0V, you shall demonstrate mesh circuit with emf, do the math to prove that I am wrong and prove that you are right:

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.

Show me the light you so often refer to. - If you can.
« Last Edit: November 14, 2018, 05:55:09 pm by ogden »
 

Offline Berni

  • Super Contributor
  • ***
  • Posts: 2422
  • Country: si
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #131 on: November 14, 2018, 05:45:19 pm »
Yes the voltmeters would read different values in a 9 to 1 ratio because the EMF voltage is in the wires used to connect them.

But along the way i found another fun way to break Kirchhoffs law without using any magnetic field:


Here the orange wire is made out of constantan(55% Cu, 45% Ni) while other wires are made out of copper. This creates a thermocouple joint at both ends of the constantan wire. If we now heat the indicated corner to 100°C while leaving all other parts of the circuit at ambient temperature we get the shown readings on both voltmeters.

3.03mV + 0.34mV = 3.37mV

And Kirchhoffs voltage law is broken again because we should be getting 0V when we sum the voltages in the loop. We could again fix it by adding a lumped thermocouple model in the corners where the missing hidden voltage source is located, but if we are not adding any lumped elements to the circuit to describe the magnetic properties of the wire then i suppose we also wouldn't add lumped elements to describe thermocouple effects. The thermoelectic effect after all does not need electric fields to push electrons around, just like the magnetic field does not (But magnetic fields do sort of create 'virtual electric fields' trough the effects of special relativity).
 

Offline free_electron

  • Super Contributor
  • ***
  • Posts: 7218
  • Country: us
    • SiliconValleyGarage
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #132 on: November 14, 2018, 06:16:44 pm »
Current induced in the measuring wires.
it would be interesting to try this with untrimmed surface mounted film resistors
Those thru-hole resistors are most likely spiral cut ( which forms inductors by itself ... )
Professional Electron Wrangler.
Any comments, or points of view expressed, are my own and not endorsed , induced or compensated by my employer(s).
 

Offline Sredni

  • Regular Contributor
  • *
  • Posts: 185
  • Country: aq
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #133 on: November 14, 2018, 06:50:03 pm »
Yes, current induced in the measuring wires. But the point is that now we read two different values from the same points.
This is not a measurement issue, it is exactly what is predicted by the theory. And in fact, the values confirm that the sum of the voltage read around the loop do not give zero. They give the value of the emf.

Even with the original Lewin circuit you had current induced in the measuring wires. It was deemed negligible because the resistances in the circuit were much lower than the multimeter's impedances. But still, despite the probes not linking the flux there was always a mesh that enclosed the whole area with variable field. In one case with a positive orientation and containing a high valued R, in the other with a negative orientation and containing a low valued R. This is what gives rise to the different reading.

When you take the probes path inside the variable flux area, you end up partitioning it and now both meshes are linking the flux with opposite orientation and variable proportion (determined by how you place the probes). You can then get a reading of all voltage values included between the extreme values of 0.9V and -0.1V depending on which of the two meshes cuts more flux. From the same two endpoints!

Now, it seems that someone would like to single out, among the infinite values of V, the one where the contribution from the two meshes cancel out, put it on a pedestal and worship it as the "true voltage". I have no problems with that. You can find that same value from the same two endpoints for an infinite number of other probe path configurations, no need to twist and shield: just jerrymander the cables so that you get the value you want.

But the point is that voltage is no longer defined in this context.
And that's exactly what the math predicted: if the integral of the E field across any closed path is not zero, then V is not determined by the endpoints only, but also by the shape of the path.
This is not even basic physics. It's basic mathematics.

All instruments lie. Usually on the bench.
 

Online ogden

  • Super Contributor
  • ***
  • Posts: 2566
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #134 on: November 14, 2018, 07:00:51 pm »
Yes, current induced in the measuring wires.

 :palm:
 

Offline Sredni

  • Regular Contributor
  • *
  • Posts: 185
  • Country: aq
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #135 on: November 14, 2018, 07:05:00 pm »
@ogden

Thou doth facepalm too much.

Solve the circuit shown in Lewin's "Science and believing are different things", do the math.
Now I am no longer convinced you will see the light, but it's still worth a shot.

EDIT to answer the post below: I am not your tutor. Lewin has already solved it in his video, I did check it on my own but I am not wasting my time translating and formatting (I even tried to upload a pic two days ago, but gave up on tinypic and its enter code here and there) it for someone who does not even want to watch a video.
Maybe next week I will tell you a fairytale about an island with a hill and an instrument to measure the energy per unit mass required to move a mass from bottom to top, with and without a curly wind. You might understand what potential really is.


edit: correct mispelled name
« Last Edit: November 14, 2018, 07:40:15 pm by Sredni »
All instruments lie. Usually on the bench.
 

Online ogden

  • Super Contributor
  • ***
  • Posts: 2566
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #136 on: November 14, 2018, 07:32:24 pm »
Solve the circuit shown in Lewin's "Science and believing are different things", do the math.
Now I am no longer convinced you will see the light, but it's still worth a shot.

Trying to shift discussion off-subject?  :-DD

I already did show simulation using lumped elements (10x0.1V EMF sources and 10x0.1R resistance). You disagreed. So now you shall solve the mesh circuit with the emf, do the math - to disprove my approach.

[edit] I remind that all I ask you - prove following words not using  :bullshit: :blah: but solving the mesh circuit with the emf, doing the math:

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.
« Last Edit: November 14, 2018, 07:48:07 pm by ogden »
 

Offline rfeecs

  • Frequent Contributor
  • **
  • Posts: 674
  • Country: us
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #137 on: November 14, 2018, 07:43:10 pm »
« Last Edit: November 14, 2018, 07:51:46 pm by rfeecs »
 
The following users thanked this post: Vtile

Online ogden

  • Super Contributor
  • ***
  • Posts: 2566
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #138 on: November 14, 2018, 07:56:01 pm »
EDIT to answer the post below: I am not your tutor. Lewin has already solved it in his video

I can't find video where he solves 1 Ohm ring loop problem where as you say - you can read any voltage between -0.5V to + 0.5V. It is not about tutoring. It's about proving that your words are not BS.
« Last Edit: November 14, 2018, 08:04:44 pm by ogden »
 

Offline Sredni

  • Regular Contributor
  • *
  • Posts: 185
  • Country: aq
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #139 on: November 14, 2018, 08:36:10 pm »
You should have read my post more carefully

"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. "

This is what I was talking about: Lewin's so much debated setup.
Your problem is even worse from the standpoint of spice simulation, because everything is distributed but still, off the top of my head, the voltage between two points forming an angle alpha with the center will be different when you measure it from one side out of the loop |V| = (alpha/2pi) * emf and from the other side of the loop |V| = (2pi - alpha)/2pi * emf (you need to fix the signs).
If you go inside you will end up will all values comprised between these two values (with adjusted signs). It all depends on how you cut the varying flux area.

EDIT
Added formulae, corrected one of the seventyfour typos and also to add:

So, your spice model shows one out of those infinite values, and that should prove what?

« Last Edit: November 14, 2018, 08:49:40 pm by Sredni »
All instruments lie. Usually on the bench.
 

Offline Berni

  • Super Contributor
  • ***
  • Posts: 2422
  • Country: si
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #140 on: November 14, 2018, 08:54:58 pm »
Yep you can get any of the infinitely many values by rearranging the wires. But then the model should capture that so that you can then solve the circuit for that particular one.

The arrangement of wires i am interested in is the one where both probing wires summed together produce a EMF voltage of 0V. This means that the end where the voltmeter is at is at the same voltage as the two points we are trying to measure since this way the probes are not affecting the measurement. This collapses the circuit down to a single solution for the voltage.
 

Offline Sredni

  • Regular Contributor
  • *
  • Posts: 185
  • Country: aq
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #141 on: November 14, 2018, 09:08:18 pm »
@Berni

Yep, and that's totally fine with me. I know that's what some called "correct probing", but you should know that you are just deceving yourself if you think that voltage has a physical meaning outside the context of the particular path you have found. As I wrote before there is no need to twist and shield your cables - just jerrymander them until the instrument shows the value you like (which for the symmetric setup of Lewin's circuit will be the average of the values, considering the signs, so... 0.4V IIRC).

The fact that that is not the only voltage is not only a matter of definition. In my fairytale about the hill, there are Conservatives and SUPERDEMOcrats that have to ship medicine to their relatives at the top of the hill. Conservatives laugh at SUPERDEMOcrats, thinking that they believe the height of the hill changes with wind, but they end up all dead because their conservative "true value" of the energy required to shoot the package up is either too high or too low to safely reach the destination when there is net wind along the path.

PS
no politics please, it's just that the names were so fitting...
« Last Edit: November 14, 2018, 09:17:53 pm by Sredni »
All instruments lie. Usually on the bench.
 

Online ogden

  • Super Contributor
  • ***
  • Posts: 2566
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #142 on: November 14, 2018, 09:31:46 pm »
Your problem is even worse from the standpoint of spice simulation, because everything is distributed but still, off the top of my head, the voltage between two points forming an angle alpha with the center will be different when you measure it from one side out of the loop |V| = (alpha/2pi) * emf and from the other side of the loop |V| = (2pi - alpha)/2pi * emf (you need to fix the signs).

Where did you lost 1 Ohm distributed resistance? You also forgot that loop we are talking about is electrically short!

Let's view at obvious case where alpha equals PI. As two identical halves of the loop we have one +0.5V and one -0.5V EMF source and two 0.5 ohm resistors connected in parallel series to each of them. How do we measure any voltage here between those two points?

- By "scientifically" manipulating voltmeter leads? ;)
« Last Edit: November 14, 2018, 10:22:37 pm by ogden »
 

Offline Sredni

  • Regular Contributor
  • *
  • Posts: 185
  • Country: aq
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #143 on: November 14, 2018, 10:03:39 pm »
Where did I lose the resistance? I didn't. If the emf is 1V, the current is 1V/1ohm = 1amp.
Why do you think the measured voltage should depend on the value of the resistance? Is it not a uniform resistance loop?
What do you expect by raising the total resistance to 2 ohms?

EDIT
also with 1V emf, and the measure points on a diameter, you will measure -0.5V on one side, from outside the loop, + 0.5V from the other side, still outside the loop, and any value you wish in between when you go inside the loop, depending on how much of the flux you cut in the two parts you are partitioning the area.
If you want to read the 0V you like so much, you place your inner voltmeter along a diameter. Or, you can go yin-yang.

EMF is still 1V, why should that change?
« Last Edit: November 14, 2018, 10:09:50 pm by Sredni »
All instruments lie. Usually on the bench.
 

Online ogden

  • Super Contributor
  • ***
  • Posts: 2566
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #144 on: November 14, 2018, 10:18:32 pm »
Where did I lose the resistance? I didn't. If the emf is 1V, the current is 1V/1ohm = 1amp.
Why do you think the measured voltage should depend on the value of the resistance? Is it not a uniform resistance loop?
What do you expect by raising the total resistance to 2 ohms?

Just realized mistake in a hurry in previous post, corrected. - Each half of the loop is 0.5V EMF generator having 0.5 Ohm internal resistance, so obviously 0.5 Ohms is connected in series to 0.5V EMF source.

Quote
also with 1V emf, and the measure points on a diameter, you will measure -0.5V on one side, from outside the loop, + 0.5V from the other side

Only if loop is not shorted. We talk about shorted loop!
 

Offline Sredni

  • Regular Contributor
  • *
  • Posts: 185
  • Country: aq
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #145 on: November 14, 2018, 10:35:10 pm »
"Each half of the loop is 0.5V EMF generator"

EMF on half loop?
What's next? Magnetic monopoles?
I am sorry, I can't help you fill those gaps.
Continue facepalming.

EDIT after your following post: my last remark, I'll leave you to emoticons after this.
When your endpoints are on a diameter you have +0.5V measuring outside on the left (whatever); -0.5 V measuring on the other side, still with voltmeter outside the loop. Total, following the signs +0.5 - (-0.5) = 1V. That's the emf. It does not show up as a lumped generator, it's distributed. Your error is trying to localize it and compute fractions of it when you consider only part of the circuit.
« Last Edit: November 14, 2018, 11:15:03 pm by Sredni »
All instruments lie. Usually on the bench.
 

Online ogden

  • Super Contributor
  • ***
  • Posts: 2566
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #146 on: November 14, 2018, 10:53:10 pm »
"Each half of the loop is 0.5V EMF generator"
EMF on half loop?

What's next? Magnetic monopoles?
I am sorry, I can't help you fill those gaps.
Continue facepalming.

LOL. Nice try. When it's convenient, you just pretend that you do not understand what I am talking about - partition (half) of the closed loop. Now suddenly your formula "|V| = (alpha/2pi) * emf" giving 0.5V with alpha=PI and emf=1V is not true anymore as well. [edit] In same post I say "we talk about closed loop".
« Last Edit: November 15, 2018, 12:14:55 am by ogden »
 

Online ogden

  • Super Contributor
  • ***
  • Posts: 2566
  • Country: lv
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #147 on: November 14, 2018, 11:27:02 pm »
Just watched "Science and believing are different things", did the math. Left meter reads -900mV, right 100mV. In both - calculation and "lumped simulation". Funny that LTspice does not scream at me ;)



[Edit] agrees to paper http://www.phy.pmf.unizg.hr/~npoljak/files/clanci/guias.pdf  as well.
« Last Edit: November 14, 2018, 11:40:58 pm by ogden »
 

Offline Sredni

  • Regular Contributor
  • *
  • Posts: 185
  • Country: aq
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #148 on: November 15, 2018, 05:28:37 am »
Just watched "Science and believing are different things", did the math. Left meter reads -900mV, right 100mV. In both - calculation and "lumped simulation". Funny that LTspice does not scream at me ;)

That's amazing! I did not know Spice could do that!
Can you please highlight on the circuit the two (2) nodes that give two different values of voltage between them?
Maybe one is GND, so what is the other node?
All instruments lie. Usually on the bench.
 

Offline RoGeorge

  • Super Contributor
  • ***
  • Posts: 1657
  • Country: ro
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #149 on: November 15, 2018, 07:43:34 am »
The explanation video promised by Prof. Walter Lewin.

 
The following users thanked this post: radioactive


Share me

Digg  Facebook  SlashDot  Delicious  Technorati  Twitter  Google  Yahoo
Smf