Author Topic: Does Kirchhoff's Law Hold? Disagreeing with a Master  (Read 60248 times)

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Offline KL27x

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #150 on: November 15, 2018, 09:39:13 am »
So... Dr. Lewin just discovered impedance.

And he dropped his two scope, mobius strip with induction battery and improper probing experiment.

What a waste of time.

I still don't get why he says Kirchoffs works when the voltmeter is attached. Unless he is saying that the voltmeter has a resistance (conductance), and thus the voltmeter is what is dropping the voltage. At T approaches zero and the inductor's impedance is higher, yeah, that could be true and you could say that the voltemeter is part of the circuit; but between T= 0 and T= Tsaturation, most of the current is through the inductor and the voltage is dropped across the inductor, and the voltmeter will be pretty much not significant part of the circuit. I don't get the point he was making with that.
« Last Edit: November 15, 2018, 10:30:13 am by KL27x »
 
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Offline bd139

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #151 on: November 15, 2018, 10:04:22 am »
Just watched all three videos.

These are all abstraction tools. Some tools you pick for some jobs. Some tools you pick for others.

One thing you do is pick the right tool for the job. KVL works fine for the use cases we use.

Two things you don't do are claim people are religious and fuck up your test environment and then wiggle out of it.
 
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Online ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #152 on: November 15, 2018, 10:14:10 am »
Can you please highlight on the circuit the two (2) nodes that give two different values of voltage between them?

So you are believer.

These are all abstraction tools. Some tools you pick for some jobs. Some tools you pick for others.

Well said. He integrate over time to disprove Kirchhoff's Law but to show special case when Kirchhoff's Law holds, he uses voltmeter with high time resolution :)

What a waste of time indeed.
« Last Edit: November 15, 2018, 11:04:21 am by ogden »
 

Offline KL27x

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #153 on: November 15, 2018, 11:18:01 am »
Kirchoff reduced his observations empirically and mathematically to produce kirchoffs laws. Later, someone studied a curious thing dubbed impedance/reactance, and Faraday was able to correlate impedance with magnetic flux, mathematically, using integrals.

Dr. Lewin used these equations of giants to re-discover impedance and has amazed several college students.
 

Offline bd139

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #154 on: November 15, 2018, 11:34:58 am »
That's basically it.

Let's look at the insignficance of the entire point too....

Avoiding worrying about this is the most powerful tool in your arsenal. Avoiding any problem is.

Extrapolating my point earlier, now I have a few minutes, the whole point of the "tree of knowledge" is that you pick a branch which allows you to solve your problem without having to understand the underlying abstractions if you can help it. Your objective is to solve the problem and move on. How many people in the modern world require knowledge of simple things like integrals, derivatives and logarithms? Not many. When it comes to applied knowledge, we almost never need to care about Maxwell's Equations or vector calculus. And thank goodness for that. They're horrid. I spent a good while scratching my head over these going through Feynman's books.

An example. Just last week I was building an opamp integrator. I couldn't be bothered to work it out as an integral (it was late and I'd had way too little coffee that day and the magic integral button on my calculator didn't work) so I picked a higher level abstraction which is a current source (input resistor to non-inverting input) and ideal behaviour of an opamp (lifts output to keep non inverting at zero here). Thus you can solve it using I = C * dv/dt ... I knew I wanted to go 1v/second and had a 1uF cap sitting there staring at me waiting to be used so I knew I. I know input voltage (saturated near rail) so picked an R for that using ohms law. Did maths in head, picked closest value, job done.

Not once did I look at integrals, maxwell, even kirchoff here. Just a rule of thumb which was dervied from this entire stack of knowledge.

At university, many hours did I spend sitting there doing page after page of kirchoff, solving simultanous equations to the point I wrote a linear equation solver on my calculator so I didn't have to do it any more because my arms were so heavy that the thought of doing more made me want to just quit and go and do a BA in arts or something.

And to make a mockery of it all, two points: I never used it. Not even once.  TAOE dedicates about 1/3 of a page to it because that's all the attention it needs in the real world.

So this is an hour+ of my life I have wasted on people arguing about technicalities which change nothing, unless you're a research physicist, which I'm not and most of us aren't.

Lewin: Look my abstraction is more right than yours!

Who gives a shit?
« Last Edit: November 15, 2018, 11:36:55 am by bd139 »
 

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #155 on: November 15, 2018, 02:41:09 pm »
The explanation video promised by Prof. Walter Lewin.



Case closed.



« Last Edit: November 15, 2018, 02:50:28 pm by bsfeechannel »
 
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Offline bd139

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #156 on: November 15, 2018, 02:46:59 pm »
Well duh  :-DD

(which is the whole point here)
 

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #157 on: November 15, 2018, 03:11:43 pm »
Duh, indeed.

 

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #158 on: November 15, 2018, 08:49:36 pm »
Can you please highlight on the circuit the two (2) nodes that give two different values of voltage between them?
So you are believer.

I am sorry, it seems you do not have sarcasm in Latvia.
Let me rephrase it:

Show the id of the two nodes (not three or four, two) whose potential difference gives both 0.1V AND 0.9V.
If you can't, because you need at least 3 nodes, or more likely 4, it's because Spice can only give you an answer with a single-valued potential.

And if your aim was to find an equivalent lumped circuit where you could see 0.1V and 0.9V you should have not bothered to add all those inductances. Just uses a lumped secondary coil giving you 1V and put the two resistor sin series. There you go: in this case you need three nodes to show two voltages.

Please re-post your schematics indicating with TWO arrows the TWO nodes (not two pairs of nodes, just two nodes) that give you two separate potential differences at the same time.
My bet is that you cannot.


Quote
What a waste of time indeed.

You simply did not understand it.
It's not a crime, after all this is very counterintuitive stuff, like everything that has its roots in relativity.
Do not give up, study a bit more and you will see that Lewin is indeed right.


PS
Wanna see something nice: put two batteries with different voltage in parallel in spice. You will find that spice will give a single-valued voltage, namely the average of their voltages. Why? Because it cannot handle multivalued potentials. So it assumes a finite internal resistance to avoid impossible systems of equations.
I guess it's written in the manual.

But nowadays, who read the... fine manual anymore?
All instruments lie. Usually on the bench.
 

Online ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #159 on: November 15, 2018, 10:29:16 pm »
And if your aim was to find an equivalent lumped circuit where you could see 0.1V and 0.9V you should have not bothered to add all those inductances.

My aim was to show opposite. -That miraculous reading of two voltages in single point is just bad probing and it can be explained using simplified model of  EMF sources and loads. BTW Berni showed more detailed model of what actually happens. Where do you see inductances? You think R1 and R2 in Dr.Lewin's circuit (picture attached) are inductances as well?

Honestly I do not see any reason to continue this forum chat with you. I agree to disagree and stop internet ink waste.
 

Offline RoGeorge

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #160 on: November 15, 2018, 11:23:17 pm »
ogden, the "miraculous reading" of two voltages in single point is NOT about bad probing.

Please look at this picture made by bsfeechannel.
bsfeechannel perfectly summarizes the whole debate:



The first equation is Kirchhoff's KVL law.  In words, "the sum of all voltages in a closed loop = zero".

In the second equation, the term on the right side of the "=" is the induced EMF.  In words, "the sum of all voltages in a closed loop = induced EMF"

Kirchhoff says the sum of voltages in a loop is zero.
Faraday says the sum of voltages in a loop is EMF.
Contradiction.  Why?

When there is no induced voltages, Kirchhoff is true.
If you have induced voltages in your circuit, Kirchhoff is false.

Q: So, Kirchhoff was stupid, and the Physics is broken?
A: No.  Kirchhoff was a genius, but when he made those laws of him, Kirchhoff was talking only about circuits with NO induced voltages.
« Last Edit: November 15, 2018, 11:39:07 pm by RoGeorge »
 

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #161 on: November 15, 2018, 11:23:43 pm »
@odgen It was a lapsus. Change inductances with resistances and the gist does not change: the lumped circuits you can model in Spice cannot give you two different voltage readings from the very same two points (edit: and that's why Spice did not scream at you). BUT the system in the real world CAN.

As for your model to show two readings, yes, I sent you there just to show where those readings came from: the different flux linkage. Where did you guys think this multivalued potential came from? An alternate universe?
It is written clearly in Faraday's law.

What you call 'bad probing' is an inherent and irremovable characteristic of the physical system. Voltage is no longer 'positional' so if your aim is to create a map of 'voltage' values on the ring, you should know you can have any kind of mapping you want depending on how you place your probes.

(Edit: If your definition of 'true voltage' is what you read when the probe paths partition the area of the loop in order to give cancelling contributes...) ...what happens when the flux is not spatially uniform? You have to follow strange paths in the ortogonal section of the loop area. And if the loop is not circular symmetric, same thing. And if the flux change is relative not only to the module of the B field but also to its spatial distribution? You have to make your probes dance with the field to cancel the contribute of the flux linked by the two inner meshes.

Oh well, I guess you have the right to remain ignorant.
And I start to see why Lewin is giving those copy-pasted comments on youtube. (edit, posted after reading a post below) He was literally drag to post video after video and now people are saying "he's just repeating himself".

edit: typos and additions, as shown
further edit: typos typos typos. They're coming out of the effing walls!
« Last Edit: November 15, 2018, 11:50:50 pm by Sredni »
All instruments lie. Usually on the bench.
 

Offline rfeecs

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #162 on: November 15, 2018, 11:27:28 pm »
So in this latest video, Dr. Lewin again says that almost all the textbooks are wrong.  One he specifically calls out is Halliday and Resnick.  So just for fun, what does it say about inductors?  From Halliday and Resnick "Fundamentals of Physics"  9th edition, section 30-8:

Quote
In Section 30-6 we saw that we cannot define an electric potential for an
electric field (and thus for an emf) that is induced by a changing magnetic flux.
This means that when a self-induced emf is produced in the inductor of Fig. 30-13,
we cannot define an electric potential within the inductor itself, where the flux
is changing. However, potentials can still be defined at points of the circuit that
are not within the inductor-points where the electric fields are due to charge
distributions and their associated electric potentials.

Moreover, we can define a self-induced potential difference VL across an
inductor (between its terminals, which we assume to be outside the region of
changing flux). For an ideal inductor (its wire has negligible resistance), the magnitude
of VL is equal to the magnitude of the self-induced emf EL.

If, instead, the wire in the inductor has resistance r, we mentally separate the
inductor into a resistance r (which we take to be outside the region of changing
flux) and an ideal inductor of self-induced emf EL. As with a real battery of emf
E and internal resistance r, the potential difference across the terminals of a real
inductor then differs from the emf. Unless otherwise indicated, we assume here
that inductors are ideal.

So it doesn't attribute anything to Kirchoff's law.  It specifically says we can't define a potential from the E field produced by the inductor's changing magnetic flux.  It says we can still define a voltage outside the changing magnetic field of the inductor and use this as the potential difference across the inductor.  And we can make a "mental" lumped model of the inductor with internal resistance as an EMF in series with a resistor.

None of that sounds criminal.

He's made his point several times over in four or five videos.  At this point he's just repeating himself.
 
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Offline Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #163 on: November 15, 2018, 11:41:11 pm »
After watching the response video I do have to apologies that I missed the point. Yes I agree that Kirchhoffs voltage law is not describing the actual underlying physics. If you need to describe that go and use maxwells equations, I'm sure they work great. The video explains the difference nicely. :-+

However I still don't agree with the demo experiment. But I do agree with some of it. There is indeed 0.9V and 0.1V on those resistors, it's Ohms law after all. Where I stop agreeing is the claim that the mid point of the circle is both at 0.9V and 0.1V at the same time. It's only at those voltages at the ends of the wire where it touches the resistors, but the voltage gradualy changes between the two as you move the measurement point along the wire. So i thing the experiment is misleading in the way it is explained.

The magnetic field is pushing those electrons down the wire because the two are moving in relation to each other. However the electrons want to equalize along the wire so they push back. At some point the two forces balance out and you end up with a certain number of extra electrons towards one end of the wire. More electrons means more charge so the voltage is higher there (actually lower since they are negative charges, but you get the point). If you provide another path for the electrons to equalize towards, such as placing a voltmeter between the two nodes in question they will happily flow towards it, the meter simply senses how eager those electrons are to equalize. Where the tricky part comes is that wires need to carry the electrons to the meter and electrons in these wires are also susceptible to being pushed around by that magnetic field, this creates the extra voltage we measure. The voltage between the two probing points is defined by the difference in charge density and can only be one number. This is the voltage that I refer to as the true voltage between the nodes. Since the probe wires are not supposed to be part of the circuit (but are a nececerry evil to be able to connect the meter) this means they should be used in a way that does not produce extra voltages. These are the voltages that push electrical current trough circuits (even if they are not the direct fault of a electric field)

SPICE is showing these voltages in its results and yes it will throw an over defined circuit error if you parallel two voltage sources due to there being no way for standard circuit analysis methods to resolve that. If it does simulate it fine then the simulator added a parasitic resistance to the voltage source. Often simulators apply some default (but overridable) parasitic values because it makes circuits act closer to what we expect in real life.

Kirchhoffs voltage and current laws are still incredibly useful tools for circuit analysis and they always work for that. It's a matter of the right tool for the job. These circuit analysts methods do a very good job of predicting what would happen in a real circuit. This is what science is about, making theories and then rigorously testing them to the limit with experiments. The ones that match experimental results are considered to be more valid and can then be of incredible use to engineers from all fields to help them predict the performance of there designs before they are built. You don't want to build a bridge just to test if it will collapse or not. Abstractions are great and all but they don't always give the whole picture even if the math works out so nicely (And yes Kirchhoffs laws ride on top of quite a few stacked up abstraction layers, it also relies on these layers to work in a certain way)
 

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #164 on: November 16, 2018, 12:01:44 am »
There is indeed 0.9V and 0.1V on those resistors, it's Ohms law after all.

Well, yes and no.
There is 0.9V if you measure with the voltmeter on one side, and there is 0.1 V if you measure with the voltmeter on the other side of the loop. Even if you place the probe across THE SAME resistor.
Look at the meshes, can you see that in one case you are linking the flux with one orientation inside the mesh with the big resistor, and in the other case you link the flux with the opposite orientation inside the mesh with the small resistor?

Quote
Where I stop agreeing is the claim that the mid point of the circle is both at 0.9V and 0.1V at the same time.

Ay, there's the rub. You still think voltage is a property of the points on the circuit. It is not. Not anymore, it's not!
It depends on the points A and B AND on the PATH.
So you can have both readings at the same time, there is no 'quantum superposition of voltage states' so to speak.

Quote
It's only at those voltages at the ends of the wire where it touches the resistors, but the voltage gradualy changes between the two as you move the measurement point along the wire. So i thing the experiment is misleading in the way it is explained.

You are still under the spell of the positional voltage.
But you are this close to see the light.
Come to the bright side!!!
We literally own the light! [note]

[note] as a matter of fact, when the magnetic flux changes it concatenates a circulating electric field - you do not even need electrons or any kind of matter. And that varying electric field concatenates a variable magnetic field, and... well, it's the bright side, no?


« Last Edit: November 16, 2018, 12:06:53 am by Sredni »
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Offline bd139

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #165 on: November 16, 2018, 12:04:17 am »
In a world of spherical cows, the path does not exist. What then?

The only reason the measurement is different is because the path is different.
 

Online ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #166 on: November 16, 2018, 02:00:47 am »
The first equation is Kirchhoff's KVL law.  In words, "the sum of all voltages in a closed loop = zero".

It is attempt to apply Kirchhoff's KVL law to superconductive loop placed in the changing magnetic flux. Kirchoff's law does not hold for such. Do not recall anybody arguing that. Interesting stuff. Excerpt: "An electric current flowing in a loop of superconducting wire can persist indefinitely with no power source."

Debate was about flawed experiment which had cut loop as EMF source and resistor(s) as load. On resistor terminals you can measure voltage and there are no two different values. Also kinda obvious that nobody tries to apply Kirchoff's law to circuit with single element, not to mention shorted battery!
« Last Edit: November 16, 2018, 02:29:30 am by ogden »
 

Offline rfeecs

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #167 on: November 16, 2018, 06:04:12 am »
However I still don't agree with the demo experiment. But I do agree with some of it. There is indeed 0.9V and 0.1V on those resistors, it's Ohms law after all. Where I stop agreeing is the claim that the mid point of the circle is both at 0.9V and 0.1V at the same time. It's only at those voltages at the ends of the wire where it touches the resistors, but the voltage gradualy changes between the two as you move the measurement point along the wire. So i thing the experiment is misleading in the way it is explained.

Maybe consider that potential or voltage is a mathematically calculated quantity that doesn't necessarily correspond to something physically real.  In this case, it is not defined at all unless you define a path over which to do the calculation.

So consider real quantities instead:

The current is real.  The current has to be the same everywhere around the loop because of the continuity equation.

The magnetic field is real.  The only magnetic field is in the middle of the loop.  There is no magnetic field anywhere else in this simple two dimensional scenario.  So there is no magnetic field in the wires or in the resistors.

The electric field is real.  The electric field is caused by the changing magnetic field.  There is no field inside the perfect (or near perfect) conducting wires.  The only field is present in the resistors.  As you said, there is a charge distribution that maintains zero field in the wires and enhances the field in the resistors.

Energy is real.  So consider the charge that moves around the loop due to the current.  When it goes through the wires it doesn't gain or lose energy because there is no field inside the wire.  Where it passes through the resistors, it loses energy due to heat, and this energy is provided by the electric field.

So how could the voltage in the wire gradually change from one end to the other?  Wouldn't that require the charge to gain or lose energy as it moved from one end to the other?

There's no contradiction here.  All the real quantities are consistent, energy is conserved, charge is conserved. 

But voltage is not a real quantity in this case.

The experiment is not saying that the midpoint of the loop is both at 0.9V and 0.1V at the same time.  It is saying one meter is reading 0.9V and the other is reading 0.1V.  This is because the meter leads follow different paths.  Simple as that.
« Last Edit: November 16, 2018, 06:06:30 am by rfeecs »
 

Offline Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #168 on: November 16, 2018, 11:07:03 am »
Sorry for the long post, but it is worth a read since i think it finds a good middle ground between Lewin and ElectroBoom


My explanations leans on this:
https://en.wikipedia.org/wiki/Electromotive_force#Formal_definitions
Quote
Inside a source of emf that is open-circuited, the conservative electrostatic field created by separation of charge exactly cancels the forces producing the emf. Thus, the emf has the same value but opposite sign as the integral of the electric field aligned with an internal path between two terminals A and B of a source of emf in open-circuit condition

If you have a open circuit length of wire in a moving field (Or vice versa) you do get a different amount of charge density on each of the ends that corresponds to the EMF voltage. The longer the wire is the more electrons there are for the magnetic field to tug along so as you go along the wire they cumulatively get pushed more and more. Much like a vertical column of water getting pulled on by gravity, the bottom ends up with more pressure than the top (Yes i know water is not the same as electricity but the idea is similar). Here this effect is called charge separation.

This works even in superconductors. At first it sounds wrong because more electrons at one end would create a electric field inside of a superconductor, but the magnetic field that shoved the electrons over to the end did so using its 'virtual electric field' (Well it is a real electric field, but its not caused by a charged object). So the two fields put together are again zero. Once you connect it into a loop they are free to move so the field disappears and the current they cause opposes the outside field so then you have no electric field induced by charge separation and the field caused by the magnetic fields cancel out to zero too. The current and field sustain each other so the current flows forever and the field stays forever. Very useful for making incredibly strong magnets and is used extensively for this in things like MRI machines and particle accelerators.

If we instead close the loop by putting a resistor in series then we get a case of both. We need an electric field to push electrons trough that stubborn resistance inside the resistor so this effect of open loop charge separation on the wire puts extra electrons on one side of the resistor so they can force themselves trough using there own electric field. But because now electrons are flowing trough the resistor we have a current in the loop so the loop makes its own opposing magnetic field. So far it looks like a closed loop superconductor again, but the resistors don't allow the electrons to flow freely so they can't make it around the loop fast enough to fill the 'electron void' on the other side of the resistor. As a result some electrons are left behind on one end of the resistor and continue to experience charge separation, thus making the wire look like it has voltage and this voltage appears as a smooth gradient across the length of the superconductor. Due to the resistor limiting the amount of current the magnetic field it creates around the loop is smaller than the outside magnetic field and so the 'virtual electric field' it creates in the loop does not fully subtract out the one caused by the outside field. The field that steps in to fill the missing part is the electric field caused by charge separation and gets the sum of fields inside the superconductor back to zero as it should be.

This means that if we connect a wire between two points on the superconductor and route it in a way that generates no EMF on the wire we get current flow that is proportional to the voltage on the two points and the wire resistance.(But only if this superconductors loop is closed with a resistor in series). This gives the two points a set voltage between them that is a single value.

So lets see the definition of voltage then:
https://en.wikipedia.org/wiki/Voltage
Quote
Voltage is the difference in electric potential between two points. The difference in electric potential between two points (i.e., voltage) in a static electric field is defined as the work needed per unit of charge to move a test charge between the two points.

Wait... ??? Yeah this is what throws the wrench in the works.

So if you integrate the total electrical fields around the path you get zero volts inside the superconductor and all the rest of the voltage on the resistor. Since the path goes trough a different resistor depending on what way around you go you also get a different voltage. So by definition of voltage it checks out. This is why this is such a big argument, Dr. Lewin is not saying anything wrong.


So where is the problem then?
We don't have a voltmeter that drags an electron around and logs the work needed to do so. Tho if someone did make one id love to see it cause it sounds really cool. So because of this we can't measure the voltage in the exact way it is defined. What we have to do instead is tap off the voltage with extra wires and bring that voltage to the voltmeters input port. When the wires are run in such a way that they don't get affected by the magnetic field only get the charge separation effect pushing electrons trough the voltmeter so the voltmeter ends up measuring electron charge density between the points. The voltmeter becomes part of the circuit and the voltage drop on the 10MOhm resistor inside the voltmeter is this voltage we see. (Doing this adds a third possible solution to the node)

TLDR starts here:

Okay our voltmeters suck... so really what is the problem?
The experiment is never explained how the voltmeter 'selects' what voltage it can see. There is no mention given to the importance of the path that the voltmeters probe wires take and why they are routed in that exact way. It just leaves you head scratching how is it possible to see two different voltages at the same point. It demolishes your intuitive notion of voltage in circuits. Many electronics engineers after university are likely still confused as to how it works.

The whole thing is explained with a schematic and using some basic circuit analysis tools. Any voltmeters in the schematics are assumed to tap off the electric field integral of the loop you want to see and ignore others. It would have been much better to explain it on the level of electric fields and electrons moving around if the goal was to show the underlying physics. Using a schematic and then simply using the ideal wire model out of circuit analysis methods and then talking about the fields inside a wire is confusing. You ether don't use  the abstraction of circuit node analysts methods at all and focus on electrons in a wire (so you can interact your magnetic fields with them), or you go all the way with circuit analysis and create an equivalent circuit that shows the magnetic effects as inductors. One or the other ways of explaining it makes sense and works great! Kirchhoffs law is a circuit analysis tool and it works (for circuit analysis), its not a law that governs how the universe works.

So why does circuit analysis not agree with physical electrons moving in wires? Because that's not the point of this abstraction. The goal of circuit analysis methods is to make it as easy as possible to predict the behavior of a circuit with as little math as possible. So to not complicate something as simple as a wire it simply cuts the concept of voltage down to the effect the voltage has on components (including the effect the voltage has on a voltmeter). In this simplified world Kirchhoffs law works perfectly and because the abstraction uses voltages that we can observe in real life means that the results of these circuits analysis methods also work in real circuits with electrons running trough them.

Many simple equations you have been taught in your first few years of physics are actually set inside an abstracted world where for example the speed of light is infinite and our atmosphere is a perfect vacuum. So are they wrong? Well... in theory yes they are wrong, but they work just fine in the abstracted world. The math is much easier and faster in this abstracted world, yet when done carefully still gives results that are very close to real ones you would get in the real world.

Please use a hammer for hammering nails rather than screws.
« Last Edit: November 16, 2018, 11:22:27 am by Berni »
 
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Offline rfeecs

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #169 on: November 16, 2018, 06:39:23 pm »
You are defining a "true" measurement as one where the fields don't interfere with the meter leads.  We are taught that the leads are not supposed to be part of the measurement.  If there is a voltage drop in the leads, that is an error.  So you have to eliminate or account for the error to make an accurate measurement.

Fair enough.

So to measure from A to B, arrange your leads vertically so they are parallel to the magnetic field and perpendicular to the E field.  Or in some other way shield the leads from the fields.  So now apply Faraday's law.  First integrate the flux over the surface of your measurement loop.  There is no net flux through the area surrounded by the measurement leads and the line from A to B, because of the way the wires are arranged or shielded.  So you can pick either surface in the current loop going from A to B.  You are splitting the loop in half, so they each have the same flux which is half the total flux.  So either way you pick, there is an EMF of 0.5V.  Then integrate around the path of the measurement leads and either half of the loop.  If you go one way, you get 0.5V-0.1V=0.4V.  If you go the other way, 0.9V-0.5V=0.4V.  OK.  So the true voltage is is 0.4V.

And yes, if you move the test points toward one of the resistors, you would "slice the pie" of the surface of the current loop differently, and the voltage you measured would continuously change and end up just the voltage across the resistor.

So your model, using lumped coupled coils or transformers, works.  It gives you the "true" answer.  I get it.  It agrees with Electroboom and it agrees with Faraday's law.  There's nothing wrong with it.

This is basically the same thing bsfeechannel did in his analysis back in reply #106.  Some of the reactions were "you are just picking a path that gives you the answer you want".

But based on the standard way that electrical engineers look at measurements, this is the correct path.  Again, fair enough.

edit:  And yes, if you choose this path, the resistive ring always measures 0V.
« Last Edit: November 16, 2018, 07:17:29 pm by rfeecs »
 
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Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #170 on: November 17, 2018, 05:19:09 am »
So to measure from A to B, arrange your leads vertically so they are parallel to the magnetic field and perpendicular to the E field.  Or in some other way shield the leads from the fields. 

So now apply Faraday's law.  First integrate the flux over the surface of your measurement loop.  There is no net flux through the area surrounded by the measurement leads and the line from A to B, because of the way the wires are arranged or shielded.

I am not sure to understand this.
You do know that the flux depends on the area orthogonal to the field that is enclosed by the contour, right?
How on earth do you plan to place or shield your leads to avoid intercepting the flux when you are partitioning a disk? You should shield the area, but then forget connecting to a circuit immersed in the field.

But let's say we have find a probe placing such that...
Quote
You are splitting the loop in half, so they each have the same flux which is half the total flux.  So either way you pick, there is an EMF of 0.5V.  Then integrate around the path of the measurement leads and either half of the loop.  If you go one way, you get 0.5V-0.1V=0.4V.  If you go the other way, 0.9V-0.5V=0.4V.  OK.  So the true voltage is is 0.4V.

Ok, let's call 'true' the voltage measured along paths that split the area in equal parts. This is not the only possibile logic choice but let's go with this.

Quote
And yes, if you move the test points toward one of the resistors, you would "slice the pie" of the surface of the current loop differently, and the voltage you measured would continuously change and end up just the voltage across the resistor.

If you call true potential the one along paths that split the area in equal parts, no. You have to gerrymander the probe to give two identical areas when your endpoints are not on a diameter.

Quote
So your model, using lumped coupled coils or transformers, works.  It gives you the "true" answer.  I get it.  It agrees with Electroboom and it agrees with Faraday's law.  There's nothing wrong with it.

Ahem, but let's suppose that we have agreed to use a certain class of possible paths (they are not unique for the same measurement - for example a measure with endpoints on a diameter can be done using the diameter as a path, or the yin-yang path, but let's brush this aside for the moment)

Quote
This is basically the same thing bsfeechannel did in his analysis back in reply #106.  Some of the reactions were "you are just picking a path that gives you the answer you want".

But based on the standard way that electrical engineers look at measurements, this is the correct path.  Again, fair enough.

I beg to differ. And here's another reason why.
The standard way that engineers look at measurements, voltage is addittive. If you want to find VAB you can find it by summing VAC and VCB.
Also, you might like the property VAB = -VBA.
Well, with path dependent measurements you have to give up both of these properties.

The latter one can easily be amended. If we agree to use the same path, only reversed, for BA and AB, then we gave to change

VAB + VBA = 0   (KVL)

in

VAB + VBA = EMF   (Faraday, or as engineers sometime call it, generalized KVL)

What happens when you want to sum two voltages? Each of them might require a different path to give the "true" potential reading. In general the three paths connecting the points A, B and C will enclose a portion of space that non necessarily encloses ALL of the flux. You you'll end up with

VAB+VBC+VCA = a fraction of emf corresponding to the varying flux linked by that area.

And what area is that? There is more than one path that can give you the 'true' potential (remember? diameter and yin yang). So, your sum of 'true potentials' is... whatever you want it to be.

Quote
edit:  And yes, if you choose this path, the resistive ring always measures 0V.

This is right. :-)


PS
I wrote this after having slept nine hours in three days. So, check it.
All instruments lie. Usually on the bench.
 

Offline GeorgeOfTheJungle

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #171 on: November 17, 2018, 08:59:30 am »
Hey guys, I've solved it for ya:



Vab = -I*R = I*9R

Are these resistors in series, in parallel, or both?
« Last Edit: November 18, 2018, 12:02:09 pm by GeorgeOfTheJungle »
int main (void) { while (1) fork(); }
 

Offline riyadh144

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #172 on: November 17, 2018, 10:43:16 pm »


Dr. Lewin just uploaded another video.
 

Offline rfeecs

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #173 on: November 17, 2018, 10:56:00 pm »
So to measure from A to B, arrange your leads vertically so they are parallel to the magnetic field and perpendicular to the E field.  Or in some other way shield the leads from the fields. 

So now apply Faraday's law.  First integrate the flux over the surface of your measurement loop.  There is no net flux through the area surrounded by the measurement leads and the line from A to B, because of the way the wires are arranged or shielded.

I am not sure to understand this.
You do know that the flux depends on the area orthogonal to the field that is enclosed by the contour, right?
How on earth do you plan to place or shield your leads to avoid intercepting the flux when you are partitioning a disk? You should shield the area, but then forget connecting to a circuit immersed in the field.

I'm thinking 3 dimensionally.  The contour is bent 90 degrees along the line from A to B.  The idea was to try to keep the test leads parallel to the magnetic field and perpendicular to the electric field as I said.  But yes, it is probably not physically possible.  For shielding, I was thinking more like coaxial shields around the test leads.

The other way is to account for the "error" introduced by the EMF induced in the loop of the test leads.  Measure the EMF separately in a calibration step, and then subtract it out the measurement.  The result is the same.

I was trying to rationalize the lumped element circuit that people are automatically coming up with as a valid interpretation of a "correct" measurement.

Personally, I think this demo is a fairly simple demonstration of Faraday's law, and trying to come up with the lumped circuit just adds unnecessary complications.

Electronics people automatically react strongly because of Dr. Lewin's use of a schematic in a confusing way and his lack of detailed explanation of what is really happening.I'm sure it was covered in more detail for his students in discussion sections and lecture supplements. And then there's his provocative "Kirchoff's law is for the birds." 

I haven't watched the latest video he just posted, yet.  We'll see.

 
 

Online ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #174 on: November 18, 2018, 02:10:18 am »
For shielding, I was thinking more like coaxial shields around the test leads.

Just put whole experiment into transformer core like this:


 
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