Author Topic: Does Kirchhoff's Law Hold? Disagreeing with a Master  (Read 53780 times)

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Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #325 on: December 04, 2018, 12:30:29 pm »
When there's paper with real arguments you talk about youtube comments??

I promised not to interact with you on this matter, but... are you asking me where is the paper stating what I put into quotes in my previous post?
Well, it's Belcher's document.

Quote
You really shall read Prof. Belcher's document

Yup. You should indeed.

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which says that Mehdi won his argument.

Of course, reading is necessary but not sufficient condition to understand what it says.
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Online ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #326 on: December 04, 2018, 12:38:26 pm »
I promised not to interact with you on this matter, but... are you asking me where is the paper stating what I put into quotes in my previous post?
Well, it's Belcher's document.

Well, then in case you missed that - same document states "KVL holds, as argued by Mehdi Sadaghdar"
 

Offline electrodacus

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #327 on: December 04, 2018, 12:47:47 pm »
You can not measure two different voltages when you measure in the exact same two points at the exact same point in time.
I will give it a go at explaining the problem but I'm to lazy to draw something so I will rely on my bad English and your capability to reconstruct the proposed experiments in your mind.

1) Imagine a copper loop say it is a ring with a resistance of 1Ohm and an emf of 1V thus creating a round 1A current trough the ring.
Now get an ideal voltmeter and probe two opposite points on the ring so that the ring is exactly split in two equal parts by this two points.
What do you think the reading will be ?  It will be 0V.
It will not make any sense to try and probe with two voltmeters there is nothing to be proven with that as they will read the exact same thing.

2) While point 1 should be enough to demonstrate my point let me get another example.
I imagine the same ring but this time half of the ring has 0.9Ohm and the other half has 0.1Ohm. (same 1V / 1A).
Same ideal voltmeter connected at the points where those two half rings intersect will read what ? :)
Answer will be the same as if you have an equivalent circuit with two 0.5V ideal sources one on each half and with a 0.1Ohm series resistor on one side and 0.9Ohm resistor on the other side.
So 400mV (the sign will also be fixed but to keep things simple I did not provided all the info).

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #328 on: December 04, 2018, 02:08:43 pm »
You can not measure two different voltages when you measure in the exact same two points at the exact same point in time.

Aaaaand... we're back to square one.

Quote
I will give it a go at explaining the problem but I'm to lazy to draw something

Let me guess, you are too lazy to read the other posts in the previous thirteen pages of this thread as well.

<sigh>


Quote from: ogden
Well, then in case you missed that - same document states "KVL holds, as argued by Mehdi Sadaghdar"

Thank you for proving my point about the disjointness of reading and understanding.
Maybe if you read it again you will see that section 11 KVL is about lumped circuits. You too should go back read all past posts.
I'm done talking with walls.
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Offline electrodacus

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #329 on: December 04, 2018, 03:52:47 pm »
You can not measure two different voltages when you measure in the exact same two points at the exact same point in time.

Aaaaand... we're back to square one.

Please point out what part of my short statement is wrong.
And if that is not wrong then what was the point of that experiment?

Offline Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #330 on: December 04, 2018, 05:40:12 pm »
You can not measure two different voltages when you measure in the exact same two points at the exact same point in time.

Aaaaand... we're back to square one.

Please point out what part of my short statement is wrong.
And if that is not wrong then what was the point of that experiment?

I do have to agree with Sredini this time.

The problem is that this ideal voltmeter still needs wires to connect to the points of interest on the circle and that's how you can get different voltages depending on what path these wires take, they are as much part of the circuit as the loop itself.

If you take the formal definition of voltage that says that the voltage is the integral of all forces acting on electrons along a path. Then you do get a different number depending on what path do you take trough the loop. By definition there are indeed two possible voltages across the two points. This is because both the electric field of charge separation and the magnetic EMF count towards the total voltage.

Voltmeters only read the charge separation part of the voltage because that's what drives current trough its internal resistance. If you connect the voltmeter with the wires taking such a path that the magnetically induced EMF is zero you get to measure that charge separation and you get the single result you expect (The result is indeed 0.4V). This is the same result that circuit mesh analysis will give (that is called KVL here, but its more than just KVL).

The voltage from charge separation is always defined as a single number for all points in any circuit. Its simply how many electrons are sitting there. More electrons more negative the voltage. Its only the magnetic EMF part of the voltage that depends on what path you take, this is because that EMF is pulling the electrons into a certain direction. This makes electrons motivated to move in that direction even if there is already the same amount of electrons there, but only in that direction.

So to conclude yes there are two voltages across A and B in Dr. Lewins circuit according to the formal definition of voltage, but this is not useful voltage that can be harnessed, its just a incomplete calculation of voltage that requires the rest of the loop to be added in too and that then gives you a single result. In your case that is a single result of voltage across the voltmeter terminals.

These multiple voltage across two points in electrical engineering do sort of the same thing as complex numbers in math. The imaginary part of complex numbers don't really physically exist, but it is there to make the math work out that otherwise wouldn't be possible(Such as square roots of negative numbers). Same goes here with Dr. Lewins example. The voltages are indeed there according to the math, but its not a real voltage you can "physically touch" in the real world. Much like a voltage of 6 V or 5.66+j2 V look the same to a voltmeter but are not the same in the math.

EDIT: Note there are no complex numbers involved in Dr. Lewins example. I was just making a comparison to math. Tho if you want you can still stick complex values of voltage and current into Kirchhoffs circuit laws and it still works(very useful when you have AC sources and reactive components)
« Last Edit: December 04, 2018, 05:46:44 pm by Berni »
 

Online beanflying

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #331 on: December 04, 2018, 05:45:54 pm »
I think that regardless of absolute right or wrong Mehdi is to be commended for questioning (politely) an academic of note, creating a reasoned hypothesis of his own, testing and working through his thoughts in a practical manner on a difficult concept in a way to not turn off those not as technically minded. :-+

Some of the comments on youtube and the initial ego driven reply from Lewin did nothing to help solve the question. There is scope for rigorous scientific experimental verifiable testing of this to maybe get closer to a resolution, living room maths on a whiteboard and playing on a kitchen bench isn't going to do it.

Musing to myself over a Beer Crazy Beard to with guests of Crazy Hair and Crazy Unibrow at some point in the very distant future on proper measurement and probing techniques would be a fun watch too 8)

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Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #332 on: December 04, 2018, 05:47:36 pm »
Please point out what part of my short statement is wrong.
And if that is not wrong then what was the point of that experiment?

Allow me to apologize for being so blunt, but I was really disappointed by Electroboom's latest video and the reaction of his fans.
The answer to your question, though, is in the many many posts of this thread. Please, browse through it and you will see that it is indeed possible for two voltmeters whose probes' tips are attached to the very same two points to read different values at the same time. There are also youtube video showing this, if you do not believe it.
The reason is in the different flux intercepted by the two meshes the circuit is partitioned into.

If you are curios, there are thirteen pages awaiting for you.
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Offline electrodacus

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #333 on: December 04, 2018, 06:27:47 pm »
I do have to agree with Sredini this time.

The problem is that this ideal voltmeter still needs wires to connect to the points of interest on the circle and that's how you can get different voltages depending on what path these wires take, they are as much part of the circuit as the loop itself.

What will be the point of a real voltmeter in this example ? They will need to be added as components to the circuit and will only complicate the demonstration but not change the result if you correctly add them.   


If you take the formal definition of voltage that says that the voltage is the integral of all forces acting on electrons along a path. Then you do get a different number depending on what path do you take trough the loop. By definition there are indeed two possible voltages across the two points. This is because both the electric field of charge separation and the magnetic EMF count towards the total voltage.

The result will be the same 0V for first example and 0.4V for the second example and it the second case not just the value will be the same but also the polarity. We are talking about a fixed moment is time and the real voltage not the indication of two real voltmeters that are part of the circuit and inside the magnetic field measuring something else than the real voltage between those two points.


Allow me to apologize for being so blunt, but I was really disappointed by Electroboom's latest video and the reaction of his fans.
The answer to your question, though, is in the many many posts of this thread. Please, browse through it and you will see that it is indeed possible for two voltmeters whose probes' tips are attached to the very same two points to read different values at the same time. There are also youtube video showing this, if you do not believe it.
The reason is in the different flux intercepted by the two meshes the circuit is partitioned into.

If you are curios, there are thirteen pages awaiting for you.

See the replay above made to Berni.  I did read most of this tread (about 70%) and seen the a few videos from both sides.
Is not about believing is about real world as we are currently able to understand and in this universe there can be only one voltage at a fix moment in time.

The reason is in the different flux intercepted by the two meshes the circuit is partitioned into.

That is what I was answering above as in you are not measuring the actual voltage at the tip of the probes.
You can even completely remove the ring made of two resistors and keep just the two voltmeter's and you will read something but not what it will be at the tip of those two voltmeters.

Online ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #334 on: December 04, 2018, 07:14:24 pm »
Maybe if you read it again you will see that section 11 KVL is about lumped circuits. You too should go back read all past posts.

I thought that this forum agreed about lumped circuit of Dr.Lewin's experiment, yet you woke up. Aaaaand... we're back to square one. To talk about Maxwell's equations make sure you know laws of nature as well. Seems, I am done with this thread disregarding what sofa experts like you say. Your copy/paste skills are not even entertaining anymore.  :=\
 

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #335 on: December 04, 2018, 08:15:38 pm »
I think that regardless of absolute right or wrong Mehdi is to be commended for questioning (politely) an academic of note, creating a reasoned hypothesis of his own, testing and working through his thoughts in a practical manner on a difficult concept in a way to not turn off those not as technically minded. :-+

Politeness is commendable, but we are engineers. We design serious stuff: buildings, bridges, cars, airplanes, life-supporting systems, telecommunication systems, power systems.

If we give up our integrity, people die.
 

Online beanflying

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #336 on: December 04, 2018, 08:23:00 pm »
I think that regardless of absolute right or wrong Mehdi is to be commended for questioning (politely) an academic of note, creating a reasoned hypothesis of his own, testing and working through his thoughts in a practical manner on a difficult concept in a way to not turn off those not as technically minded. :-+

Politeness is commendable, but we are engineers. We design serious stuff: buildings, bridges, cars, airplanes, life-supporting systems, telecommunication systems, power systems.

If we give up our integrity, people die.

Not maintaining Integrity? Explain please?

I see NO where he didn't maintain his integrity at the highest level which has little or nothing being proven correct or incorrect in the long run. If English is not your first language then best you check the definition of the word. Your attacking of the mans integrity and not debating the technicalities of the subject is out of order!
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Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #337 on: December 04, 2018, 09:24:31 pm »
I think that regardless of absolute right or wrong Mehdi is to be commended for questioning (politely) an academic of note, creating a reasoned hypothesis of his own, testing and working through his thoughts in a practical manner on a difficult concept in a way to not turn off those not as technically minded. :-+

Politeness is commendable, but we are engineers. We design serious stuff: buildings, bridges, cars, airplanes, life-supporting systems, telecommunication systems, power systems.

If we give up our integrity, people die.

Not maintaining Integrity? Explain please?

I see NO where he didn't maintain his integrity at the highest level which has little or nothing being proven correct or incorrect in the long run. If English is not your first language then best you check the definition of the word. Your attacking of the mans integrity and not debating the technicalities of the subject is out of order!

The first thing I did was to analyze his latest video from the technical standpoint. And I've been discussing the technical points of Lewin's experiment from my post #1 in this thread.

I've been an engineer for long enough to know what integrity means and I don't need the English language to tell me that.
 

Online beanflying

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #338 on: December 04, 2018, 09:47:37 pm »
Clearly you do need a dictionary and some basic decency and manners. Attacking the integrity of the person and not the ideas as you have just done is what in Australia we would call a c... act amongst others! In particular where the person in question is not here to defend themselves. You clearly can't show where Mehdi has compromised his 'integrity'!

interesting

A quick check shows you have maybe 40 posts in this thread so clearly you have a major bee in your bonnet and in a chunk of those posts you have decided to play the man not the subject. I will leave others to decide what this says about your character.
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Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #339 on: December 04, 2018, 10:27:33 pm »
Please point out what part of my short statement is wrong.

All of it.

Suppose that we have the circuit of the schematic below.



All the components are lumped, and we only have batteries. The real value of the voltages and resistances are not important. However, to simplify our calculations, let's suppose the two resistors have the same value (it could be any other known proportion). Connected at the same two points in the circuit we have five voltmeters.

We agree that the five voltmeters will measure exactly the same voltage.

Now let's remove the batteries and apply a magnetic field that rises linearly in intensity with time, points towards you and is confined in the total area of the circuit, i.e., A1 + A2 + A3 + A4. 



According to Maxwell, this field will generate a constant EMF that will be proportional to the area enclosed by a circuit and whose polarity will be defined by the corkscrew rule. In short, the topology now counts. To simplify our calculations, let's suppose that the EMF generated by the loop enclosed by all four areas is 1V.

The voltages seen by the voltmeters will now be like that (in volts):

 #1
-0.5

 #2
( A1 - ( A2 + A3 + A4 ) ) / ( 2 * ( A1 + A2 + A3 + A4 ) )

 #3
( ( A1 + A2 ) - ( A3 + A4 ) ) / ( 2 * ( A1 + A2 + A3 + A4 ) )

 #4
( ( A1 + A2 +  A3 ) - A4 ) / ( 2 * ( A1 + A2 + A3 + A4 ) )

 #5
+0.5

So, what voltmeter is measuring the "correct" voltage?
« Last Edit: December 04, 2018, 10:33:58 pm by bsfeechannel »
 

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #340 on: December 04, 2018, 10:40:14 pm »
Clearly you do need a dictionary and some basic decency and manners. Attacking the integrity of the person and not the ideas as you have just done is what in Australia we would call a c... act amongst others! In particular where the person in question is not here to defend themselves. You clearly can't show where Mehdi has compromised his 'integrity'!

You're young. You need a hero. I'm old. I do not have time for this kind of nonsense.

Quote
A quick check shows you have maybe 40 posts in this thread so clearly you have a major bee in your bonnet

I like to contribute. Is that a problem?

Quote
and in a chunk of those posts you have decided to play the man not the subject. I will leave others to decide what this says about your character.

I'm not the subject of this thread. Mehdi and his (now debunked) claims are.
 

Online BrianHG

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #341 on: December 04, 2018, 10:47:04 pm »
All the components are lumped, and we only have batteries. The real value of the voltages and resistances are not important. However, to simplify our calculations, let's suppose the two resistors have the same value (it could be any other known proportion). Connected at the same two points in the circuit we have five voltmeters.

We agree that the five voltmeters will measure exactly the same voltage.

Now let's remove the batteries and apply a magnetic field that rises linearly in intensity with time, points towards you and is confined in the total area of the circuit, i.e., A1 + A2 + A3 + A4. 



According to Maxwell, this field will generate a constant EMF that will be proportional to the area enclosed by a circuit and whose polarity will be defined by the corkscrew rule. In short, the topology now counts. To simplify our calculations, let's suppose that the EMF generated by the loop enclosed by all four areas is 1V.

The voltages seen by the voltmeters will now be like that (in volts):

 #1
-0.5

 #2
( A1 - ( A2 + A3 + A4 ) ) / ( 2 * ( A1 + A2 + A3 + A4 ) )

 #3
( ( A1 + A2 ) - ( A3 + A4 ) ) / ( 2 * ( A1 + A2 + A3 + A4 ) )

 #4
( ( A1 + A2 +  A3 ) - A4 ) / ( 2 * ( A1 + A2 + A3 + A4 ) )

 #5
+0.5

So, what voltmeter is measuring the "correct" voltage?
With your example, you are either claiming that both resistor values in you example are equal in value, or identical current is flowing on each side like a perfect mirror, otherwise the EMF field will be lop-sided due to the different load of the resistors.

What I would like to see is having modify the original loop so that it bends inwards to the center at the measurement point, insert a tiny 6 pin sot-23 MCU right on a watch battery directly in the middle sampling the voltage at 10msps, not connection anywhere else, and optically feeding out the readings.
« Last Edit: December 04, 2018, 10:54:01 pm by BrianHG »
__________
BrianHG.
 

Offline sectokia

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #342 on: December 04, 2018, 11:15:39 pm »

With your example, you are either claiming that both resistor values in you example are equal in value, or identical current is flowing on each side like a perfect mirror, otherwise the EMF field will be lop-sided due to the different load of the resistors.

What I would like to see is having modify the original loop so that it bends inwards to the center at the measurement point, insert a tiny 6 pin sot-23 MCU right on a watch battery directly in the middle sampling the voltage at 10msps, not connection anywhere else, and optically feeding out the readings.

Have a look at this:
http://www.hep.princeton.edu/~mcdonald/examples/lewin.pdf

See equation 17 on page 5.

Here you will get the answer as per doing what you describe above.

There is actually no need for the meter to be 'in the middle' of the magnetic field. Remember its the loop path enclosing the magnetic flux that matters. Thus putting the meter in the middle of the magnetic field and having the probes divide it is no different to the meter being outside the magnetic field and having one probe divide it up the middle, as shown in the link above on page 5 circuit diagram.

In the case of Lewins original circuit you will read 0.4V so long as the area containing the flux is enclosed equally on both sides. As you move the meter from left to right you will read -0.1 to +0.9 simply depending on the ratio of the flux encircled on each side. It really is that simple.
« Last Edit: December 04, 2018, 11:26:30 pm by sectokia »
 

Online beanflying

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #343 on: December 04, 2018, 11:33:52 pm »
You're young. You need a hero. I'm old. I do not have time for this kind of nonsense.

Seriously Not young at all and yet you decide this based on what? As to 'nonsense' you have time to attack Mehdi's integrity but not tell us where didn't maintain his integrity? Still waiting?

Quote
I like to contribute. Is that a problem?
I'm not the subject of this thread. Mehdi and his (now debunked) claims are.

When you descend into criticism of the person or others here  you open yourself up to becoming a subject of discussion for your behavior and personal attacks.

And Mehdi's claims are 'debunked' as decided by you? WOW your verbosity and frequency of posting must have made it so awesome work you have me instantly convinced....
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Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #344 on: December 04, 2018, 11:37:35 pm »
Quote
With your example, you are either claiming that both resistor values in you example are equal in value, or identical current is flowing on each side like a perfect mirror, otherwise the EMF field will be lop-sided due to the different load of the resistors.

EMF has nothing to do with the value of the resistors according to Maxwell, only with the varying field and the geometry of the circuit. If you change the proportion of the resistors, the voltmeters will read different values from what you see above, but they will all show different values among themselves.

In particular, if you choose the make the left resistor nine times greater than the right resistor, you will have the following voltages.

 #1
-0.9

 #2
( A1 - 9 * ( A2 + A3 + A4 ) ) / ( 10 * ( A1 + A2 + A3 + A4 ) )

 #3
( ( A1 + A2 ) - 9 * ( A3 + A4 ) ) / ( 10 * ( A1 + A2 + A3 + A4 ) )

 #4
( ( A1 + A2 +  A3 ) - 9 * A4 ) / ( 10 * ( A1 + A2 + A3 + A4 ) )

 #5
+0.1

Quote
What I would like to see is having modify the original loop so that it bends inwards to the center at the measurement point, insert a tiny 6 pin sot-23 MCU right on a watch battery directly in the middle sampling the voltage at 10msps, not connection anywhere else, and optically feeding out the readings.

So you are suggesting to change the TOPOLOGY of the circuit to measure what you would like to see? What if you can't do that?

What if the wires are in fact the traces of a PCB? Or any other kind of installation where you cannot move them?
« Last Edit: December 04, 2018, 11:39:55 pm by bsfeechannel »
 

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #345 on: December 05, 2018, 12:39:35 am »
Seriously Not young at all and yet you decide this based on what?

You looked young to me.

Quote
As to 'nonsense' you have time to attack Mehdi's integrity but not tell us where didn't maintain his integrity? Still waiting?

He tried to pass the impression that professor Belcher endorsed his claims against Lewin's demonstration. He even invoked the "ghost" of Feynman to bless his "honest questioning".

We read the literature Mehdi recommended and found nothing to discredit Lewin. Actually we realized that Lewin employs the same terminology and concepts Feynman uses. And that everything we discussed here that proves that Lewin's demonstration is right is in perfect accordance to what Feynman describes. No wonder. We, Feynman and Lewin studied Maxwell.

The comment section of his videos is full of people making personal attacks against Lewin. This means nothing because Youtube comments mean precisely dick (I don't know what this means but I found the phrase cute).

However, Mehdi congratulates with those who post that kind of comment and even capitalizes on all the brouhaha (Is that an English word?).

The title of his latest video on the subject is absolutely misleading. "Kirchhoff vs. Faraday"? Can we have a "Newton vs. Einstein?" Or perhaps a "Euclid vs. Hilbert"?

So you judge his integrity. You're right. I do not have time for that.

Quote
When you descend into criticism of the person or others here  you open yourself up to becoming a subject of discussion for your behavior and personal attacks.

Fair enough. Mehdi and his "integrity" are targets, because he decided to target Lewin personally. He cleverly does not do that in his videos, but his comment section betrays him.

Quote
And Mehdi's claims are 'debunked' as decided by you?

No. Maxwell.

Quote
WOW your verbosity and frequency of posting must have made it so awesome work you have me instantly convinced....

Good to read that.
 

Offline electrodacus

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #346 on: December 05, 2018, 06:36:04 am »
So, what voltmeter is measuring the "correct" voltage?

Correct voltage in your example assuming those two resistors are equal and circuit is symmetrical (uniform magnetic field from the middle of the circuit) will be 0V same as in my example number 1.
And if you want you can add those voltmeters to the circuit but then they will be part of the circuit. Each will measure a different thing but not the voltage between those two points of interest except for the one in the middle number 3 that could read the correct value of 0V assuming it was shielded since a real voltmeter will likely not be perfectly symmetrical in internal construction.

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #347 on: December 05, 2018, 03:17:26 pm »
Each will measure a different thing but not the voltage between those two points of interest except for the one in the middle number 3 that could read the correct value of 0V assuming it was shielded since a real voltmeter will likely not be perfectly symmetrical in internal construction.

Excellent. So let's get rid of the other voltmeters and stick to the voltmeter #3. Let's suppose that its display has seven digits (pretty common on modern bench multimeters) and let's suppose it is properly "shielded", etc.

I hadn't defined the areas, but now let's suppose that their common side is 10cm, and that x+y = 20cm like in the picture below.



So, V3 = ( ( A1 ) - ( A2 ) ) / ( 2 * ( A1 + A2 ) ), x + y = 0.2 m and, since A1 = x * 0.1 and A2 = y * 0.1, and A1 + A2 = 0.02m², we will have that V3 in volts:

V3 = ( ( x * .1 ) - ( y * .1) ) / ( 0.04 ) = ( x - y)/0.4

If x = y, then V3 should read 0.000000V.

But now, let's imagine for a moment that this circuit doesn't have the voltmeter and then suddenly someone decides to connect it.

Look what happens if the unfortunate engineer assigned with that task misses the exact point of connection and places the voltmeter 200nm to the right. 200 nanometers. In that case x = 0.1000002 m, and y = 0.0999998 m.

V3 = (0.1000002 - 0.0999998) / 0.4 = 0.000001 V

So it's affecting the precision of the measurement. It's no biggie in this case because we know what to expect ( 0V). But what if the resistors didn't have exactly the same value? How do I know that this error is not due to a resistor mismatch? Because I could nanometrically place the voltmeter in the "correct" spot, adjusting it so that its voltage read the expected 0.000000V. But what if a mismatch in the value of the resistors is giving me a false reading?
 

Offline electrodacus

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    • electrodacus
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #348 on: December 05, 2018, 03:30:04 pm »
Each will measure a different thing but not the voltage between those two points of interest except for the one in the middle number 3 that could read the correct value of 0V assuming it was shielded since a real voltmeter will likely not be perfectly symmetrical in internal construction.

Excellent. So let's get rid of the other voltmeters and stick to the voltmeter #3. Let's suppose that its display has seven digits (pretty common on modern bench multimeters) and let's suppose it is properly "shielded", etc.

I hadn't defined the areas, but now let's suppose that their common side is 10cm, and that x+y = 20cm like in the picture below.



So, V3 = ( ( A1 ) - ( A2 ) ) / ( 2 * ( A1 + A2 ) ), x + y = 0.2 m and, since A1 = x * 0.1 and A2 = y * 0.1, and A1 + A2 = 0.02m², we will have that V3 in volts:

V3 = ( ( x * .1 ) - ( y * .1) ) / ( 0.04 ) = ( x - y)/0.4

If x = y, then V3 should read 0.000000V.

But now, let's imagine for a moment that this circuit doesn't have the voltmeter and then suddenly someone decides to connect it.

Look what happens if the unfortunate engineer assigned with that task misses the exact point of connection and places the voltmeter 200nm to the right. 200 nanometers. In that case x = 0.1000002 m, and y = 0.0999998 m.

V3 = (0.1000002 - 0.0999998) / 0.4 = 0.000001 V

So it's affecting the precision of the measurement. It's no biggie in this case because we know what to expect ( 0V). But what if the resistors didn't have exactly the same value? How do I know that this error is not due to a resistor mismatch? Because I could nanometrically place the voltmeter in the "correct" spot, adjusting it so that its voltage read the expected 0.000000V. But what if a mismatch in the value of the resistors is giving me a false reading?

I do not disagree with anything you mentioned in you last replay. Correct measurement will read a single value and that is the real 0V no multiple voltages at a single defined point in time.
Of course if you move the measurement point you get a different reading that is normal in any circuit not just this particular case.

Offline Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #349 on: December 05, 2018, 05:46:03 pm »
Hence why understanding how your probing works is important. Especially when expecting that sort of accuracy from the measurement.

It is why i thought it would be important for Dr. Lewin to explain the path the voltmeter wires take. His experiment only gives these results when the voltmeter wires hug the circuit loop wires tightly. You can get nearly any other voltage you want if you move the voltmeter wires around.

Once you start regularly using oscilloscopes for things >10MHz or with fast changing large currents you encounter all sorts of probing anomalies. Its part of the engineers job to determine if the measurement they did is accurate enough to be useful.

Yes according to the math there are two voltages across A and B, but this is not a voltage that can be used for anything until wires are run to it to connect it to something(Such as a voltmeter). Once that is done the path is defined completely and the resulting voltage is only a single well defined voltage appearing across the voltmeter you connected it to. What sort of EMF the voltmeters wires experience is totally up to you, but usually its most usefully that there is zero EMF because you can then consider the wire as being an ideal connection between two circuit nodes.
 


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