Author Topic: Does Kirchhoff's Law Hold? Disagreeing with a Master  (Read 50644 times)

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Online beanflying

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #425 on: December 08, 2018, 10:52:57 am »

But why did you quote Bsfeechannel exasperation post to tell him that?  ;D

Not even close to funny.

Being a bully in person or online isn't ok ever and when done under the guise of 'education and learning' it is just pathetic and nothing will be learned.

Like I posted yesterday attack the persons ideas or challenge a theory by all means people can learn from this sort of discourse.
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Offline electrodacus

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #426 on: December 08, 2018, 11:21:05 am »

And this is why it is nonsense to believe that you can locate a field big enough to give 0.25V voltage (integral of E.dl along the conductor) in the copper parts of the loop.
You would break the constitutive equation in copper. You would count twice the effects of the induced field.


I do not remember if I answered this concern you have in a direct way so I will do that now.
You asked several times about the size of the loop and I probably mentioned that is irrelevant for the problem so let me explain a bit better (I think).
emf = B * l * v
In the examples we were discussing the emf was a given none of the others after the equal sign where needed to calculate the potential difference between two points that was the topic of interest.
Now you are thinking on real world examples and somehow imagine a loop of a few centimeter (nowhere I ever mentioned anything about absolute length and neither what that the case in Lewin's example).
But looking at the formula you can understand that you can manipulate any of the 3 parameters and get the same emf (magnetic field, wire length and speed).

Any length of copper (or other conductor) will also be an inductor thus a energy storage device sort of the reverse of a capacitor but for magnetic field.
So using the analog of a battery that has an emf and an internal resistance is the correct way to view and simplify a piece of wire in a variable magnetic field when you consider this for a fixed moment in time else you need to integrate.

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #427 on: December 08, 2018, 11:40:41 am »
Quote from: beanflying
Not even close to funny.

It wasn't meant to be funny. It was a sarcastic grin.
I suggest you go back reading ogden's posts in this thread and revise your judgment. All he did was to facepalm, LOL, belittle and tell other people they were trolls, to stick it where it hurts and so on.
But no, we should put up with that shit and give him an award for just showing up and saying bullshit.
Sorry, not all posters in this forum align to the politically correctness craze that is so endemic to the US.
I tried not to respond to provocations, but to fault bsfeechannel for telling him to go study, is in my eye a bit excessive.
He is the incarnation of the people who do not know, but think they know Lewin was talking about.
Enough is enough.


Quote from: electrodacus
You asked several times about the size of the loop and I probably mentioned that is irrelevant for the problem

Physical size is relevant to get the E field and current density j values.
I am beginning to suspect you think those are mytical quantities. Like unicorns.
I'll try to be more specific: if I know the current and I want to find the current density, I have to divide by the area of the wire's section. Also, voltage along a path is the integral of the electric field, do you think that knowing how long the path you are integrating on can have some relevance?



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Online beanflying

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #428 on: December 08, 2018, 12:22:54 pm »

It wasn't meant to be funny. It was a sarcastic grin.
I suggest you go back reading ogden's posts in this thread and revise your judgment. All he did was to facepalm, LOL, belittle and tell other people they were trolls, to stick it where it hurts and so on.
But no, we should put up with that shit and give him an award for just showing up and saying bullshit.
Sorry, not all posters in this forum align to the politically correctness craze that is so endemic to the US.
I tried not to respond to provocations, but to fault bsfeechannel for telling him to go study, is in my eye a bit excessive.
He is the incarnation of the people who do not know, but think they know Lewin was talking about.
Enough is enough.

Nowhere have I defended any of you including Ogden that have resorted to name calling and petty or even malicious taunts. This has nothing to do with political correctness so don't use that as an excuse to continue disrespectful conduct! Do you behave like this professionally or just here behind a keyboard?

Stick to the subject and there may be hope!
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Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #429 on: December 08, 2018, 12:39:55 pm »

It wasn't meant to be funny. It was a sarcastic grin.
I suggest you go back reading ogden's posts in this thread and revise your judgment. All he did was to facepalm, LOL, belittle and tell other people they were trolls, to stick it where it hurts and so on.
But no, we should put up with that shit and give him an award for just showing up and saying bullshit.
Sorry, not all posters in this forum align to the politically correctness craze that is so endemic to the US.
I tried not to respond to provocations, but to fault bsfeechannel for telling him to go study, is in my eye a bit excessive.
He is the incarnation of the people who do not know, but think they know Lewin was talking about.
Enough is enough.

Nowhere have I defended any of you including Ogden that have resorted to name calling and petty or even malicious taunts. This has nothing to do with political correctness so don't use that as an excuse to continue disrespectful conduct! Do you behave like this professionally or just here behind a keyboard?

Stick to the subject and there may be hope!

Disrespectful conduct?
What do you expect people to do? Hold hands and sing kumbaya when people tell you to stick it where it hurts, wasting space with useless comments that only expose their ignorance? The least you can expect is for those remarks to be sent back to the sender.

I was trying to stick to the subject, but I tell you - paraphrasing a certain Roy - I feel pretty much unappreciated.
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Offline electrodacus

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #430 on: December 08, 2018, 12:42:35 pm »
Physical size is relevant to get the E field and current density j values.
I am beginning to suspect you think those are mytical quantities. Like unicorns.
I'll try to be more specific: if I know the current and I want to find the current density, I have to divide by the area of the wire's section. Also, voltage along a path is the integral of the electric field, do you think that knowing how long the path you are integrating on can have some relevance?

I have a hard time understanding your questions.
None of your questions are relevant to the problem (as far as I can see).
It sort of seems to suggest you disagree with the result's I got on those experiments. Is that the case ?
Or do you agree with the results but want to make some sort of point that I fail to see ?

Offline Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #431 on: December 08, 2018, 10:07:21 pm »
That's just an engineering problem at this point.

1) Make a length of wire that would fit across those two points as if the pesky barrier was not there
2) Make a rigid wire structure that goes around the barrier as needed and connects to a voltmeter on the other end
3) Make another copy of the structure from 2, but short it using the piece of wire from 1
4) Place the structure from 2 onto the circuit to tap the voltage and place the structure from 3 anywhere near by
5) Subtract the readings of the voltmeters.

The compensation structure from 3 can be used multiple times to ensure the field is indeed uniform all around so that we know the placement of the structure has shown valid readings.

Alternatives are to just calculate the voltage of the compensation structure if you already know the exact properties of the field, or in that case if you know the properties of the field and the path of the wire you want to measure you can just apply Faradays law to the whole thing.

Remember im not trying to disprove anything about Faraday or Maxwell. Just saying that Kirchhoffs laws are convenient to use in some cases (And they do work when used correctly).

After trying to follow your instructions, I am not sure if I understand what you mean without a drawing. So I decided to simplify the challenge. Let's suppose that the whole internal area of the loop is completely occupied by the obstruction.



Perhaps that way it becomes easier for me to understand.

Same procedure still works, just needs longer probe wires to get around the larger obstruction.

Explaining it more simply you just measure your probe wires in the same field first by shorting them, then remove the short and actually connect them to the circuit, measure again and subtract out the first measurement to get the result. That way probe wires are compensated out. If you want to double check you can repeat the whole thing with probe wires taking a different path and the result will be the same.

Tho a lot of this thread has seam to have devolved into insults (Nothing towards me but towards others) rather than creative discussion so il probably stop participating in it if this continues.
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #432 on: December 08, 2018, 11:15:35 pm »
Tho a lot of this thread has seam to have devolved into insults

From very beginning "Kirchoff for the birds" fans arrogantly insulted nearly everybody who disagree with them. Later rather than sooner it resulted in opposite reaction. I was part of it and not proud about it. Apologies to anyboody who got hurt in the process. Most likely I shall stay away from this thread which is/was just baseless "arguing against the nonexistent strawman who is apparently suggesting that Farady's law is incorrect, and Kirchoffs is always correct", like broken record.
 

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #433 on: December 09, 2018, 12:58:33 am »
Same procedure still works, just needs longer probe wires to get around the larger obstruction.

Thank you for your reply. However I'm still not sure what you mean, perhaps due to my limited fluency in English. When you say the probe goes around the obstruction, where exactly do you place your voltmeter? I understand that, in the area of the obstruction, it'll be impossible.

Quote
Tho a lot of this thread has seam to have devolved into insults (Nothing towards me but towards others) rather than creative discussion so il probably stop participating in it if this continues.

Apparently, encouraging people to get smarter is offensive for some.
 

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #434 on: December 09, 2018, 02:39:15 am »
From very beginning "Kirchoff for the birds" fans arrogantly insulted nearly everybody who disagree with them.

There's no such thing as "Kirchhoff for the birds" fans. I myself was considered a king of circuit analysis when I was in college. I got my first job as an engineer because of that ability.

However, I know darn well that Kirchhoff can only be applied to lumped circuits and it doesn't work for anything under a varying electromagnetic field. Because, under that condition, a circuit becomes an "antenna", and antennas are everything but lumped. You'll have induction everywhere so it will be impossible to solve a circuit using Kirchhoff which does not provide the tools for either designing or probing your circuits.

How do I know that? By my own experience. When I got my first job as an engineer, digital circuits were leaving the domains of Kirchhoff and foraying into the realm of electromagnetism. Frequencies were such that the wavelengths were becoming as short as the size of the PCBs.

We had problems with the companies that designed our PCBs at the time, because they were kirchhoffalwaysholders (the cute name I decided to bestow upon those who think that Kirchhoff always holds) not by choice, but because they were used to low frequency circuits, where Kirchhoff is good enough.

We had to invite the help of our colleagues from the radio division, for whom Maxwell was second nature, and use their experience to instruct our designers how to properly design a PCB for higher frequencies, which is commonplace today.

This kirchhoffalwaysholdery sucks, therefore, because not only it is a pseudo-scientific doctrine, it is an encumbrance for any serious electronics engineering today.

And apparently it is now taking the shape of a religious movement where its converts get very offended when it is shown in theory or practice that their nonsense dogma is false.

As Lewin says, "this tells you something about them".
« Last Edit: December 09, 2018, 09:01:58 am by bsfeechannel »
 
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Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #435 on: December 09, 2018, 02:56:41 am »
I'm afraid of you being a troll and wasting my time.

Fear not. According to the dictionary, a troll is someone who posts deliberately inflammatory articles on an internet discussion board. I may rant or show I am indignant at some absurd fact sometimes, but I never troll. I don't have the time or the motivation for hanging around in forums posting provocative assertions just to see the world burn.

I'm just questioning your claims. That's what forums are for.

Quote
Read the replay above as it may be relevant if you are not a troll.

I did, thank you.

Quote
Shape of the loop will make no difference as long as B-flux is uniform but if that is not the case then you will not be able to calculate that with just pen and paper (you may be able to approximate something) but you will need a computer simulation tool to solve that and of course all details to scale.
And even if flux is uniform you will need to know the total length of that loop and the length between the two points you want to make the measurement then calculation is the same as for the simple ring model as shape alone makes no difference.   

Thanks again for your reply. I thought of many other "challenges", but I won't tire you with an infinite list of questionings. I just wanted to understand the basic principles of your claim.
« Last Edit: December 09, 2018, 09:03:25 am by bsfeechannel »
 

Offline rfeecs

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #436 on: December 09, 2018, 10:17:15 am »

So how does a voltmeter tell the difference if it only shows the field caused by electron density and not the magnetic EMF? (Hint: It does show some EMF too but not where you would typically want)

If voltmeters treat the two separately, why should we treat them as the same thing? We are trying to calculate what the voltmeter would show after all.

I'm not trying to pick sides here, or say anything negative about anyone. To me it seams that most people in this thread are not saying anything wrong for the most part, but the disagreement seams to stem from using a slightly different definition of things and more rarely a bit from just having a different thought process about this thing.

I'm not clear what you are talking about.  So the two voltmeters in this experiment are not affected by the EMF?  Then why do they read different voltages?

Anyway, voltmeters don't read the field caused by electron density.  They don't read the electrostatic potential.  Take the example of a PN junction diode.  It clearly has different electron densities in the P and N depletion region.  There is an electrostatic potential difference due to the charge separation.  But a voltmeter measures zero volts when connected to the leads of the diode.

If you integrate E dot dl through an unbiased diode, you get a voltage!  Diodes violate KVL!



Obviously using integral of E dot dl has a problem.  Circuits with diodes would be another KVL fail according to Dr. Lewin's definition.
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #437 on: December 09, 2018, 11:07:16 am »
If you integrate E dot dl through an unbiased diode, you get a voltage!  Diodes violate KVL!

Indeed. Chemical battery violates KVL as well. Resistor and any other lone component violates KVL. Kirchoff's Circuit Laws requires closed Circuit. As some insist that Dr.Lewin's loop cannot be split into lumped elements, then all this conversation is futile. When we agree that 1/4 of the Dr.Lewin's experiment (inner) loop receives EMF/4 and can be treated as lumped element meaning Berni model is correct - then there's ground for conversation. Honestly I do not see such time coming.

Some (you know who) can prove me wrong  :popcorn:
 

Offline rfeecs

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #438 on: December 09, 2018, 11:26:27 am »
If you integrate E dot dl through an unbiased diode, you get a voltage!  Diodes violate KVL!

Indeed. Chemical battery violates KVL as well. Resistor and any other lone component violates KVL. Kirchoff's Circuit Laws requires closed Circuit.

That's not my point.  You can make a closed circuit by connecting the diode leads together with a wire.  Go around the loop and calculate the integral of E dot dl.  For the wire it is zero, for the diode, it is not zero.  So the integral of E dot dl around the loop is not zero.  According to Dr. Lewin's definition, Kirchoff's Loop Rule is violated.  So are Maxwell's equations?  The world is flat.

But everyone knows the voltage across the diode is zero, and the voltage across the wire is zero.  So KVL is OK, and the world is not flat.
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #439 on: December 09, 2018, 12:38:37 pm »
You can make a closed circuit by connecting the diode leads together with a wire.  Go around the loop and calculate the integral of E dot dl.

No. Kirchoff's Circuit Laws requires closed circuit of lumped elements. So diode is one lumped element of the circuit and wire supposedly with it's internal resistance - another. So two elements. When you do circuit analysis - you don't go around the loop integrating everything in the path. This is not how CIRCUIT analysis shall be done. You do E dot dl over diode and make it lumped element - black box with two terminals, you can name it as voltage source when subject to light for example. Then you do E dot dl with wire and again make it black box with two terminals, name it load. Both are lumped elements of our circuit. Connect those together and *then* check against Kirchoff's Laws using voltage/current measurements.

Shall I repeat? Pay close attention to term "lumped element":

When we agree that 1/4 of the Dr.Lewin's experiment (inner) loop receives EMF/4 and can be treated as lumped element meaning Berni model is correct - then there's ground for conversation.

[edit] Better stick to electromagnetism.
« Last Edit: December 09, 2018, 01:51:04 pm by ogden »
 

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #440 on: December 09, 2018, 01:40:35 pm »
Take the example of a PN junction diode.  It clearly has different electron densities in the P and N depletion region.  There is an electrostatic potential difference due to the charge separation.  But a voltmeter measures zero volts when connected to the leads of the diode.

The reason for that is that to connect to the silicon you have to create ohmic contacts (non-rectifying contacts) and...
Nah, I'll use the first link.
https://www.quora.com/Why-we-cant-measure-the-barrier-potential-existing-across-a-p-n-junction-by-connecting-voltmeter-across-the-p-n-junction
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Offline rfeecs

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #441 on: December 09, 2018, 02:21:25 pm »
Take the example of a PN junction diode.  It clearly has different electron densities in the P and N depletion region.  There is an electrostatic potential difference due to the charge separation.  But a voltmeter measures zero volts when connected to the leads of the diode.

The reason for that is that to connect to the silicon you have to create ohmic contacts (non-rectifying contacts) and...
Nah, I'll use the first link.
https://www.quora.com/Why-we-cant-measure-the-barrier-potential-existing-across-a-p-n-junction-by-connecting-voltmeter-across-the-p-n-junction

Don't believe everything you read on the internet.

The voltmeter reads the difference in Fermi levels between the two contacts:

https://en.wikipedia.org/wiki/Fermi_level

Quote
Sometimes it is said that electric currents are driven by differences in electrostatic potential (Galvani potential), but this is not exactly true.[2] As a counterexample, multi-material devices such as p–n junctions contain internal electrostatic potential differences at equilibrium, yet without any accompanying net current; if a voltmeter is attached to the junction, one simply measures zero volts.[3] Clearly, the electrostatic potential is not the only factor influencing the flow of charge in a material—Pauli repulsion, carrier concentration gradients, electromagnetic induction, and thermal effects also play an important role.

In fact, the quantity called voltage as measured in an electronic circuit has a simple relationship to the chemical potential for electrons (Fermi level). When the leads of a voltmeter are attached to two points in a circuit, the displayed voltage is a measure of the total work transferred when a unit charge is allowed to move from one point to the other. If a simple wire is connected between two points of differing voltage (forming a short circuit), current will flow from positive to negative voltage, converting the available work into heat...

Maybe this is straying too far off topic.  The point was supposed to be that a voltmeter doesn't measure electrostatic potential, and what a voltmeter actually measures and what is the definition of potential and voltage are more complicated than we usually think.
 

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #442 on: December 09, 2018, 03:01:25 pm »
The reason for that is that to connect to the silicon you have to create ohmic contacts (non-rectifying contacts) and...
Nah, I'll use the first link.
https://www.quora.com/Why-we-cant-measure-the-barrier-potential-existing-across-a-p-n-junction-by-connecting-voltmeter-across-the-p-n-junction
Don't believe everything you read on the internet.
The voltmeter reads the difference in Fermi levels between the two contacts:
https://en.wikipedia.org/wiki/Fermi_level

Well, wikipedia is on the Internet, so I shouldn't believe it. But maybe you misquoted it.
Besides, the difference in Fermi levels is the barrier potential (if we agree on how to treat the sign). I guess you were the one saying that you cannot read it with a voltmeter.

But if you want a reference that is not on the Internet, you might want to read page 242 of "Semiconductor Physics and Devices" by Donald Neamen.

"This potential difference across the junction cannot be measured with a voltmeter because new potential barriers will be formed between the probes and the semiconductor that will cancel V_bi"

This is the first book I took off my shelf, but I'm pretty sure I could find something along the same line on Sze, or on Streetman, or on Muller Kamins.
Oh my.
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Offline rfeecs

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #443 on: December 09, 2018, 05:00:49 pm »
The reason for that is that to connect to the silicon you have to create ohmic contacts (non-rectifying contacts) and...
Nah, I'll use the first link.
https://www.quora.com/Why-we-cant-measure-the-barrier-potential-existing-across-a-p-n-junction-by-connecting-voltmeter-across-the-p-n-junction
Don't believe everything you read on the internet.
The voltmeter reads the difference in Fermi levels between the two contacts:
https://en.wikipedia.org/wiki/Fermi_level

Well, wikipedia is on the Internet, so I shouldn't believe it. But maybe you misquoted it.
Besides, the difference in Fermi levels is the barrier potential (if we agree on how to treat the sign). I guess you were the one saying that you cannot read it with a voltmeter.
With no bias, in equilibrium, the Fermi levels on both terminals are equal, so zero voltage.  Refer to figure 7.3 of Neamen.
Quote

But if you want a reference that is not on the Internet, you might want to read page 242 of "Semiconductor Physics and Devices" by Donald Neamen.

"This potential difference across the junction cannot be measured with a voltmeter because new potential barriers will be formed between the probes and the semiconductor that will cancel V_bi"
You have to wonder if "potential barrier" is the right phrase for an ohmic contact, which should have little or no barrier.
Quote
This is the first book I took off my shelf, but I'm pretty sure I could find something along the same line on Sze, or on Streetman, or on Muller Kamins.
Oh my.
So what's your point?  Are you saying that there is no net electrostatic potential across the diode terminals?
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #444 on: December 09, 2018, 08:04:41 pm »
"This potential difference across the junction cannot be measured with a voltmeter because new potential barriers will be formed between the probes and the semiconductor that will cancel V_bi"

Right. This is where I fully agree with you :)

So what's your point?

What's *your* point to talk about semiconductors in discussion about Dr.Lewin's lecture explaining electromagnetism?

Observer effect of quantum theory while looking at bare PN junction is way too huge stretch off the rails of said discussion. We shall stick to Dr.Lewin's original experiment conditions where circuit element terminals/lugs are made out of conductor (not semiconductor) and voltmeter measures potential difference using Ohms law - by running current through it's internal resistance.

So if such "voltmeter vulgaris" measure 0V on lumped element terminals, we say it's E dot dl is zero and disregard quantum or chemical or whatever phenomena inside it.
 

Offline Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #445 on: December 09, 2018, 08:45:41 pm »
I'm not clear what you are talking about.  So the two voltmeters in this experiment are not affected by the EMF?  Then why do they read different voltages?

Anyway, voltmeters don't read the field caused by electron density.  They don't read the electrostatic potential.  Take the example of a PN junction diode.  It clearly has different electron densities in the P and N depletion region.  There is an electrostatic potential difference due to the charge separation.  But a voltmeter measures zero volts when connected to the leads of the diode.

If you integrate E dot dl through an unbiased diode, you get a voltage!  Diodes violate KVL!



Obviously using integral of E dot dl has a problem.  Circuits with diodes would be another KVL fail according to Dr. Lewin's definition.

This diagram shows a diode with a current flowing trough it. In such a case all semiconductors show a voltage drop that can indeed be measured with a voltmeter.

In a rest state any voltage created on the junction is subtracted back out once the semiconductor connects to the copper pins. If a diode was to create a voltage in such a conduction this would mean i will also have to be capable of pushing current in that direction of voltage. Once you have both you have power being output from the diode and this would violate conservation of energy. That being said it is possible to use a diode to generate a voltage. If you are to heat up one end of a diode and cool the other you can get a strong thermocouple effect that converts some of that heat into output power on the pins. Additionally if this is a glass encapsulated diode or a LED then shining the right wavelengths of light on the diodes junction will also cause it to operate like a solar cell and produce power. In all of these cases external energy had to be put in to make it do that.

Again KVL has no way of dealing with diodes as that's not part of its job. But circuit analysis theory makes it work by having lumped models for all these semiconductor devices (Diodes, BJTs, FETs, IGBTs, SCRs...). There lumped models often contain parametric current sources and volt/amp meters inside of them and they vary in complexity depending on how accurate you need it to be. This allows for circuit analysis to be used on circuits with active components without any issues.

So what does a voltmeter measure? Well it actually measures the current trough its internal resistance and then displays what voltage it takes to push such a current. Notice how in Dr. Lewins example the voltage across the resistors is always defined as a single value. In the same way it is defined to have a single value across the terminals of a voltmeter. Since an ideal resistor has zero physical dimension, means that it is impossible to generate any magnetic EMF across it (It can't be part of a surface area edge as it has no length) and a external electrostatic field can't produce a gradient sharp enough to pull electrons along. So the only "electron pusher" that remains to convince electrons to flow trough the resistor is the difference in charge density on the resistors terminals. The crowded electrons on one end want to get trough to the not as crowded electrons on the other end. Hence why the voltmeter ends up showing a difference in charge density across its terminals.

But it is possible to have EMF generated on a resistor that has physical length. Its basically the combination of a wire and a resistor (And can be lump modeled as such if desired). In the same way a voltmeter that's longer than zero will read EMF across itself. But its only the EMF induced in the section that the voltmeters size occupies. So the larger the voltmeter the more EMF it will show on the display. This just makes things more confusing so we say voltmeters have zero size so they don't read any EMF.
 

Offline rfeecs

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #446 on: December 10, 2018, 04:24:13 am »
This diagram shows a diode with a current flowing trough it. In such a case all semiconductors show a voltage drop that can indeed be measured with a voltmeter.

In a rest state any voltage created on the junction is subtracted back out once the semiconductor connects to the copper pins. If a diode was to create a voltage in such a conduction this would mean i will also have to be capable of pushing current in that direction of voltage.

Actually, the diagram is in equilibrium, meaning no net current is flowing.

I fully agree that the diode contacts, which are not shown on the diagram, will have their own potential difference, E-field and charge that compensates for the voltage across the diode junction.  Maxwell's equations still work.

So I guess the only point of bringing up the diode is to point out that E field is not the only "pusher" of charge.  The concentration gradient at the diode junction is another.  You have electrochemical potential as well as electrostatic potential and induced EMF that can all move charge.  The voltmeter can't tell the difference between them.

Quote
So what does a voltmeter measure? Well it actually measures the current trough its internal resistance and then displays what voltage it takes to push such a current. Notice how in Dr. Lewins example the voltage across the resistors is always defined as a single value. In the same way it is defined to have a single value across the terminals of a voltmeter. Since an ideal resistor has zero physical dimension, means that it is impossible to generate any magnetic EMF across it (It can't be part of a surface area edge as it has no length) and a external electrostatic field can't produce a gradient sharp enough to pull electrons along. So the only "electron pusher" that remains to convince electrons to flow trough the resistor is the difference in charge density on the resistors terminals. The crowded electrons on one end want to get trough to the not as crowded electrons on the other end. Hence why the voltmeter ends up showing a difference in charge density across its terminals.

But it is possible to have EMF generated on a resistor that has physical length. Its basically the combination of a wire and a resistor (And can be lump modeled as such if desired). In the same way a voltmeter that's longer than zero will read EMF across itself. But its only the EMF induced in the section that the voltmeters size occupies. So the larger the voltmeter the more EMF it will show on the display. This just makes things more confusing so we say voltmeters have zero size so they don't read any EMF.

So you are still saying that EMF is located at specific segments of the loop.  A zero length voltmeter won't show any EMF on the display?  I thought it would show the EMF of the whole loop that includes the voltmeter and the test leads and the path connecting the two points you are measuring.

We're all just stuck in an endless loop now.  break;
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #447 on: December 10, 2018, 07:20:39 am »
You have electrochemical potential as well as electrostatic potential and induced EMF that can all move charge.  The voltmeter can't tell the difference between them.

What does it prove? - That all electrons are equal and you can't mark them?

So you are still saying that EMF is located at specific segments of the loop.

Seems, you are alone denying this here. Remember time when this thread talked about resistive ring (introduced by you BTW)?

http://www.eevblog.com/forum/chat/does-kirchhoffs-law-hold-disagreeing-with-a-master/msg1961348/#msg1961348
« Last Edit: December 10, 2018, 05:12:10 pm by ogden »
 

Offline radioactive

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #448 on: December 10, 2018, 07:24:36 am »
[edit] remove un-necessary comment, as I believe it was taken wrong way  (thread died).
« Last Edit: December 20, 2018, 02:22:27 pm by radioactive »
 

Offline radioactive

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #449 on: December 20, 2018, 03:27:01 pm »
I probably shouldn't dig this thread back up, but wanted to (humbly) share the sim that I worked on such as it is.  I've seen some of the comments referring indirectly to my previous posts as trolls, idiot, etc. First off, let me start by saying that I'm not trying to *prove* anything here.  I don't care who is right or wrong.  I'm just someone who had some time recently to play with openEMS and see if I could reproduce something similar to the Romer experiment and thought I would share it in case someone else might see how cool FDTD sims are.  I feel bad for screwing up the initial sims and posting them.  It may still have issues, but I think I figured out what was wrong in the first couple of posts.  The mesh you see in the geometry viewer is not necessarily how the materials get discretized before being passed on to the FDTD engine  (I had forgotten about this).  It is really important to use the --debug-PEC option to view the mesh in Paraview in order to verify materials didn't disappear too much after the engine matches materials up on the mesh sub-volumes.   This sim is still not the right scale for Romer because it is just too much to discretize a meshed coil with sub-mm wire radius.   I still think it is worth sharing.  If for no other reason, then just to give some more exposure to the really nice open/free software called openEMS  http://www.openems.de/start/index.php

As for the crazy voltages / currents in my previous post that some obviously found funny,  I should have explained more on that.   I was testing several different voltages for the excitation with highly scaled up geometry to try and see if I got a log response (I did).  I noticed the thing with the feed coil and posted some images while the voltage was something like 10e6 or something crazy like that while trying to make an unrelated point (unconvincingly).   Part of the issue may have been with the mesh on that one... or it may have been find at that point... not sure.  Anyway, hopefully that makes sense.  I understand some of the comments.  Sorry if I wasted your time or offended in previous posts. 

The images and videos are from a sim with a triangular waveform (like Romer), but with a DC bias (not like Romer).  This gives a slow rise DC component in the beginning of the excitation.  Note that due to the time it takes to simulate a complex geometry like this, the excitation is at a higher frequency (above SRF and much higher than Romer).  I also ran the sim for 1 cycle below SRF (still much shorter wavelength than Romer) and will post voltage measurements for that.   The videos are generated with Paraview.   Also another very nice free software.

SRF of the coil was determined by excitation with a bandwidth-limited Gaussian pulse which generates a nice flat stimulus in the frequency domain  (think using a wideband noise source with an FFT spectrum analyzer).   The simulation generates time domain voltages / currents in a text file.  If you process that with FFT, then you can analyze arbitrary materials / structures in the frequency domain.   

Another much-easier-to-simulate-with-amazing-accuracy example would be a planar microstrip filter made of copper, vias, dielectric, airbox, etc.

Time steps in the videos should match time steps in the plots.





Another thing that someone might find interesting...  was talking to a guy who does almost nothing but RF sims.   He told me about another method.  MoM  (method of moments) that can be used to mesh geometries and convert them to equivalent reactive components for simulating in a standard analytical circuit simulator  (e.g. Spice).  I couldn't find any open-source projects that looked like something worth investigating.  Would be very interested this if someone is aware of any.
 
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