Author Topic: Does Kirchhoff's Law Hold? Disagreeing with a Master  (Read 52326 times)

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Offline electrodacus

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #350 on: December 05, 2018, 05:59:53 pm »
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.

What is the math for the above example and what will those two different voltages be ?  -0V and +0V :)
From my understanding of physics in this universe (going to assume there is no parallel universe with same experiment running at the exact same time) there will always be a single voltage between any two points at a fixed moment in time no mater if there is a measurement device in there or not.

Online Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #351 on: December 05, 2018, 06:22:39 pm »
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.

What is the math for the above example and what will those two different voltages be ?  -0V and +0V :)
From my understanding of physics in this universe (going to assume there is no parallel universe with same experiment running at the exact same time) there will always be a single voltage between any two points at a fixed moment in time no mater if there is a measurement device in there or not.

This is the reason why its so confusing.

First of all this is not a weird quantum mechanics thing like Schrodingers cat thought experiment that the cat is both dead and alive simultaneously until you look at it. Instead it is actually the fault of how voltage is defined in textbooks.

The more common way of thinking about voltage is that more electrons there are in one place the more negative that point is, connect this area with lots of electrons to an area with few electrons and you get current between them as the charges want to even out.

The way voltage is actually formally defined is "An integral of all forces working on charges along a path between two points". These forces include the electric field that bunched together electrons create, but it also includes the magnetic forces pushing electrons around (Any charged particle is affected by a changing magnetic field). This force is dependan't on where you are in the magnetic field. This results in a different result of the integral of the force and hence a different voltage.

In Dr. Lewins example you get the magnetic EMF adding to the resistors voltage if you go around one way, but subtracting from the resistors voltage if you go around the other way. Hence the integral is different and there is different voltage. Its just how the math ends up working out. The actual electron charge density at the points A and B is always a single well defined number. The voltage you measure by connecting a voltmeter as you shown will measure this electron density. Hence why the voltmeter shows one voltage.

The two different voltages result from Dr. Lewins example could be sort of a incomplete result of the voltage so far, you need to include the rest of the circuit to properly define the voltage. Think of the two voltages as sort of like complex number math, they don't necessarily exist in the real world but they make the math work.
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #352 on: December 05, 2018, 07:13:36 pm »
This is the reason why its so confusing.

It's made to be confusing as presented by Dr.Lewin and his cultists, but it's not. It confuses only those who can't understand that voltmeter wire can be EMF source. The same "path dependent voltage" mind tricks can be played using other EMF sources like chemical batteries or photovoltaic cells, but those are not so confusing because "such batteries cannot be easily hidden in the wires" :)
« Last Edit: December 05, 2018, 07:49:19 pm by ogden »
 

Offline electrodacus

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #353 on: December 05, 2018, 07:47:58 pm »
This is the reason why its so confusing.

First of all this is not a weird quantum mechanics thing like Schrodingers cat thought experiment that the cat is both dead and alive simultaneously until you look at it. Instead it is actually the fault of how voltage is defined in textbooks.

The more common way of thinking about voltage is that more electrons there are in one place the more negative that point is, connect this area with lots of electrons to an area with few electrons and you get current between them as the charges want to even out.

The way voltage is actually formally defined is "An integral of all forces working on charges along a path between two points". These forces include the electric field that bunched together electrons create, but it also includes the magnetic forces pushing electrons around (Any charged particle is affected by a changing magnetic field). This force is dependan't on where you are in the magnetic field. This results in a different result of the integral of the force and hence a different voltage.

In Dr. Lewins example you get the magnetic EMF adding to the resistors voltage if you go around one way, but subtracting from the resistors voltage if you go around the other way. Hence the integral is different and there is different voltage. Its just how the math ends up working out. The actual electron charge density at the points A and B is always a single well defined number. The voltage you measure by connecting a voltmeter as you shown will measure this electron density. Hence why the voltmeter shows one voltage.

The two different voltages result from Dr. Lewins example could be sort of a incomplete result of the voltage so far, you need to include the rest of the circuit to properly define the voltage. Think of the two voltages as sort of like complex number math, they don't necessarily exist in the real world but they make the math work.

I know it has nothing to to with Schrodingers cat :) but it sounds like that is what you are trying to say.
I asked what those two different voltages are, at a fixed moment in time and I do not see that in your replay.

Online Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #354 on: December 05, 2018, 11:40:36 pm »
Ah okay you want to know the voltages. Given that the total EMF is 1V the result is -0.5V on one side and + 0.5V on the other side. This is because wires show up as having no voltage while the resistors show a voltage according to Ohms law. Since the current trough the loop is identical everywhere this means both resistors have to show the same drop (given they are the same value). Sine the current is flowing upwards in one resistor and down the other you get opposite voltage polarity. So you do get two voltages.

To be honest tho such a result is not very useful when you are trying to understand what the circuit does. Dr Lewins math is not wrong about this (Its wrong when it comes to KVL)

The more useful way of analyzing this circuit is using lumped circuit mesh analysis. This is thought in every EE Highschool and gives more useful results. Here all lengths of wire are modeled as coupled inductors. Since we are interested in the voltage at only one point in time we can calculate the voltage using Faradays law and just pretend there is a battery there. With this, the loop on the left and right of the voltmeter get a 0.5V battery (each loop is half of the whole loop). Because the resistors also have a 0.5V drop the result is 0V for both halves and you get the result of the voltmeter reading 0V (Not +0V and -0V, just a single well defined 0V). If you want to connect the voltmeter to other points you have to recalculate the loop areas accordingly.KVL holds just fine here.

In the experiment he just connects the scope with wires in just the right way to expose the two voltages. If you cosider his scope wires as part of the circuit you will get the same result with regular circuit mesh analysis. The two scopes are connected to two different parts of the circuit since each connects via its own wire. The EMF in those wires added up is the exact amount the readings on the two scopes differs. The actual voltage on the two points stays the same (It's the average of the two scopes when probing the middle like that)
 

Offline alanb

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #355 on: December 06, 2018, 12:08:50 am »
I don't know if someone has previously posted this link. Its worth watching.

https://youtu.be/JpVoT101Azg
 

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #356 on: December 06, 2018, 12:22:12 am »
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.

Well, in case of the first circuit supplied by batteries with no varying magnetic flux, you agreed with me that all five voltmeters would read the same, i.e. the real position of the voltmeters didn't matter. But let's forget that for the moment.

Because I agree with you that in this particular case, any misplacement of the voltmeter will reflect on the precision of the measuring.

But my question remains unanswered.

How do I know that this imprecision is due to the voltmeter not being exactly in the place it should be, and not due to a resistor mismatch?

You know, resistors change their values, either with temperature, or age, or both, etc.
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #357 on: December 06, 2018, 02:23:08 am »
I don't know if someone has previously posted this link. Its worth watching.

Yes indeed it was posted.
Next time please do read the thread or at least use search.
 

Offline electrodacus

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #358 on: December 06, 2018, 07:11:24 am »
Ah okay you want to know the voltages. Given that the total EMF is 1V the result is -0.5V on one side and + 0.5V on the other side. This is because wires show up as having no voltage while the resistors show a voltage according to Ohms law. Since the current trough the loop is identical everywhere this means both resistors have to show the same drop (given they are the same value). Sine the current is flowing upwards in one resistor and down the other you get opposite voltage polarity. So you do get two voltages.
You can only read -0.5V and +0.5V if you cut the circuit in half on those two measurement points and then measure each of the half circuits but then that is a completely different circuit and it is no different from a circuit where you have batteries. A real battery has the internal impedance distributed trough the battery is not like there is an ideal source inside and then a separate impedance as it is represented in a diagram.

Wires do have a resistance unless they are supercondutors and my original example specifically had only wires that had a stated resistance was 1Ohm in both examples just that first had a loop made of same type of wire half the ring was 0.5Ohm and the other half 0.5Ohm while the second example had a ring made of two half rings soldered together one with 0.1Ohm and the other with 0.9Ohm resistance.
My examples where meant to get rid of the confusion of having wires and separate resistors but any example will work the same.
Also my example allows you to imagine swiping the virtual voltmeter probes around the ring similar to a potentiometer.

You can use KVL to find out voltage between any two points on those example rings.

If you make an infinitely small cut in any of those two example rings so that there is no current then using the same potentiometer method to measure any two points you will get the same result for both example rings as resistance no longer plays a role.


Well, in case of the first circuit supplied by batteries with no varying magnetic flux, you agreed with me that all five voltmeters would read the same, i.e. the real position of the voltmeters didn't matter. But let's forget that for the moment.
Because I agree with you that in this particular case, any misplacement of the voltmeter will reflect on the precision of the measuring.
But my question remains unanswered.
How do I know that this imprecision is due to the voltmeter not being exactly in the place it should be, and not due to a resistor mismatch?
You know, resistors change their values, either with temperature, or age, or both, etc.

Not sure what sort of point you want to make. The discussion here is that there is only one voltage between the two defined points at a fixed moment in time and that KVL applied correctly will give you the correct result.
Of course real circuits are not perfect and that is why you have tolerances but you add those tolerances in your calculation and your result will have a margin of error proportional with those tolerances.

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #359 on: December 06, 2018, 08:10:08 am »
My examples where meant to get rid of the confusion of having wires and separate resistors but any example will work the same.
[...]
You can use KVL to find out voltage between any two points on those example rings.

Ok, let's take a loop made of two big resistors - physically big - and some copper wire. Let's say the resistors are shaped into an arc spanning 45 degrees. One is 0.1 ohm, the other one is 0.9 ohm. Emf in the loop is 1V.
What is the real and only voltage across the resistors?
What is the real and only voltage across the remaining two portions of wire (say it's copper)?

Edit: added plurals.
« Last Edit: December 06, 2018, 08:13:11 am by Sredni »
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Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #360 on: December 06, 2018, 09:07:15 am »
Not sure what sort of point you want to make.

I'm just trying to understand what you said in your first post. Of course if you please.

Quote
The discussion here is that there is only one voltage between the two defined points at a fixed moment in time and that KVL applied correctly will give you the correct result.

OK. That's exactly what I am discussing too.

Quote
Of course real circuits are not perfect and that is why you have tolerances but you add those tolerances in your calculation and your result will have a margin of error proportional with those tolerances.

Fine. So let's suppose now that we don't know the value of the resistors. We don't and we can't know ( for some reason, doesn't matter). The "correct" voltage will obviously be measured when we find the exact right place that in our case demands a nanometric precision.

Let's suppose that the voltmeter now indicates something like 0.437582 V. How can we be sure that this is the correct value? Because now, since the resistances and their relation are unknown, we don't know what to expect. How can we know that we are not a couple of hundreds of nanometers off?
 

Offline electrodacus

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #361 on: December 06, 2018, 09:23:17 am »
Ok, let's take a loop made of two big resistors - physically big - and some copper wire. Let's say the resistors are shaped into an arc spanning 45 degrees. One is 0.1 ohm, the other one is 0.9 ohm. Emf in the loop is 1V.
What is the real and only voltage across the resistors?
What is the real and only voltage across the remaining two portions of wire (say it's copper)?

The copper wire is also a resistor so you need to provide the resistance of those in order for me to be able to give you the exact answer.
But all you did was split the ring in to 4 equal parts and have 4 resistors two with equal low resistance (the copper wires) and the other two with higher resistance 0.1Ohm and 0.9Ohm
If you assume the wires are made of superconducting material then result is same as in my original experiment 0.4V across the same two opposite points (middle of each copper wire).
In case you measure across the quarter of the ring you will measure say +650mV across the 0.9Ohm resistor and then -150mV across the 0.1Ohm resistor this values will be slightly different if those are not superconductors but copper wires.
The equivalent circuit to make the calculations will be made of 4 voltage sources each equal with 0.25V and each with a series resistance corresponding to whatever that quarter resistance is.

Still there is no point in your slightly more complex example as the voltage on any of the points will be clearly defined and you will never have two different values at a fixed moment in time.

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #362 on: December 06, 2018, 09:41:48 am »
But all you did was split the ring in to 4 equal parts

Not equal (360 - 2x45 = 270; 270/2 = 135 != 45), but that's immaterial.

Quote
and have 4 resistors two with equal low resistance (the copper wires) and the other two with higher resistance 0.1Ohm and 0.9Ohm

Correct. So, let's make the resistors span 90 degrees instead of 45 so we can use your numbers.

Quote
you will measure say +650mV across the 0.9Ohm resistor and then -150mV across the 0.1Ohm resistor

The only and true voltage across the 0.9 ohm resistor is .65 V. Ohm would say there's a current of  .65/.9 = .72 amps
The only and true voltage across the 0.1 ohm resistor is .15 V. Ohm would say there's a current of .15/.1 = 1.5 amps

Now, .65 + .15 = .80 V. EMF is 1V, I suppose you want to put that into 'modified KVL', so did you bungle the calculation or are you assuming there are 0.20 voltage drop on the copper wires? If we assume 0 resistance that would mean ohm would think there's infinite current.
But let's leave ohm alone for the moment because you might want to put some EMF here and there.

Can you put KVL into a formula with numbers? Please make the (possibly corrected, if required) numbers of all the 'true' voltage drops in the loop and show that KVL balance.

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Offline electrodacus

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #363 on: December 06, 2018, 10:01:59 am »
But all you did was split the ring in to 4 equal parts

Not equal (360 - 2x45 = 270; 270/2 = 135 != 45), but that's immaterial.

Quote
and have 4 resistors two with equal low resistance (the copper wires) and the other two with higher resistance 0.1Ohm and 0.9Ohm

Correct. So, let's make the resistors span 90 degrees instead of 45 so we can use your numbers.

Quote
you will measure say +650mV across the 0.9Ohm resistor and then -150mV across the 0.1Ohm resistor

The only and true voltage across the 0.9 ohm resistor is .65 V. Ohm would say there's a current of  .65/.9 = .72 amps
The only and true voltage across the 0.1 ohm resistor is .15 V. Ohm would say there's a current of .15/.1 = 1.5 amps

Now, .65 + .15 = .80 V. EMF is 1V, I suppose you want to put that into 'modified KVL', so did you bungle the calculation or are you assuming there are 0.20 voltage drop on the copper wires? If we assume 0 resistance that would mean ohm would think there's infinite current.
But let's leave ohm alone for the moment because you might want to put some EMF here and there.

Can you put KVL into a formula with numbers? Please make the (possibly corrected, if required) numbers of all the 'true' voltage drops in the loop and show that KVL balance.

Sorry I confused your 45 degree with 90 degree but I guess that still makes the point you wanted to make.

The formula is very simple each quarter of the ring (90 degree) will see a quarter of that 1V so 0.25V
Thus depending on the direction of the current you have +0.9V across the 0.9Ohm resistor but you subtract 0.25V thus +650mV
On the other side of the ring (again my assumption the other resistor is on the other half) you have 0.1V on the resistor and subtract 0.25V so -150mV
The copper wires are treated the same as the resistors so a 0.25V source and whatever resistance those wires have say is 1mOhm for each quarter segment in series
Then total ring resistance will be 1.002Ohm and thus the current will be smaller 0.998A thus the values calculated before will be slightly influenced by this but not much.

If you disagree with this please provide your numbers and how you got to them.
 
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Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #364 on: December 06, 2018, 10:24:11 am »
Mehdi posted a follow-up video:

Well, thank you for coming with old news. Thread is "boiling" about it few days already ;) Maybe next time read the thread please?
 

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #365 on: December 06, 2018, 10:37:55 am »
The formula is very simple each quarter of the ring (90 degree) will see a quarter of that 1V so 0.25V
Thus depending on the direction of the current you have +0.9V across the 0.9Ohm resistor but you subtract 0.25V thus +650mV
On the other side of the ring (again my assumption the other resistor is on the other half) you have 0.1V on the resistor and subtract 0.25V so -150mV
The copper wires are treated the same as the resistors so a 0.25V source and whatever resistance those wires have say is 1mOhm for each quarter segment in series

Ok, so you are Mabilde-like.
Hence you believe that the emf is located inside the wire in the form of a distributed voltage generator.
Can we summarize your KVL balance as

net voltage drop across 0.9 ohm + voltage gain across copper + net voltage drop across 0.1 ohm + voltage gain across copper = 0

(-0.9 +0.25) + 0.25 + (-0.1 + 0.25) + 0.25 = 0

Does that mean that there is an electric field inside the copper conductor? (either perfect or as you wish with 1mohm resistance per arc)?

Quote
If you disagree with this please provide your numbers and how you got to them.

I do not believe in a 'true' voltage (actually it's not a belief, the formulas tell me). In this case it all depends on how you measure it - I can get your numbers by suitably partitioning the disk with the probes, or many other values. But I tell you how I can balance Faraday here:
 EMF = 1V, total loop resistance 1 ohm, current 1 amp
 integral of E dl in 0.9 ohm + integral of E dl in 0.1 ohm + nearly nothing in copper = EMF
 0.9 + 0.1 + 0 = 1

(signs come about when you consider the correct conventions)

My ohm's law still work. I had to give up uniqueness of voltage between two points, though.
Most importantly, there is practically no field inside the copper conductor, as predicted by Maxwell's equations.

In your case, well, what is the field inside the copper parts, if you have 0.25 V across each of them?
« Last Edit: December 06, 2018, 10:39:51 am by Sredni »
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Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #366 on: December 06, 2018, 10:56:56 am »
I do not believe in a 'true' voltage (actually it's not a belief, the formulas tell me). In this case it all depends on how you measure it - I can get your numbers by suitably partitioning the disk with the probes, or many other values. But I tell you how I can balance Faraday here:
 EMF = 1V, total loop resistance 1 ohm, current 1 amp
 integral of E dl in 0.9 ohm + integral of E dl in 0.1 ohm + nearly nothing in copper = EMF
 0.9 + 0.1 + 0 = 1

Right. Summary E field in resistors including wire resistance is equal to EMF of the loop, 1V. You shall not ignore EMF in the big, long resistors or we can even name them segments of resistive wire. This is where you add and subtract 0.25V EMF in them accordingly.

Quote
In your case, well, what is the field inside the copper parts, if you have 0.25 V across each of them?

If superconductive, then E field is 0.0V. In this case it is 0.002 V because wire is specified as 0.001 Ohms per segment.

[edit] We don't measure E field or EMF between two points of the circuit. We usually measure potential difference.
« Last Edit: December 06, 2018, 11:01:33 am by ogden »
 

Offline electrodacus

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #367 on: December 06, 2018, 11:18:58 am »
Quote
If you disagree with this please provide your numbers and how you got to them.

I do not believe in a 'true' voltage (actually it's not a belief, the formulas tell me). In this case it all depends on how you measure it - I can get your numbers by suitably partitioning the disk with the probes, or many other values. But I tell you how I can balance Faraday here:
 EMF = 1V, total loop resistance 1 ohm, current 1 amp
 integral of E dl in 0.9 ohm + integral of E dl in 0.1 ohm + nearly nothing in copper = EMF
 0.9 + 0.1 + 0 = 1

(signs come about when you consider the correct conventions)

My ohm's law still work. I had to give up uniqueness of voltage between two points, though.
Most importantly, there is practically no field inside the copper conductor, as predicted by Maxwell's equations.

In your case, well, what is the field inside the copper parts, if you have 0.25 V across each of them?

So are you saying that voltage across the 0.9Ohm resistor in the current example is 0.9V ? because that sure is not the case unless the resistor has an infinitely small size not a quarter of the ring (90 degree arc).
The copper parts are no different from the resistors as they are basically all resistors so same rule will apply.

You may want to think on what voltages you will read if the ring was open no current.
Maybe think on the same ring (no matter if the resistors are there or not) and you make a small cut (so no current) and insert there the world smallest voltmeter with almost infinite internal impedance. What do you think the reading will be ?
If you guest approximately 1V you will be right. That will be the case even if the ring was all made of superconducting material so how can you explain the measured 1V (assuming you agree).

So across the 0.9Ohm resistor will be 0.650V and not 0.900V and also as important there will be only one voltage there 0.650V and no multiple voltages at the same point in time as implied in the Lewin experiment.

If superconductive, then E field is 0.0V. In this case it is 0.002 V because wire is specified as 0.001 Ohms per segment.

[edit] We don't measure E field or EMF between two points of the circuit. We usually measure potential difference.

Thanks I'm worse on expressing myself :)  so your replay is shorter and more to the point.

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #368 on: December 06, 2018, 11:22:50 am »
I'd like to kindly ask the Kirchhoff experts what tools do Kirchhoff rules give me to calculate the "right" voltage of the loop below. Except for the irregular perimeter, all conditions are the same as for my rectangular loop above.



I need to know the exact points where to connect the voltmeter and its precise location. Any reply will be appreciated.
« Last Edit: December 06, 2018, 11:35:36 am by bsfeechannel »
 
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Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #369 on: December 06, 2018, 11:36:33 am »
So are you saying that voltage across the 0.9Ohm resistor in the current example is 0.9V ?
If I measure from outside, and without crossing the flux varying region, yes.
Note that I did not use 'voltage drop' but the integral of E.dl in my balance.

Quote
because that sure is not the case unless the resistor has an infinitely small size

No, no, that's the case and it is due to the fact that charge accumulate at the resistors end, where there is a gradient in conductivity. This makes the field strong along the resistors, and nearly zero in the copper conductor, just as expected by Maxwell's equations, the equation of continuity and the constitutive relation in the conductor.
Basically, all the EMF falls across the resistors.
It's a matter of charge displacement and the ensuing superposition of field.

Quote
If superconductive, then E field is 0.0V. In this case it is 0.002 V because wire is specified as 0.001 Ohms per segment.

[edit] We don't measure E field or EMF between two points of the circuit. We usually measure potential difference.

Thanks I'm worse on expressing myself :)  so your replay is shorter and more to the point.

So, you measure electric field in volts?
You need to brush up on basic physics.

Please, humor me and try again: what is the field inside the copper conductor and how do you justify - with formulas - that there is a 0.25 volts drop difference across it?
« Last Edit: December 06, 2018, 11:45:47 am by Sredni »
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Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #370 on: December 06, 2018, 11:40:36 am »
I'd like to kindly ask the Kirchhoff experts what tools do Kirchhoff rules give me to calculate the "right" voltage of the loop below. Except for the irregular perimeter, all conditions are the same as for my rectangular loop above.



I need to know the exact points where to connect the voltmeter and its precise location. Any reply will be appreciated.

bsfeechannel, you dirty bastard!  :-DD

All instruments lie. Usually on the bench.
 
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Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #371 on: December 06, 2018, 11:55:06 am »
So, you measure electric field in volts?
You need to brush up on basic physics.

I did not mention any units at all. You need to check your vision.

Quote
Please, humor me and try again: what is the field inside the copper conductor and how do you justify - with formulas - that there is a 0.25 volts drop across it?

You really need to check your vision. I said there is 0.25 volts EMF and 0.002V drop (in 0.002 ohms conductor). Don't you know Ohms law? BTW you did use it yourself to calculate drop across 0.1 and 0.9 Ohms resistors.
 

Offline electrodacus

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    • electrodacus
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #372 on: December 06, 2018, 12:14:16 pm »
So are you saying that voltage across the 0.9Ohm resistor in the current example is 0.9V ?
If I measure from outside, and without crossing the flux varying region, yes.
Note that I did not use 'voltage drop' but the integral of E.dl in my balance.

V = IR - EMF  so voltage at the terminals of that resistor is as mentioned 0.650V (for the 0.9Ohm resistor) The EMF will only be zero for a infinitely small resistor and as that is not the case in our example where resistor is a quarter of the loops size thus 0.250V
Same of course apply to a section of copper say 0.001Ohm with 0.001V drop thus voltage measured will be -0.249V
I sort of feel like I deal with trolls and I hope that is not the case.
« Last Edit: December 06, 2018, 12:18:58 pm by electrodacus »
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #373 on: December 06, 2018, 12:19:36 pm »
I sort of feel like I deal with trolls

Me too. What a coincidence :-DD
 

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #374 on: December 06, 2018, 12:52:37 pm »
So, you measure electric field in volts?
You need to brush up on basic physics.

I did not mention any units at all. You need to check your vision.

You agreed with ogden, who expressed electric field in volts. You should have read better, before saying that his reply was more to the point.
But, anyway, ok.
I know you specified the emf, but I am asking you about the electric field in the copper.
Sorry I used 'voltage drop', I corrected my self shortly after with 'difference', but you were already answering and quoted my previous sentence.
So, let me ask you again, because this is important:

What is the field inside the copper conductor and how do you justify - with formulas - that there is a (0.25-0.002) volts difference across it?

Assume standard conductivity for copper, say 5.8 10^7 mhos per meter and a copper section of 1 mm in diameter (or any real world value you can attribute to a circuit similar to those shown by Lewin, Mehdi or Mabilde - it's about 10 cm diameter loop, suppose half of it is allocated by the big resistors, but it's not important).

I am asking for the electric field E inside the conductor - to be more precise, the tangential component that contribute to the integral of E.dl .
I can tell you that in my case it would be in the mV/m range.
What value do you get in your case?
All instruments lie. Usually on the bench.
 


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