Author Topic: Does Kirchhoff's Law Hold? Disagreeing with a Master  (Read 16717 times)

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Offline In Vacuo Veritas

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #400 on: December 07, 2018, 02:21:47 am »
Does Ohm's Law still work? I've got this LED I have to turn on and I need to know which side of the LED to put the resistor...
 
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Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #401 on: December 07, 2018, 02:41:52 am »
Does Ohm's Law still work? I've got this LED I have to turn on and I need to know which side of the LED to put the resistor...

LOL. You can rest assured - complex laws do not invalidate basic laws of physics and nature easily.

Thou in case you want detailed answer - you better start new thread in "n00bs" section ;)
 

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #402 on: December 07, 2018, 03:42:32 am »
The field inside the copper conductor is the sum of E.coloumb with E.induced, you said (and I agree). How do you think the copper can tell which is which?
What are you smoking?

ogden, I promised not to interact with you, but you are making this promise very hard to keep. If you interfere with my exchanges with other posters I have to reply to you as well.

Quote from: ogden
Integral of E.dl where E = E.coloumb + E.induced. EMF of wire segment is EMF.total/4 (because segment is 1/4 of loop) = 1/4V and voltage drop due to current is 0.001Ohm*1A = 0.001V. So, this sum is 0.25+(-0.001) Volts. What's the point to ask question so many times?

Because you guys keep telling me the voltage. I want to know the electric field.

Quote from: Sredni
Please, indulge me. Give me the number in V/m (volts per meter).
Quote from: ogden
:-// With same success you can ask me weight of the wire used in experiment. Before asking V/m, make sure you give enough data to calculate such  :palm:

I gave all the data, and also allowed for some freedom in choosing the parameters. Go back and read what I wrote in Reply #379, yesterday at 12:52:37 pm

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?

And still, no answer.


Edit: typo, added slant and bold.
« Last Edit: December 07, 2018, 03:57:04 am by Sredni »
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Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #403 on: December 07, 2018, 04:05:12 am »
Quote from: ogden
Integral of E.dl where E = E.coloumb + E.induced. EMF of wire segment is EMF.total/4 (because segment is 1/4 of loop) = 1/4V and voltage drop due to current is 0.001Ohm*1A = 0.001V. So, this sum is 0.25+(-0.001) Volts. What's the point to ask question so many times?
Because you guys keep telling me the voltage. I want to know the electric field.

I said (E = E.coloumb + E.induced). Are you satisfied now?

Quote
Quote from: Sredni
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?

And still, no answer.

You can either provide solution yourself and tell what you want to say with it or stick that tangential component where it hurts. I do not see the point of solving your tasks. "Trail of the troll" was at least funny.
 

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #404 on: December 07, 2018, 04:16:58 am »
Because you guys keep telling me the voltage. I want to know the electric field.
I said (E = E.coloumb + E.induced). Are you satisfied now?

No, I want to know the value in V/m (or in J/C if you prefer).

Quote
You can either provide solution yourself and tell what you want to say with it or stick that tangential component where it hurts. I do not see the point of solving your tasks. "Trail of the troll" was at least funny.

Ok, you have no idea on how to compute the electric field inside a conductor. It's not a crime. Maybe all that facepalming has interfered with your mental processes but, fine.
Any other Kirchhoffian who believes that the 'real' voltage across the 0.9 ohm resistor is 0.65 V and the real voltage across one of the two arcs of copper is 0.25-0.001 V care to tell us what the electric field is inside said copper?
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Offline SiliconWizard

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #405 on: December 07, 2018, 04:32:03 am »
Does Ohm's Law still work? I've got this LED I have to turn on and I need to know which side of the LED to put the resistor...

This question is actually a lot more relevant than it appears here (due to its ironic nature), as the basic Ohm's law links voltage, resistance and current. Now what is voltage again? ;D
Incidentally, Kirchhoff (not him again!) reformulated Ohm's law as: J = sigma.E
So, may be on to something.
 

Offline In Vacuo Veritas

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #406 on: December 07, 2018, 04:45:49 am »
Does Ohm's Law still work? I've got this LED I have to turn on and I need to know which side of the LED to put the resistor...

This question is actually a lot more relevant than it appears here (due to its ironic nature),

Thanks, at least someone appreciates my work here.
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #407 on: December 07, 2018, 04:58:07 am »
Ok, you have no idea on how to compute the electric field inside a conductor. It's not a crime. Maybe all that facepalming has interfered with your mental processes but, fine.
Any other Kirchhoffian who believes that the 'real' voltage across the 0.9 ohm resistor is 0.65 V and the real voltage across one of the two arcs of copper is 0.25-0.001 V care to tell us what the electric field is inside said copper?

Yes, I have to dig into it to solve it. So what. Original Dr.Lewins experiment assumed that conductors have no resistance, so no coloumb E-field. Now you are modifying it to prove what exactly?- That your debate opponents cannot calculate something during time they are willing to spend, so this is proof that you are right? BTW this is typical tactic of internet trolls - derail discussion into personal attacks.

Better tell your E-field number and make your point. Educate Kirchoff believers, don't let them compute what you can do in a snap.
 

Online Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #408 on: December 07, 2018, 05:03:46 am »
Does Ohm's Law still work? I've got this LED I have to turn on and I need to know which side of the LED to put the resistor...

This question is actually a lot more relevant than it appears here (due to its ironic nature), as the basic Ohm's law links voltage, resistance and current. Now what is voltage again? ;D
Incidentally, Kirchhoff (not him again!) reformulated Ohm's law as: J = sigma.E
So, may be on to something.

Or maybe its just using metric volts instead of imperial volts. We already crashed a probe into mars because of this shit. ;D
(You do have a valid point there tho)
 

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #409 on: December 07, 2018, 05:33:36 am »
Ok, you have no idea on how to compute the electric field inside a conductor. It's not a crime. Maybe all that facepalming has interfered with your mental processes but, fine.
Any other Kirchhoffian who believes that the 'real' voltage across the 0.9 ohm resistor is 0.65 V and the real voltage across one of the two arcs of copper is 0.25-0.001 V care to tell us what the electric field is inside said copper?

Yes, I have to dig into it to solve it. So what. Original Dr.Lewins experiment assumed that conductors have no resistance, so no coloumb E-field.


No, so no resultant E field at all.

Quote
Now you are modifying it to prove what exactly?- That your debate opponents cannot calculate something during time they are willing to spend, so this is proof that you are right? BTW this is typical tactic of internet trolls - derail discussion into personal attacks.

Coming from the person who constantly facepalms, calls other posters trolls and asks what they smoke, this is hilarious.
I've been extremely restrained with you, but you are like one of those small dogs that keep jumping and barking when their owners are trying to have a conversation.

Quote
Better tell your E-field number and make your point. Educate Kirchoff believers, don't let them compute what you can do in a snap.

I already told you my number. It's in the mV/m range. With 1 amp in a 1mm diameter copper wire it's about 30 mV/m; with 10 mA as in the original Lewin experiment is about 30 uV/m. And it is perfectly consistent with the constitutive equation j = sigma E.

Also, I already told what my aim is in one of my previous post: to show that since there could not be a significant electric field inside the copper (it is zero in a perfect conductor) it is nonsense thinking that you can still have an induced field capable of producing a 0.25V voltage at the extremes (much more in the case of Lewin's experiment, since there the resistors were much smaller).
The induced E field inside the conductor is compensated by the field caused by redistribution of charge. All the resultant electric field is located in the resistor region, with nothing left in the conductor.

This is the point.

Now, dig in and try to compute that field. Then try to explain why you have to give up on j = sigma E as well.

Edit: added "resultant electric", and some plurals I had missed
« Last Edit: December 07, 2018, 05:42:43 am by Sredni »
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Offline electrodacus

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #410 on: December 07, 2018, 05:42:08 am »
Ok, you have no idea on how to compute the electric field inside a conductor. It's not a crime. Maybe all that facepalming has interfered with your mental processes but, fine.
Any other Kirchhoffian who believes that the 'real' voltage across the 0.9 ohm resistor is 0.65 V and the real voltage across one of the two arcs of copper is 0.25-0.001 V care to tell us what the electric field is inside said copper?

Have you read my replay with the servo motor or transformer experiment.  Please do that experiment as there your measurement device will be outside of the changing magnetic field and so you will get the correct result.
Even just reading that proposed experiment you should know that those will be the results that you will get and that they use the same logic that resulted in 0.65V on the 0.9Ohm resistor and 0.250-0.001V on the two 90 degree copper arcs situated in an uniform magnetic flux.
A piece of wire is a low value resistor so you introducing that in my simplified example did not helped with anything other than complicating the example.
Maybe is worth noting that if you shield the 0.9Ohm resistor the EMF at the same point in time will drop from 1V to 0.75V and since it is uniform it will be distributed equally on the other 3 quarters of the circuit. I leave it to you to calculate what will be the voltage on the 0.9Ohm in this case.
Don't care what you use to solve this as long as the result is correct.
« Last Edit: December 07, 2018, 05:46:03 am by electrodacus »
 

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #411 on: December 07, 2018, 05:58:14 am »
Have you read my replay with the servo motor or transformer experiment.

Sorry, I am not allowing mutatio controversiae. Let's stick to the ring and keep it as simple as possible, because heaven knows what excuses you people could come up with with slightly more complex systems.

Quote
Please do that experiment as there your measurement device will be outside of the changing magnetic field and so you will get the correct result.

So, when the voltmeter is outside the loop in the two resistor experiment, and measures 0.9V across the 0.9ohm resistors, that is the correct result?
Outside the loop there is no changing magnetic field.
Your rule does not apply here?

And, just to be clear, I know that if you measure the voltage on the ring following radial path as Mabilde did, the  voltmeter will read the value you say. What I am trying to tell yuo is that such value is perfectly compatible with the application of Faraday's law and does not require for any resultant E-field inside the conductor.
On the other hand, you need that field to justify 0.25 volts across 7-8 cm of copper wire.

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Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #412 on: December 07, 2018, 07:01:27 am »
Also, I already told what my aim is in one of my previous post: to show that since there could not be a significant electric field inside the copper (it is zero in a perfect conductor) it is nonsense thinking that you can still have an induced field capable of producing a 0.25V voltage at the extremes

Wait... What you just said? - That in ideal conductor can't be EMF (induced field)? I am speechless to be honest. We are back to square one where you stop posting and go watch videos of Dr.Lewin. He is brilliant teacher BTW.

Quote
The induced E field inside the conductor is compensated by the field caused by redistribution of charge.

This is exactly what I was telling multiple times already, E = E.coloumb + E.induced.
 

Online Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #413 on: December 07, 2018, 07:18:36 am »
Also, I already told what my aim is in one of my previous post: to show that since there could not be a significant electric field inside the copper (it is zero in a perfect conductor) it is nonsense thinking that you can still have an induced field capable of producing a 0.25V voltage at the extremes

Wait... What you just said? - That in ideal conductor can't be EMF (induced field)? I am speechless to be honest. We are back to square one where you stop posting and go watch videos of Dr.Lewin. He is brilliant teacher BTW.

Quote
The induced E field inside the conductor is compensated by the field caused by redistribution of charge.

This is exactly what I was telling multiple times already, E = E.coloumb + E.induced.

Yes i noticed the two being thrown into the same basket and considered as one thing all too often in this thread. The real electric field caused by charge separated electrons is a different thing that the apparent electric field that the electrons feel due to the magnetic interaction with them. The two have very different underlying mechanisms behind them.

Its especially important because voltmeters can only see the first kind of "electron pusher".

Oh and user electrodacus seams to be using the electron charge density kind of voltage everywhere. His claims are correct when you consider that. Tho i think he should look into what the other magnetic EMF part of the voltage looks like to see the whole picture in this thread.
 

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #414 on: December 07, 2018, 08:30:03 am »
Also, I already told what my aim is in one of my previous post: to show that since there could not be a significant electric field inside the copper (it is zero in a perfect conductor) it is nonsense thinking that you can still have an induced field capable of producing a 0.25V voltage at the extremes

Wait... What you just said? - That in ideal conductor can't be EMF (induced field)?

No, I said that in a perfect conductor there cannot be a non-zero resultant E field. While in copper you get a small field compatible with j = sigma E.
And as you keep repeating:

Quote
E = E.coloumb + E.induced

So, what it E (and, if you will, what are E.coloumb and E.induced, but I content myself with the resulting E field.

Quote
I am speechless to be honest.

Yes, I know. It makes that effect.

Quote
Quote from: Sredni
The induced E field inside the conductor is compensated by the field caused by redistribution of charge.
This is exactly what I was telling multiple times already, E = E.coloumb + E.induced.

And still you can't see.
What is E, then?

---

Quote from: Berni
Yes i noticed the two being thrown into the same basket and considered as one thing all too often in this thread.

Yes, and you should ask yourself why.

Quote
The real electric field caused by charge separated electrons is a different thing that the apparent electric field that the electrons feel due to the magnetic interaction with them. The two have very different underlying mechanisms behind them.

Do you think the copper can tell the difference?
Or will it just experience the superposition of both fields?
I gave up my hopes on ogden, but you might make it.

Here's a hint, from electrostatics.
The field produced from a point charge is radially directed and goes as 1/r^2.
Now put a piece of copper near it.
Will the field inside the piece of copper be still radially directed from the source?
Or maybe, the free charges in the conductor will distribuite themselves in such a way to compensate for that radial field, so that the resulting field will be zero inside the conductor?

Does it matter the underlying mechanism that produced the various contributions to the total field?
Where is it written that superposition of electric fields only works for... 'same mechanism origin' fields only?

Edit: fixed minor parts
« Last Edit: December 07, 2018, 08:33:16 am by Sredni »
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Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #415 on: December 07, 2018, 09:35:54 am »
Also, I already told what my aim is in one of my previous post: to show that since there could not be a significant electric field inside the copper (it is zero in a perfect conductor) it is nonsense thinking that you can still have an induced field capable of producing a 0.25V voltage at the extremes

Wait... What you just said? - That in ideal conductor can't be EMF (induced field)?

No, I said that in a perfect conductor there cannot be a non-zero resultant E field. While in copper you get a small field compatible with j = sigma E.

This is not what you said. You correctly say that E.coloumb in ideal conductor is zero, then you imply that it means that it is nonsense to have 0.25V induced field (EMF). Read your own words for god's sake: "since there could not be a significant electric field inside the copper (it is zero in a perfect conductor) it is nonsense thinking that you can still have an induced field capable of producing a 0.25V voltage at the extremes".

Quote
And still you can't see.

What I shall see? Enlighten me.

Quote
What is E, then?

E is sum of two fields, E.induced + E.coloumb - you can do the math and calculate (do integral over E.dl) potential difference at the ends of wire segment that is subject to both E-fields. This part is explained by Lewin himself BTW.
 

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #416 on: December 07, 2018, 09:56:30 am »
Also, I already told what my aim is in one of my previous post: to show that since there could not be a significant electric field inside the copper (it is zero in a perfect conductor) it is nonsense thinking that you can still have an induced field capable of producing a 0.25V voltage at the extremes

Wait... What you just said? - That in ideal conductor can't be EMF (induced field)?

No, I said that in a perfect conductor there cannot be a non-zero resultant E field. While in copper you get a small field compatible with j = sigma E.

This is not what you said. You correctly say that E.coloumb in ideal conductor is zero, then you imply that it means that it is nonsense to have 0.25V induced field (EMF). Read your own words for god's sake: "since there could not be a significant electric field inside the copper (it is zero in a perfect conductor) it is nonsense thinking that you can still have an induced field capable of producing a 0.25V voltage at the extremes".

When there is the primary coil only, say in vacuum, you can find the induced E field in all space. It is directed in circles and has a module that grows with the distance r from the center inside the coil, while it decreases as 1/r outside of it.
Ok, there you can see the field, in vacuum.

Now you put your copper loop with the two resistors.
Do you still think that inside the copper there will be the same induced field? No, charges will be displaced, producing a coloumbian field that will compensate this field, actually erasing it inside the conductor. All that remains is zero in a perfect conductor and that tiny little field compatible with j = sigma E in a real conductor.
Your induced field is no mas. Obliterated by the field produced by the displaced charges - that will accumulate at the resistors ends and in general wherever there are gradients in conductivity and permeability.

So I confirm my words: "no significant electric field inside the copper (it is zero in a perfect conductor)" and that if you think that you can still have your unaltered induced field inside the copper you are thinking nonsense.
Just like thinking that you can have a radially direct E-field inside a conductor placed nearby a point charge.

Quote
E is sum of two fields, E.induced + E.coloumb - you can do the math and calculate (do integral over E.dl) potential difference at the ends of wire segment that is subject to both E-fields. This part is explained by Lewin himself BTW.

So, after all this posts you still have to answer what is the value of the resulting E field in V/m in copper.


Edit: added the gradient eps and gradient sigma parts.
« Last Edit: December 07, 2018, 10:06:30 am by Sredni »
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Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #417 on: December 07, 2018, 10:40:45 am »
Do you still think that inside the copper there will be the same induced field?
if you think that you can still have your unaltered induced field inside the copper you are thinking nonsense.

I don't think so and never did. You just discovered EMF? We are long over this part, kid.

Quote
So, after all this posts you still have to answer what is the value of the resulting E field in V/m in copper.

If you want value of field *inside* copper segment, then this is all you get: (I*R)/length. Better just watch Dr.Lewin's famous video where he explains those questions
 

Offline electrodacus

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #418 on: December 07, 2018, 10:52:41 am »
So, after all this posts you still have to answer what is the value of the resulting E field in V/m in copper.

You always ask silly questions that have no relevance to the problem. Showing that you do not understand the mechanisms behind all this.

Original problem had this as data

A closed loop made of two different resistors connected with some pieces of wire (nit that relevant as wire is also a resistor with lower value).
EMF = 1V (I think this was the same in my example and in Lewin example).
Resistor values 0.9Ohm and 0.1Ohm in my example not sure but I think it was 1000x more in Lewin example (not relevant).

Noting else was given and there is no need for anything else to find out what the voltage on those two symmetrically opposed points will be.

So let me explain to you what a 1V EMF means as you do not seem to understand (likely my fault as my examples should have been very clear).

You can have a wire open circuit no matter what shape how long or what type of conductor as long as you mention that EMF is 1V then you will measure 1V across that wire (this is a snapshot in time as of course there will be a variable magnetic field that generates this). There is also the assumption that magnetic field is uniform across the loop meaning that you will read 0.5V if you measure two points at half the wire length between them anywhere on the wire.

Say you close this wire in a loop (again not relevant what shape the loop has and is the exact same moment in time thus EMF=1V) then there will be a current generated in this loop that will be result of EMF divided by loop resistance.
Keep in mind that this is the case because the EMF = 1V was given as input for the closed loop meaning that the magnetic field was strong enough to generate that 1V EMF no matter how much current went trough the closed loop.

So in my example with a total loop resistance of around 1Ohm current was 1A and I think Lewin experiment had 1mA but there is no difference for the explanation.

Now if the loop was all made of copper wire with 1Ohm resistance it seems you do not have a problem with the fact that 1A will travel trough this wire as EMF is defined as 1V (am I correct in assuming that).

Your problem seems to be understanding why if say a quarter of this loop has 0.999Ohm and the remaining 3 quarters have 0.001Ohm

a) If this new loop is open circuit you will agree that you will read 1V between ends and not only that but you will read 0.5V on any half part of the loop.
b) If I cut this loop in two open parts one quarter size 0.999Ohm it will read 0.25V and the other 3 quarter size 0.001Ohm will read 0.75V
c) Each of this two separately can be equivalent with a battery with the respective internal resistance

Now can you understand why is simple to calculate the voltage in any two points on this close loop made of basically two resistors (you can add 10 different resistors in the loop if you want).
And you do not need to know anything else to calculate the voltage between any two points.

All your unrelated questions are useless and just show you do not understand what EMF is.
Just hope this made it clear to you and others as I love sharing and acquiring knowledge.  If I'm not able to explain it so anyone can understand it means I do not understand it.

Online beanflying

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #419 on: December 07, 2018, 11:40:35 am »

By the choice of your words you sense that there is probably something wrong with your "probing technique". It's not sponsored by any electronics engineering fundamentals which pretty much describes tried and true experimental phenomena along the past two centuries up to this day. You only rely on a couple of 10 min or so videos on the internet without even questioning their content. Any serious trade like ours upon which the lives of people depend deserves a little more rigor.

There is no real rigor left in this thread what you have all descended into is tit for tat point making among 5 or 6 of you and then arguing who has the biggest probe er I mean best resolution to the point then disagreeing on that point and going around again in a circle!

Attacking the person for having a different opinion is not ever going to solve anything! Attack the technicalities just might get a resolution but in this threads case I doubt it very very much.

'Any serious trade like ours upon which the lives of people depend deserves a little more rigor.' Yep because projects like the Manhattan one 'saved lives' with Physics and Engineering. You have used this same statement a few times and while it can be true it is also a logical falsehood.
« Last Edit: December 07, 2018, 11:49:44 am by beanflying »
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Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #420 on: December 07, 2018, 11:52:56 am »
There is no real rigor left in this thread

What kind of rigor can you expect from people who wriggle like eels and refuse to compute the field in their circuit?
I have seen lots and lots of words to go around that simple question. And the reason is that they will end up with inconsistent results.
They believe they can have .25 V across a piece of copper 7cm long, 1mm diameter with nearly zero field inside. Or non-negligible field (much much higher than that allowed by the constitutive equation) inside a good conductor. What rigor can you expect?

This is why Lewin stopped answering questions about this matter. Flat-earthers always come up with new excuses, no matter what.
Now there is 'apparent electric field'.
All instruments lie. Usually on the bench.
 

Offline electrodacus

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    • electrodacus
Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #421 on: December 07, 2018, 12:06:19 pm »
There is no real rigor left in this thread

What kind of rigor can you expect from people who wriggle like eels and refuse to compute the field in their circuit?
I have seen lots and lots of words to go around that simple question. And the reason is that they will end up with inconsistent results.
They believe they can have .25 V across a piece of copper 7cm long, 1mm diameter with nearly zero field inside. Or non-negligible field (much much higher than that allowed by the constitutive equation) inside a good conductor. What rigor can you expect?

This is why Lewin stopped answering questions about this matter. Flat-earthers always come up with new excuses, no matter what.
Now there is 'apparent electric field'.

Have you even read my last replay?  Do you still do not understand the problem and the parameters given ?
Nothing else is needed to solve the problem and nothing else was specified in my examples or Lewin's
There was no mention anywhere about the loop size, diameter of the wires or magnetic field as they are not required to solve the problem and you can have an infinite combination of those to get the spec EMF but just the EMF was needed.
Please make sure you read that else it makes no sense for me to replay to you.
« Last Edit: December 07, 2018, 12:30:25 pm by electrodacus »
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #422 on: December 07, 2018, 12:13:03 pm »
Flat-earthers always come up with new excuses, no matter what.

When they are out of arguments in existing discussion, they invent new useless challenges - just like you.
 

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #423 on: December 07, 2018, 12:18:37 pm »
Electrodacus
I read it and it shows that you have a high school student mentality. You see it as an exercises with 'givens' from the prof. and you have to find a value that coincides with the prof. result.
But getting the right number does not mean you are getting the theory behind it. In fact, Mabilde is getting the same number I would expect based on Faraday, the nearly zero field inside the copper and no little generators dispersed along the wire.

What I am doing here is trying to point out inconsistencies in your (erroneous) view of the phenomena.
If you compute the darn field inside your copper conductor you will find that you either have to give up j = sigma E or you have to renounce having that 0.25V located there.
And I have to resort to field theory (EDIT: to do that, and that should not be a problem: my view is perfectly consistent with that, can you say the same for yours?).

I can show you exactly how the charge distribution is affected by gradients in conductivity and in permeability, but that would require vector calculus and if you do not understand as basic a concept as the superposition of fields, what hope is there that you will understand that?
Stop making up other examples where you can find the right number. Focus on the underlying theory and principles and show me you do not find inconsistencies. My view has no inconsistency whatsoever: the field inside the copper is perfectly abiding j = sigma E. Can you say the same?

Also, did you answer my question about the voltmeter measuring 0.9V from outside the loop? No, you didn't.
You keep making up new examples to avoid facing the inconsistencies of your view.
While I have always been focused on the two resistor loop of Romer-Lewin.
Try to do the same.
« Last Edit: December 07, 2018, 12:25:49 pm by Sredni »
All instruments lie. Usually on the bench.
 

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #424 on: December 07, 2018, 12:25:04 pm »
When they are out of arguments in existing discussion, they invent new useless challenges - just like you.

A challenge is what distinguishes a professional from a wannabe. The versed from the amateur. The authentic from the impostor.
« Last Edit: December 07, 2018, 12:31:42 pm by bsfeechannel »
 


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