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Guru Vishnu
Guru Vishnu
05:45
Can anyone tell what is wrong with the following method of computing the effective capacitance between two points (marked with red circles)? :
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I got the effective capacitance as 2C when I considered the system to be a combination of balanced Wheatstone network and a capacitor connected in parallel with the terminals. And it's the correct answer as per the book.
 
3 hours later…
John Rennie
John Rennie
09:01
@GuruVishnu I can't see why this is valid?
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If you look at the voltages either side of the vertical capacitor:
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The symmetry suggests they will be equal $V_a = V_b$ so that vertical capacitor can be removed without affecting the circuit.
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Then what's left adds up to $2C$.
But I don't see how your second circuit is equivalent to the first
Guru Vishnu
Guru Vishnu
I understood your method. Let me explain why I think the second is equivalent to the first so that it would be easier to spot the error.
John Rennie
John Rennie
Ok ...
Guru Vishnu
Guru Vishnu
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Here A, B and C are of same potential.
So it's not an issue to merge these points.
So the first two resistors from the left on the top are in parallel. Ok so far, sir?
John Rennie
John Rennie
I don't think B is a junction. I think it's just an artefact of the drawing.
The capacitors are on the edges of a tetrahedron.
Guru Vishnu
Guru Vishnu
Hm. Let me think about it. However it wasn't mentioned that this is a tetrahedron.
If it were not a junction, I think the best way to show that is a $\Omega$ symbol.
John Rennie
John Rennie
09:16
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Like that
Guru Vishnu
Guru Vishnu
Ok sir.
But the other method I used (based on Wheatstone network) assumed this to be a junction rather than a tetrahedron. In a later step, I disassembled the junction to give a balanced network and a capacitance C in parallel across the red circle terminals.
So, I think both tetrahedron and junction are equivalent.
To avoid any other confusion, this is a direct quote of the question:
> If the capacitance of each capacitor is C, then the effective capacitance of the shown network across any two junctions is
> (a) 2C
> (b) C
> (c) C/2
> (d) 5C
(Correct answer: 2C)
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John Rennie
John Rennie
Any two junctions strongly suggests it is a tetrahedron since for the tetrahedron the symmetry makes all corners identical.
Guru Vishnu
Guru Vishnu
Ok sir. Could you tell whether the following assertion is true in capacitor circuit under electrostatic condition? :
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This is why I think both tetrahedron and a planar junction are equivalent.
But this is not valid the other way round i.e., we cant call a bridge a junction.
John Rennie
John Rennie
In general a crossover would be drawn as you have drawn it, or with gaps as I drew it. If neither of these are present it would be a junction not a crossover.
But I am 99.9% sure the question is just using a badly drawn diagram and it does mean a crossover not a juction.
That's based on seeing many such questions over the years.
Guru Vishnu
Guru Vishnu
09:31
Ok sir. I agree.
That must have been an $\implies$ symbol rather than $=$.
So can you please tell whether the following is true? :
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I think yes.
Because that was how I converted the planar network to the tetrahedral one and came to the correct answer.
John Rennie
John Rennie
In general intersecting lines means a junction not a bridge.
So strictly speaking in your diagram B is a junction i.e. the wires join together there.
But I'm sure this is just a badly drawn circuit and it is not meant to be a junction.
Guru Vishnu
Guru Vishnu
@JohnRennie Yes sir. I understand. I'm just asking whether it's ok to split the two lines which intersect in electrostatics. If yes then I think the method I showed you in the beginning has some serious fault which I don't know what it is :-(
John Rennie
John Rennie
@GuruVishnu no it is not.
Guru Vishnu
Guru Vishnu
@JohnRennie Ok sir. Let's shift to electrodynamics for a bit. Do you agree with this? :
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In this first junction, the current entering the junction is equal to the current exiting it as per Kirchhoff's law.
We're just converting the junction to a bridge network.
I did the above manipulation for a lot of problems in Current electricity.
And it worked really well!
John Rennie
John Rennie
09:48
Ah, OK, I had misunderstood what you were asking.
I guess the two are equivalent if you know the currents are as you have marked.
Guru Vishnu
Guru Vishnu
Yes sir. But the reverse is not true, as we'll not be sure whether the potential of the horizontal branch is same as the potential of the vertical branch.
Ok switching to electrostatics...
17 mins ago, by Guru Vishnu
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Now, do you agree with this? :
In electrostatics these rules are slightly relaxed. We're free to split the junction in any way we wish.
John Rennie
John Rennie
If you know the potentials are identical in the two wires then you can join them because if the potentials are identical no current will flow between the wires. Whether this would be a useful thing to do is another question.
Guru Vishnu
Guru Vishnu
Ok sir. That's why the first row is correct and the second row is not always correct and that's why the $\implies$ is cancelled. Now we're only interested in the first row.
And that's how I converted the planar network:
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to a balanced Wheatstone network:
John Rennie
John Rennie
But the potentials in the two wires are not the same at B. If you connected them there a currnt would flow between them.
Guru Vishnu
Guru Vishnu
I understand sir. I was just mentioning this for the validity of converting the planar network in the bad question to the tetrahedral network suggested by you not the other way round.
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or without words, this is what I'm trying to say:
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Moral of the story: Whether it's a planar network or (converted to) a tetrahedral network, the result must remain the same.
@JohnRennie: Did you understand this idea sir?
John Rennie
John Rennie
10:06
If the point B really is a junction then you don't get total capacitance of 2C. So if you change the junction to a bridge the total capacitance changes.
Guru Vishnu
Guru Vishnu
Could you tell why must it change? It seems similar to the electric current analogy.
John Rennie
John Rennie
You did the calculation with B as a junction, and got a total capacitance of 8C/3. So the result is different.
Guru Vishnu
Guru Vishnu
Ok sir. Now, I understand there's not a mistake in the first method. But there's some error in this:
9 mins ago, by Guru Vishnu
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I don't see why this above must be invalid.
Could you please tell the reason sir?
John Rennie
John Rennie
Because you can only convert the junction to a bridge if the potential of the two wires is equal. And in this case the potentials are not equal.
Guru Vishnu
Guru Vishnu
Ah. Ok sir. So in this question, I problem with my first method was assuming the tetrahedral network as a planar network. Or simple I was doing what I was not supposed to do i.e., the second cancelled $\implies$ statement. Did I understand this sir?
John Rennie
John Rennie
10:19
Yes, the two errors cancelled each other out and you got the right answer anyway :-)
Guru Vishnu
Guru Vishnu
Ok sir. Now understood. Thank you for your time and help :-)
John Rennie
John Rennie
:-)
Guru Vishnu
Guru Vishnu
If you wish, could you please transfer only our conversation from this room to Particle accelerator for de-referencing? But I guess it'll be quite tedious as our conversation is too large to click one by one.

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