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Homework Help: Capacitance on cubic circuit

  1. Nov 28, 2012 #1
    1. The problem statement, all variables and given/known data

    A potential electrostatic difference is stablished between points M and Q in the cubic circuit of identical capacitors shown in the image. What is the potential difference between points N and P?

    2. Relevant equations



    3. The attempt at a solution

    I tried opening up the cube to make it look like something that I am used to, but it still looked confusing.
     

    Attached Files:

  2. jcsd
  3. Nov 28, 2012 #2

    mfb

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    There are two unnamed points equivalent to N, and two unnamed points equivalent to P. You can add (imaginary) connections between them, this allows to draw a simple, equivalent planar diagram.
     
  4. Nov 28, 2012 #3
    Would it be correct to assume that whatever current I have flowing into M it gets split up in three? And then in two at the next node? Reuniting at the end to deliver the same current to node Q?

    I tried to draw an equivalent diagram, but I can't get it to look right.. I think there's something I am missing.
     
  5. Nov 28, 2012 #4

    mfb

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    Right. And you can draw this as planar graph as well.
     
  6. Nov 28, 2012 #5
    I struggled a lot with it.. Than I came across this article that explained a similar problem but with resistors!

    In the article, it said that since the first three nodes from the source of potential difference have the same current going though them. I can assume they are somehow equivalent points. And by using the same assumptions with the remaining nodes you can arrive at a schematic where you have two sets of three capacitors in parallel in series with six capacitors in parallel. Like M--(3 parallel)---(6 parallel)---(3 parallel)--Q.

    The thing is, I don't get how you could arrive at this. How can I just say that those three first nodes could be represented by one?

    Also trying to flatten the cube will get you nowhere. I tried all possibilities.. Either you have to understand the assumption above or I am really missing something. Because I just couldn't do it..
     
  7. Nov 28, 2012 #6

    ehild

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    The problem of cube of resistors was discussed in Physicsforums several times. Try to follow this thread: https://www.physicsforums.com/showthread.php?t=557461

    The method of solution is based on symmetry. (See the thread from Post #6) Equivalent points are at the same potential. Points on same potential can be connected with a wire, it changes nothing in the electric circuit. The points connected with a wire of zero resistance represent a single node.

    ehild
     
  8. Nov 29, 2012 #7
    Thanks! I was also wondering something else: is the voltage across the whole cube the same? If I measured the electric potential difference between any two points will they all give back the same value? Because I was thinking; if I measured the voltage from point M to point N and from point N to point P, they would differ. Because I have the same resistance but half the current, and by Ohm's law V = IR, the voltage of between M and N would be twice the voltage between N and P. Is that correct?
     
  9. Nov 29, 2012 #8

    ehild

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    I drew the equivalent points by coloured dots. The voltage is the same between identical pairs: between the black dot and the red ones, (that is between M and N , M and K, M and L). It is the same also between the blue ones and the orange one: PQ, RQ, SQ. But it is different for the other pairs.
    The voltage between M and N is not the same as the voltage between N and P. It is a bit easier to follow the potential of the nods. M is at zero potential and Q is at V potential. The potential increases with the same amount from M to N, from M to K and from M to L, and the same is the increase from P to Q, from R to Q and from S to Q. But you get different change of potential from N to P, N -->R, L-->S, L-->P, K-->R, K-->S.

    ehild
     

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  10. Nov 29, 2012 #9
    Okay, but I am still confused as to why the potential difference is different from:
    N --> P, N -->R, L-->S, L-->P, K-->R, K-->S.

    If I had three amperes flowing into M. Would't that current split into three currents of one ampere each flowing towards N, L and K? Then at those nodes they would split yet again into two currents of 1/2 ampere? So that you would have flowing into P 1/2 A from N and 1/2 A from L? And into S; 1/2 A from K and 1/2 A from L?
     
  11. Nov 29, 2012 #10

    CWatters

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    I think you missunderstood what he said. He meant...

    Vblack to Vred = Vblue to orange <> Vred to Vblue

    There are three caps between black and red and between blue and orange but six caps between red and blue.
     
  12. Nov 29, 2012 #11

    ehild

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    These are all equal, but different from VMN (half of it).

    ehild
     
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