1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    2016 Award

    Staff: Mentor

    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

    User Avatar
    2016 Award

    Staff: Mentor

    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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

    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
     

    Attached Files:

  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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Homework Helper
    Gold Member

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

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Capacitance on cubic circuit
  1. Capacitance in circuit (Replies: 15)

  2. Capacitive Circuit (Replies: 1)

  3. Capacitance Circuit (Replies: 5)

Loading...