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Confusion:collapsing 3 points with equal potential in a cubical resistor network

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

    The question is to find Equivalent resistance between opposite ends of a cubical resistor network . each resistor of resistance R .

    I am referring to this website here
    I am halfway through it but I am stuck at a point . Where it says
    "Now imagine shrinking these wire loops down to a single point. This
    collapses together the points BDE and CFH. The circuit now looks like
    this (when flattened out):"

    --R-- /--R--\ -R-
    / \ /---R---\ / \
    \ / \---R---/ \ /
    --R-- \--R--/ -R-

    2. Relevant equations

    Only equation I believe is used will be V= IR and Kirchoff's law maybe I think it will use only the formulas for connection of resistors in parallel and series

    3. The attempt at a solution

    I have tried using Kirchoff's law that is enetring a current I from A then distributing it , it gives the right answer but its quiet very lengthy . Just stuck there . if some one can help it would be awesome
  2. jcsd
  3. Jun 1, 2012 #2
    http://img69.imageshack.us/img69/2443/p1010017gh.jpg [Broken]
    Last edited by a moderator: May 6, 2017
  4. Jun 2, 2012 #3


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    If points in an electric circuit are connected with wires (of zero resistance) these wires can be replaced by a single node. Points of equal potential always can be connected, it does not alter anything in the circuit.


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  5. Jun 2, 2012 #4
    vsupply=Itotal . Rtotal (1)

    vsupply=I1/3 . R1/3 (2)
    vsupply=I(1/3R + 1/6R + 1/3R) . (G->H ->D->A)

    Subt. in (1)
    I(5/6R)=Itotal . Rtotal
  6. Jun 2, 2012 #5
    Sorry but I am still not getting how to collapse the three points together . Sorry for bring dumb but please help me out.
  7. Jun 2, 2012 #6
    Say current I enters point G.
    Then equally it is distributed to 3 branches.
    Makes each branch current 1/3I(pt G)
    Next points, incoming 1/3I, branchout to 2 branches which means each outlet=(1/3)/2I=1/6I.(pt H)
    finally 2 branches join and equal I(1/6+1/6)=I(1/3).(pt. E)
    All above from Kirchoff's Current Law(KCL)

    Next Kirchoff's Voltage Law(KVL)

    If you follow any route(from G to A,from higher potential to lower potential), the total voltage equal to supply voltage.
    Last edited: Jun 2, 2012
  8. Jun 2, 2012 #7
    I know how to do it by applying Kirchoff's law but I am looking forward to solve it by the method of connecting points of equal potentials . I understand that the points with equal potentials can be connected without affecting the circuit but I just wanted a more clear image of how this has been done . I mean how did they connect the three points , did they make all the point one single point ? :uhh:
  9. Jun 2, 2012 #8


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    See the figure in #3. They make it a single point at the end.

  10. Jun 2, 2012 #9
    @echild Thanks but can you please show me the picture of how the three points are made one in the cubical network . Thanks
  11. Jun 2, 2012 #10
    Also in the same cubical network If I were to finto resistance between points B and D , why will the current bit flow through the branches CG AND AE
  12. Jun 2, 2012 #11


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    You can distort the square, nothing happens when you do not change the connection between the nodes. Just press the dots representing the corners B,D,E together. The wires which were the edges, will bend but their resistance does not change.


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  13. Jun 2, 2012 #12
    :surprised alright now I know how they can be connected thank you sooo much ! but how are the remaining 6 resistors in parallel ?
    Also can I solve for equivalent resistance between any other two points using the same approach?
  14. Jun 2, 2012 #13
    Re: Confusion:collapsint 3 points with equal potential in a cubical resistor network

    1st Point - G
    2nd Point all resistor with voltage drop of RI(1/3) =3 resistor in parallel
    3rd Point alll resistor with voltage drop of RI(1/6+1/3) = 6 resistors in parallel
    4th point all resistor with voltage drop of RI(1/6+1/3+1/3) =3 resistor in parallel
    Point A

    1st total resistance. =R(1/3)
    2nd. total resistance. =R(1/6)
    3rd. total resistance. =R(1/3)

  15. Jun 2, 2012 #14
    Re: Confusion:collapsint 3 points with equal potential in a cubical resistor network

    Ok alright :biggrin: Thanks . One thing thats confusing me is , as in this cube the cell was connected between A and G and then the 3 points with the same potential are easily identified . But in case we connect the cell between points A and C then why not B D and E are at the same potential ?
    /| /|
    D-----C |
    | E---|-F
    |/ |/
  16. Jun 2, 2012 #15


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    Check the symmetry, with the cell included. Now the configuration has got a vertical mirror planedefined by A and C. B and D are equivalent points with respect to it, but E is not. In the first configuration, it was a threefold axis between A and G.
  17. Jun 2, 2012 #16
    Thanks for the reply guys , I have no idea about symmetry and how to find points with equal potentials . I am cracking my head since hours , please if you will solve my dilemma I will be very very grateful to you , Please help me
    Please answer this post
    Thank you so much for all the help !
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