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In Cube of resistances how to calculate current in one edge?

  1. Jul 18, 2013 #1
    1. The problem statement, all variables and given/known data

    Twelve resistors of equal resistance are connected so as to form a
    cube ABCDEFGH shown. Equivalent resistance between points A
    and G is 5/3 ohm. If a battery of voltage 100 V is connected across
    the points A and G, find the current (in ampere) flowing through the
    resistor CD.

    ewzas.png

    2. Relevant equations
    V=IR
    1/Rparallel = 1/R1 + 1/R2 + ...
    Rseries = R1 + R2 + ...

    3. The attempt at a solution
    V=IR
    So 100=i*5/3 ( between A and G )
    So current between A and G is i=60 .
    But what now ? How to calculate current between C and D ?
     
  2. jcsd
  3. Jul 18, 2013 #2

    ehild

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    Use the symmetry and use KCL. I=60A current flows into A. It has three equivalent ways to flow: Towards B, D, and E.

    The current flowing into D has two choices to flow further....


    ehild
     
  4. Jul 18, 2013 #3
    Whats KCL? Also by symmetry do you mean that some of the currents are equal or what ?
     
  5. Jul 18, 2013 #4

    ehild

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    KCL is Kirchhoff's Current Low. By symmetry, I mean that equal currents flow along equivalent paths.

    ehild
     
  6. Jul 18, 2013 #5

    CWatters

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    KCL = kirchhoff current law.

    I would draw the equivalent circuit first. Do you know how to calculate the equivalent resistance between A and G? if you do then it might be easier.
     
  7. Jul 18, 2013 #6
    Nope I don't know how to find equivalent resistance between A and G . Also by drawing equivalent circuit do you mean to convert this 3d cube into 2d circuit ? If so how to do that ?
     
  8. Jul 18, 2013 #7

    ehild

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    It is not needed. You are given the equivalent resistance. So you know the current flowing into the node A. How much current flows out of A along each resistor connected to A?

    ehild
     
  9. Jul 18, 2013 #8
    Follow Echild's advice. 60 amps flow out of node G and 60 amps flows into node A. By symmetry, the current flowing out of node G splits evenly between legs GH, GC, and GF. Also, by symmetry, the same 60 amp current converges evenly from BA, DA, and EA into node A. So, the current in all 6 of these legs is 20 amps. Now all you need to do is to use symmetry to figure out how the 20 amp current coming into node C from node G splits between legs CB and CD.

    Chet
     
  10. Jul 18, 2013 #9
    So the current splits evenly between CB and CD and so the current through CD is 10 amps . The answer is right but can you link me to a good online video/tutorial/guide/book which explains this "symmetry " thing ?
     
  11. Jul 19, 2013 #10

    ehild

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  12. Jul 19, 2013 #11

    CWatters

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  13. Jul 19, 2013 #12
    Thanks for the links guys !
     
  14. Jun 10, 2014 #13
    For a cube with 12 identical resistors on each edge, I remember the resistance across corners as (5/6)*R, not (5/3)*R.

    Claude
     
  15. Jun 10, 2014 #14

    ehild

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    Claude,

    The equivalent resistance was said 5/3 ohm, not 5/3 R.


    ehild
     
  16. Jun 10, 2014 #15
    Well, if the edge resistance is R, then the diagonal resistance is 5/6 of R. With 1.0 ohm resistors on each edge, the catecorner resistance should be 5/6 ohm, not 5/3 ohm. I'm pretty sure I remember it right.

    Take the 1st corner, draw a shorting plane through the vertices of the adjacent corners. Do the same at the diagonally opposite corner. What is the net resistance. If one corner is A, the other G, we compute Rag as follows. From A to plane 1 we have 3 edges in parallel for 1/3 ohm. From plane 1 to plane 2 we have 6 edges in parallel for 1/6 ohm. Then from plane 2 to G we have 3 edges in parallel for 1/3 ohm.

    Rag = (1/3) + (1/6) + (1/3) = 5/6 ohm.

    Claude
     
    Last edited: Jun 10, 2014
  17. Jun 10, 2014 #16
    Where does it say that the resistors are 1 Ohm?
     
  18. Jun 10, 2014 #17
    The OP stated that all 12 edges were equal in resistance. Call this value "R". I stated that the value of Rag = (5/6)*R. Then I was told that it was 5/3 ohm, which is incorrect. If each edge has R= 1.0 ohm, then Rag = (5/6) ohm.

    You're right, 1.0 ohm is not stated in the OP. It is merely 12 edges of equal value "R". So the diagonal resistance Rag would be (5/6)*R. Is this clear?

    Claude
     
  19. Jun 10, 2014 #18
    Clear, bot not relevant for this problem.
    The problem states that the resistance is 5/3 ohm. So you don't need the general formula.

    An interesting results to remember, though.
     
  20. Jun 11, 2014 #19
    Yes, but why remember it when it's so easy to calculate it?
     
  21. Jun 11, 2014 #20
    So am I correct in understanding that 12 edges have identical R, with the diagonal value Rag being 5/3 ohm? Is that right?

    Just for the record, in that scenario, the R value for each edge would be 2.00 ohms. Then 5/6 of 2.00, gives 10/6 ohm, or equivalently 5/3 ohm.

    Claude
     
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