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Find the equivalent resistance between AB

  1. Jan 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the equivalent resistance between AB.

    Capture.PNG

    2. Relevant equations


    3. The attempt at a solution
    I could not find any Wheatstone bridge or series parallel combination. Should I use star delta transformation ? If yes, how exactly should I use it ?
     
    Last edited by a moderator: Jan 27, 2016
  2. jcsd
  3. Jan 23, 2016 #2

    berkeman

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    Yoiks! There might be a simpler way, but I'm not seeing it right off the bat.

    Instead, I'd just put a 1V voltage source between points A & B, and write all the KCL equations for the nodes in the circuit. It will be a lot of equations, but just solve them simultaneously, and then solve for the overall current flowing in the 1V source. That will get you to the equivalent resistance.

    BTW, are all the resistors of the same value?
     
    Last edited by a moderator: May 7, 2017
  4. Jan 23, 2016 #3

    SammyS

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    There is some symmetry to this circuit. In the configuration that it's present, you can see a little of that symmetry.

    capture-png.94692.png
    I suggest labeling the resistors (or the nodes if you prefer).

    Redraw (and redraw and redraw...) emphasizing triangles formed by the resistors, until you get a very symmetric looking array.

    Make maximum use of the symmetry

    In the end, I found it handy to split that center resistor into two resistors, R/2, in series.

    You can perform many Y-Δ transformations combining a few resistors each time. (After drawing many Δs I finally saw that the original circuit has lots of symmetry.)
     
  5. Jan 23, 2016 #4
    Yes, the resistors are of same value.
     
  6. Jan 23, 2016 #5
    How to use symmetry to simplify circuit ?
     
  7. Jan 23, 2016 #6

    SammyS

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    I made some suggestions. Have you tried them?
     
  8. Jan 28, 2016 #7

    NascentOxygen

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    Solving by symmetry is a valuable technique to master. You may not be taught it, and just have to pick it up by practice. There are probably many ways to solve your particular problem, I'll start you off with the one I used.

    You can see symmetry here by noting how a rotation of 180 degrees around the red central dot causes each element and each wire to superimpose on one identical to itself in an identical location, as though each has a twin. This symmetry means that each element carries an identical current and voltage as its twin.

    To help you I have identified one set of twins by colouring them green. For convenience, I'll assume VB is at zero volts.

    resistors2.jpg

    Suppose there is a voltage drop of V1 across each green resistor, then symmetry lets us label the voltage at two nodes, V1 and VA-V1. I also did the same for nodes around another set of twins I coloured purple.

    That's the hard part finished with! You will notice how we now have every node labelled with its voltage.

    Now all that's needed are two equations in two unknowns to eliminate V1 and V2. See whether you can apply Kirchoffs Current Law to do this, and finally obtain the circuit's resistance. (= VA / current).

    Does your textbook give the answer....so that I can check my working?
     
  9. Jan 28, 2016 #8

    cnh1995

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    I ran a simulation and got Req=0.714R Screenshot_2016-01-28-19-36-16.png
     

    Attached Files:

    Last edited: Jan 28, 2016
  10. Jan 29, 2016 #9

    NascentOxygen

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    That's my answer, too. Thanks.
     
  11. Jan 29, 2016 #10

    SammyS

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    Using symmetry, lots of symmetry, I get (5/7)R .
     
  12. Jan 29, 2016 #11

    cnh1995

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    That's same as 0.714R. Cool! I'll try working it out by symmetry.:smile:
     
  13. Jan 29, 2016 #12

    SammyS

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    (So you are pursuing this thread. Good.)


    Here is a start.
    upload_2016-1-29_21-11-35.png
    Resistors are numbered. The six nodes are circled and marked A through F, using the original A & B .

    I suggest placing nodes A, E, F, and B roughly spaced equally in a line. Place nodes C & D on opposite sides of that line on its perpendicular bisector.
    Connect the nodes with the resistors.

    Code (Text):

           D

    A   E     F    B

           C
     
     
  14. Feb 4, 2016 #13

    SammyS

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    Here is the circuit element drawn showing more of the symmetry.

    upload_2016-2-4_18-17-22.png

    Resistors and nodes as in the previous post.
     
  15. Feb 5, 2016 #14

    NascentOxygen

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    Using a bit of symmetry, and a Y-∆ transformation followed by a ∆-Y transformation, I again find resistance to be 5/7 Ω. (Though I wouldn't expect anyone to memorize these transforms for unbalanced networks.)
     
  16. Feb 5, 2016 #15

    SammyS

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    From Post #3 ("that center resistor" being the #6 resistor.):
    Call the node between these two resistors Node G.

    Then nodes C, D, & G can be collapsed into one node. The whole mess can be dealt with using only series/parallel analysis.
     
  17. Feb 7, 2016 #16

    NascentOxygen

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    Now I can see how you did it. Yes, that would be the easiest. It's always good to have two methods, so you can confirm the answer to the first.
     
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