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Equivalent resistance

  1. Jul 28, 2012 #1
    1. The problem statement, all variables and given/known data
    http://i50.tinypic.com/w0ggl.png

    2. Relevant equations



    3. The attempt at a solution
    I am stuck at part (a) (i haven't yet tried the part (b)).
    I really have no clue on how should i go on finding the equivalent resistance, i don't see a way to reduce it to simpler parallel and series combinations. Any help is appreciated.
     
  2. jcsd
  3. Jul 28, 2012 #2

    SammyS

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    Try redrawing the circuits.

    Label each node and resistor.



    Use any symmetry you can find to your advantage.
     
    Last edited: Jul 28, 2012
  4. Jul 29, 2012 #3
    I see that i can apply delta star transformation, but i don't think it would be a good choice.
    I don't see any symmetry here, can i have some hints?
     
  5. Jul 29, 2012 #4

    I like Serena

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    Are you aware of Kirchhoff's circuit laws?
     
  6. Jul 29, 2012 #5
    Yes. :)
     
  7. Jul 29, 2012 #6

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    So... did you apply them (making use of any symmetries that are apparent)?
     
  8. Jul 29, 2012 #7
    I thought of doing so but i cannot find the appropriate path.
    And, still i don't see any symmetries. :(
     
  9. Jul 29, 2012 #8

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    That's not much of an effort... yet. ;)
     
  10. Jul 29, 2012 #9
    Effort? I don't have an idea on how should i start. :)
    I suppose i should post a pic of what i am thinking, the current in the resistors R1 and R2 should be the same as coming out of R3 and R4 due to symmetry, is that right?
     

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  11. Jul 29, 2012 #10

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    Through each resistor some unknown current Ik flows.
    At each node between resistors you have some unknown Vk.

    Some of those Ik might be the same due to symmetry.
    Some of those Vk might be the same due to symmetry.

    Set up a system of equations?
     
  12. Jul 29, 2012 #11
    I still don't get it. I can make the equation but i am not sure if i took the right use of symmetry. Can you review the attached pic? I have mentioned the current flowing through each resistor and took use of the symmetry of circuit, is it right?
     

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  13. Jul 29, 2012 #12

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    Your use of symmetry looks overall correct. :)

    However, in your leftmost I2-I3 you seem to have ignored the current through the bottom most resistor?
     
  14. Jul 29, 2012 #13
    Oops, sorry, i forgot about, i was focusing on the middle part. :P

    I think i will have to again label the currents, i haven't sent any current in the topmost resistor too.
     
  15. Jul 29, 2012 #14

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    Yeah well, if you designate the bottom current, the topmost one will be the same due to symmetry! :)
     
  16. Jul 29, 2012 #15
    Yes, but i will have to reassign the currents to the diagonally placed resistor.
     
  17. Jul 29, 2012 #16

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    Yep.
     
  18. Jul 29, 2012 #17
    Can you help me finding the right loop please? I have attached the figure with nodes marked.
    I attached the battery. I could find one loop useful, AYHIJEBA, from here i get an equation,
    V-xR-I2R=0, where x=I-I1-I2.
    V=IR+I1R
    Now which loop should i select? I am stuck here. :(

    EDIT:Sorry, i forgot to mention I in the pic, its the current sent by the battery.
     

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  19. Jul 29, 2012 #18

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    That looks correct! :)

    Why would you select a single loop?
    Pick any loop you like.

    For instance pick the one through the top.
    The one through the bottom will be the same, so you can leave that one out.

    Now keep picking loops until you have as much variables as you have equations (avoiding those that will duplicate existing equations).
     
  20. Jul 29, 2012 #19
    Thank you for all the help! :smile:
    I think i can do it now, this is the worst question i have ever done on finding the equivalent resistance. Is their any easier way to do this? Any trick?
     
  21. Jul 29, 2012 #20

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    Ah well, usually it starts looking as daunting.
    And then you simply set up all equations using as many symmetries as you can.
    And then it turns out you can solve it.
    Afterwards, you'll see a couple of symmetries or some such that you missed.
    But by then it doesn't matter anymore since you have the solution (assuming that you do).
     
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