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Equivalent capacitance odd circuit

  1. Nov 16, 2013 #1
    I'm having a time figuring out how to solve this one, hope someone can help me out

    1. The problem statement, all variables and given/known data
    "Calculate the equivalent capacitance between X and Y on the circuit in the figure. C1=C3=C4=C5=8uF, C2=20uF"


    2. Relevant equations



    3. The attempt at a solution
    I can't figure out a reduced circuit straight away. I tried to see what will be current at the ends of each capacitor and each of them as different ones so I can't consider any of them to be in parallel.
     

    Attached Files:

  2. jcsd
  3. Nov 16, 2013 #2

    gneill

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    Staff: Mentor

    Sometimes redrawing the circuit can help you to visualize a solution. Try moving terminal X to the top (centered over C2) and terminal Y to the bottom (centered below C2). Nodes 1 and 2 remain where they are.
     
  4. Nov 16, 2013 #3
    I simplified the circuit for you here. Hope this helps!
     

    Attached Files:

  5. Nov 16, 2013 #4

    rude man

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    Considering the symmetry of the network, what can you say about the ratio of potentials at nodes 1 and 2 if you "ground" Y (define V_Y = 0) and apply sinusoidal voltage V to X ?

    Use that simplification to compute I = jwCV where I is the current from the source V and C is your equivalent capacitance.
     
  6. Nov 16, 2013 #5

    gneill

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    Staff: Mentor

    Do not do the Original Poster's homework for them. This is a Forum rule. Your first redrawing is okay, but you should let the OP draw conclusions regarding how the capacitors might be combined.
     
  7. Nov 17, 2013 #6
    Thank you very much for your answers! I feel so stupid for not being able to redraw the circuit myself :P
     
  8. Nov 17, 2013 #7
    OK, so this is a "Wheatstone Bridge". The more you know...
    Now since C1/C5=C4/C3 the voltage at points 1 and 2 (both sides of C2) is equivalent, thus there is not current flowing through C2 and I can just ignore it. Then I end up with C1 and C5 on one side (1/Ceq_a=1/C1+1/C5) and C4 and C3 on the other (1/Ceq_b=1/C4+1/C3) and the total Ceq=Ceq_a+Ceq_b.
     
  9. Nov 17, 2013 #8

    gneill

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    There ya go! :approve:

    Actually you have a choice of methods. Since those two potentials are equal you can either remove the capacitor C2 or replace it with a short. No current will flow in either case :wink:
     
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