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Capacitance Problem

  1. May 25, 2014 #1
    1. The problem statement, all variables and given/known data

    1279_figure.JPG

    In the above circuit, find the potential difference across AB.


    2. Solution Online


    "C34 = 4μf
    C2,34 = 12μf
    CEQ = 4.8μf
    q = CEQ x V
    the q on 1 is 48μC, thus
    V1 = q/c
    V1 = 6v
    VPQ = 10 - 6 = 4v
    By Symmetry of 3 and 4 , VAB = 2v
    "

    3.My question is -
    (a).What is Symmetry and how is it used in this case?
    (b).How are they combining C2,34 with C1 in Series?


    Thanks
     
  2. jcsd
  3. May 25, 2014 #2

    BvU

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    How come you ask about C2, 34 in series with C1 but you do not ask about C34 ?

    The symmetry is that if you swap C3 and C4, you get the same situation, so the charge on each must be the same.
     
  4. May 25, 2014 #3
    How is same charge possible on both C3 and C4 and potential difference between B and A is different and P and A is different?
    and how is VAB = 2v using this ?
     
  5. May 25, 2014 #4
    C3 and c4 are in series, so the current through them will always be the same. If they start out with the same charge.m they will always have the same charge across them.

    Why do you think that the potential difference between P and A and the potential difference between A and B are different?
     
  6. May 26, 2014 #5

    BvU

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    Charge on right plate of capacitor 3 + charge on top plate of capacitor 4 = 0
    Q = C V, Q same, C same, therefore V same. Vpb = 4 Volt so Vpa = Vab = 2 V

    You know about calculating equivalent capacitance of capacitors in series ?
     
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