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Capacitors in Series and Parallel

  1. Feb 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Fig. 25-39 shows a 12.0 V battery and four uncharged capacitors of capacitances C1 = 1.00 µF, C2 = 2.00 µF, C3 = 3.00 µF, and C4 = 4.00 µF. If only switch S1 is closed, what is the charge on (a) capacitor 1, (b) capacitor 2, (c) capacitor 3, and (d) capacitor 4? (figure at http://imgur.com/xOR2AU4)

    xOR2AU4.png [edit: Image inserted by moderator]

    2. Relevant equations
    q=CV
    V is same for capacitors in parallel, P is same for capacitors in series

    3. The attempt at a solution
    a) C1 and C2 are parallel, as are C3 and C4.
    Ct (capacitance of total) = 1/(C1+C2) + 1/(C3+C4) = 0.48
    Q = (Ct*Vt) = 0.48*12 = 5.71 µC

    Q1+Q2 = 5.71 (because both come from positive end of battery)
    Q1/C1=Q2/C2
    Q1/(1µF) = 2Q1/(2µF)
    3Q1 = 5.71µC
    Q1 = 1.9 µC

    The correct answer is 9 µC, and I think I went wrong with the calculation of the initial capacitance, but I'm really not sure.
     
  2. jcsd
  3. Feb 23, 2016 #2

    ehild

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    Gold Member

    Only switch 1 is closed. C1 and C2 are not parallel, neither are C3 and C4.
    Also, you used wrong formula for the series equivalent of capacitors.
     
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