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RC Circuit

  1. Mar 4, 2006 #1
    A capacitor, a resistor, and another capacitor is connected in series. The first capacitor, C_1, has an initial charge of Q_0, and C_2 is initially uncharged. The switch is flipped at t=0, what is the charge on each capacitor as a function of time?

    So I thought
    Q_1 = Q_{0}e^{-t(C_{1}+C_{2})/RC_{1}C_{2}}

    V_1 = Q_{0}e^{-t(C_{1}+C_{2})/RC_{1}C_{2}}/C_{1}

    Q_2 = C_{2}/C_{1}(Q_{0}e^{-t(C_{1}+C_{2})/RC_{1}C_{2}})(1-e^{-t(C_{1}+C_{2})/RC_{1}C_{2}})
    but this is apparently wrong

    where am I messing up?
    Last edited: Mar 4, 2006
  2. jcsd
  3. Mar 4, 2006 #2


    User Avatar

    Staff: Mentor

    Is the configuration ---C1----R---C2 ---- ?

    as opposed to

    Vin-----C1---+---C2--- Vout

    The charge diminishes on C1, but increases on C2, so the solution cannot be exp(-At), which would go to zero at t = infinity. The solution must go to some constant, and the charge would be partitioned according to the relative capacitances.
    Last edited: Mar 4, 2006
  4. Mar 4, 2006 #3
    this is the diagram given

    so for Q_2 what would be the voltage drop V that charges C_2 in the equation
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