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Homework Help: 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
    [tex]
    Q_1 = Q_{0}e^{-t(C_{1}+C_{2})/RC_{1}C_{2}}
    [/tex]

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

    [tex]
    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}})
    [/tex]
    but this is apparently wrong

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

    Astronuc

    User Avatar
    Staff Emeritus
    Science Advisor

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

    as opposed to


    Vin-----C1---+---C2--- Vout
    |
    R
    |
    ----------+---------​

    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
    http://www.columbia.edu/~kqc2101/circuit.gif [Broken]

    so for Q_2 what would be the voltage drop V that charges C_2 in the equation
    [tex]
    Q=CV(1-e^{-t/\tau})
    [/tex]
    ?
     
    Last edited by a moderator: May 2, 2017
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