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Discharging capacitor in a grounded circuit

  1. Jun 12, 2014 #1
    1. The problem statement, all variables and given/known data

    In the circuit below, the capacitors labeled C1 and C2 both have capacitance C.
    They are connected through a switch and a resistor of resistance R. Capacitor
    C1 is initially charged with a potential V1(0) at time t = 0, while capacitor C2
    is uncharged at time t=0.
    At time t = 0, the switch is closed and the charge on C1 begins to move to C2.
    Write a differential equation for the time dependence of V1(t) and V2(t). Solve
    for the time dependence of the voltage across C2 for times after t = 0 when the
    switch is closed

    The circuit picture can be seen here: http://imgur.com/uy36j0z


    2. Relevant equations

    Kirchoff's laws
    Ohm's Law
    Q = CV for a capacitor



    3. The attempt at a solution

    Obviously, charge is conserved so [itex]Q_1(t) + Q_2(t) = Q_1(0)[/itex]. This also means that [itex] \frac{d Q_1}{dt} = -\frac{d Q_2}{dt} = 0 [/itex].

    I tried to use Kirchoff's laws and got:
    [tex] 0 = V_1(t) + IR + V_2(t) = \frac{Q_1(t)}{C} + R \frac{dQ_1(t)}{dt} + \frac{Q_2(t)}{C} = \frac{Q_1(0)-Q_2(t)}{C} - R \frac{dQ_2(t)}{dt} + \frac{Q_2(t)}{C} [/tex].

    However, I'm a little concerned here due to two things: Solving this differential equation gives something linear in time instead of exponential, as I intuitively know that it must be, and I haven't taken into account the portion of the circuit that is grounded. What am I doing wrong? (Once I find the charge it is trivial to get V from that)
     
  2. jcsd
  3. Jun 12, 2014 #2

    berkeman

    User Avatar

    Staff: Mentor

    That last part at the end "=0" looks incorrect.

    I believe you have a sign error in the loop equation that you wrote. The signs of the two capacitor voltages should be opposite, IMO.

    But I would approach the question a bit differently. You know the current is the same around the loop, so I would write the KCL at the two nodes (two sides of the resistor) and solve...
     
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