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tourjete
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Homework Statement
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
Homework Equations
Kirchoff's laws
Ohm's Law
Q = CV for a capacitor
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)