# Discharging capacitor in a grounded circuit

1. Jun 12, 2014

### tourjete

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 $Q_1(t) + Q_2(t) = Q_1(0)$. This also means that $\frac{d Q_1}{dt} = -\frac{d Q_2}{dt} = 0$.

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

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. Jun 12, 2014

### 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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