Solution

1. Find the initial condition. Just before the switch is flipped (t=0-), the circuit looks like:

The voltage over the cap is the same as the voltage over the 3Kohm resistor (because they are in parallel). To find the voltage over the 3K, use a voltage divider (since all the resistors are in series and we know the total voltage across the combination):

V3 = 3k/(1k+2k+3k)*6V = 3V

Now the voltage after the switch flips will stay the same on the capacitor (for an instant), so the voltage at t=0+ is also 3V.

2.Find the final condition. A long time after the switch has flipped, the circuit looks like:

The capacitor has gone back to looking like an open because we are in steady-state. This time, no current can flow through the 3k, so the voltage over it (and also over the cap) must be zero.

3. Find the time constant. The circuit in between the initial and final time looks like:

The time constant is RC, where R is the Thevenin resistance. The capacitor does not see the 1k because it has been disconnected by the switch opening. The capacitor does not see the 2k either, since no current could flow from the capacitor through the 2k. However, it does see the 3k. Thus Rth = 3k, and tau, the time constant is:

tau = RC = (3k)(1u) = 3ms.

Note that the unit of tau is seconds.

4. Write the formula. Just fill in the blanks:

Vc(t) = 0 + (3 - 0) e^(-t/3ms) = 3e^(-t/3ms)