An RLC circuit with the capacitor connected in parallel Consider the following circuit, where the switch is left open for a long time:

What happens with the switch is closed?  There's certainly a potential difference across the capacitor, which also means that there's that same potential difference across the wire.  I thought that would be a short circuit.  What exactly happens?
$$V=6 \, \text{Volts}, \quad R=3\,\Omega, \quad L=8\,\text{mH}, \quad C=0.4\,\mu\text{F} \, .$$
 A: After a long time, the capacitor charges up to the voltage $V_{0}$, with no current flowing.  (It is easy to verify that the sum of the voltages around the loop is zero in this case.)  When you short out the capacitor, it discharges instantly.  Then the capacitor voltage, which has been holding back the current flow across the other elements, is no longer present.  Therefore, current will begin to flow.  Since the capacitor is shorted out, what you have is effectively an RL circuit starting from zero current, and $I(t)$ follows the standard solution for that system.
A: The capacitor is a red herring and is irrelevant to the question.  When the switch is open for a long time, there is no current thru the resistor and inductor, and no voltage across them.  The capacitor is charged to the full supply voltage.
When the switch is closed, the capacitor is discharged instantly.  That can't happen exactly that way in real life because there will always be at least some resistance, but either way the capacitor discharges "quickly".
Now we simply have a resistor and inductor connected in series to a voltage source.  The only relevant initial condition is that the inductor current is zero.  The current will now ramp up, assymptotically approaching the voltage divided by the resistance:
$$I = (1 - e^{-tR/L})  V_o / R \, .$$
A: Before the switch is closed, the capacitor would have been charged to $V=6$ volts and there would be no current in the circuit.
When the switch is closed, the capacitor will be short circuited. There will be a large current in the switch branch (it will increase rapidly from 0 continously ; there won't be a jump suddenly). This current will consist of a DC component $V/R = 6/3 = 2 A$ and an oscillating component exponentially decaying to 0. The oscillating current will go to 0 and the capacitor will discharge. After a long time the current will be $2 A$ in the switch branch and $0$ in the capacitor branch.
