Step 1:

At , the switch is closed and capacitor acts as a open circuit.

The resistance is short circuited.

The resistor and are in series.

The current flowing in the circuit is .

Substitute , and .

Therefore current flowing in the circuit is .

Step 2:

At , the switch is open and capacitor starts discharging.

Apply KVL for the first loop.

Apply KVL for the second loop.

Apply Laplace transform to the equation (1)

Laplace transform of constant .

Laplace transform of integral function .

Apply Laplace transform to the equation (2)

Substitute in equation (3).

Apply inverse laplace transform.

Inverse Laplace transform : .

.

Since the current direction is in opposite direction, .

(2)

Step 1:

The current flowing in the circuit is .

Apply KVL for the first loop.

Substitute in the above equation.

At , the current flowing in the circuit is .

Substitute in the equation.

The current in the circuit is .

Solution :

The current in the circuit is .