Step 1:

nitially at , the switch is open and inductor is replaced by short circuit.

The resistance is short circuited.

Find the current in the circuit.

Initial current in the circuit : .

Substitute , and in the above formula.

Initially current flowing in the circuit is .

Step 2:

After , the switch is closed.

The resistance and are in parallel.

.

The resistance and are in series and is connected to inductor in series.

The voltage across each component is equal to the total circuit voltage.

Substitute , , and .

This is first order linear differential equation.

Solution of first order linear differential equation is , where .

Solution of the differential equation : .

Substitute corresponding values.

Initial at the value of current is .

Therefore substitute in equation (1).

Therefore current flowing in the circuit is .

Solution :

Current flowing in the circuit is .

Step 1:

Find the voltage across inductor .

Voltage across inductor is .

Substitute and in .

Voltage across inductor is .

Solution :

Voltage across inductor is .