Step 1:
\Initially at
, the switch is closed 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 open.
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
.

Solution :
\Voltage across inductor is
.