Find the current through the battery
(i) just after the switch is closed.
(ii) long after the switch has been closed.

Explanation

(i) Just after closing the switch, inductor offers infinite resistance. So, the circuit will become as

So, the current will be `I=(V)/(r)=(12)/(4)=3A`
(ii) After long time, inductor offers zero resistance. So, the equivalent circuit will be

Net resistance, `R_(eq)=(4xx4xx4)/(4+4+4)=(64)/(12)=5.33`
`therefore"Current",I=(V)/(R)=(12)/(5.33)or(12xx12)/(64)=(144)/(64)=2.25A`