Find the values of `i_(1)` and `i_(2)`

i) immediately after the switch S is closed.
ii) long time later, with S closed
ii) immediately after S is reopende

Explanation

Just after closing the switch the inductor will be treated as open circuit. So the circuit would be like as shown in figure

`i_(1)=i_(2)=(100)/(30)=(10)/(3)A`
After a long time of closing the switch, the inducotr is treated as short circuit. So the circuit would be like as shown in figure.

Solving above circuit will get
`i_(1)=(50)/(11)A, i_(2)=(30)/(11)A`
and current through inductor is `(20)/(11)`A.
Just after reopening the switch `i_(1) = 0` and as inductor tries to oposse any variation in flux reopening the switch would be same.
so `i_(1)=(20)/(11)A`
After a long time of reopening the switch the current is zero.