In a circuit `S_(1)` remains closed for a long time and `S_(2)` remain open. Now `S_(2)` is closed and `S_(1)` is opened. Find out the `di//dt` in the right loop just after the moment.

Explanation

Before `S_(2)` is closed and `S_(1)` is opened current in the left part of the circuit `=epsilon/R`
Now when `S_(2)` closed `S_(1)` opened current through the inductor can not charge suddenly current `epsilon/R` will continue to move in the inductor.
Applying `KVL` in loop `1`. ?
`L(di)/(dt)+epsilon /R(2R)=4epsilon=0`
`(di)/(dt)=-(6epsilon)/L`