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 in the left part of the circuit `= varepsilon//R`. Now when `S_(2)` is closed and `S_(1)` is opened, current through the inductor cannot change suddenly, current `varepsilon//R` will continue to flow through the inductor.

Applying `KVL` in loop 1,
`L(di)/(dt) + (varepsilon)/(R ) (2R) + 4 varepsilon = 0 rArr (di)/(dt) = - (6 varepsilon)/(L)`