Question

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.

Answer 1

Before `S_(2)` is closed and `S_(1)` is closed current in the left part of the circuit`=(epsi)/( R)`. Now when `S_(2)` closed `S_(1)` opened, current through the inductor can not change suddenly, current `(epsi)/( R)` will continue to move in the inductor.

Applying KCL in loop 1.
`l(di)/(dt)+(epsi) /(R)(2R)+4epsi=0`
`(di)/(dt)=-(6epsi)/(L)`