In the following circuit (Fig.) the switch is closed at `t = 0`. Find the current `i_(1),i_(2),i_(3)` and `di_(3)//dt` at `t = 0` and at `t = oo`. Intitially, all current are zero.

Explanation

At `t = o, i_(3)` is zero since current cannot suddenly change due to the inductor.
`:. i_(1) = i_(2)` (from `KCL`)
Applying `KVL` in the part `ABEF`, we get `i_(1) = i_(2) = (varepsilon)/(2R)`.
`varepsilon - i_(R) - L(di_(3))/(dt) = 0`
`rArr (di_(3))/(dt) = (varepsilon - i_(R))/(L) = (varepsilon - (varepsilon//2R)R)/(L) = (varepsilon)/(2L)`
At `t = oo, i_(3)` will become constant and hence potential difference across the inductor will be zero. It is just likea simple wire and the circuit can be solved assuming it to be equivantent to the circuit diagram shows below.

`i_(2) = i_(3) = (varepsilon)/(3R)`, `i_(1) = (3varepsilon)/(2R),(di_(3))/(dt) = 0` at `t = 0`