In the circuits shows in `S_(1)` and `S_(2)` are switches. `S_(2)` remains closed for a long time and `S_(1)` is opened. Now `S_(1)` is also closed. Just after `S_(1)` is closed, find the potential difference `(V)` across `R` and `di//dt` (with sign) in `L`.

Answer

`(varepsilon)/(3),(2varepsilon)/(3L)`

Explanation

Before closing `S_(1)`, current in the inductor is `I = varepsilon//2R`. Just after closing `S_(1)`, current in inductor will remain same, and other currents are as shows

For left loop: `varepsilon = i_(1) R + L(di)//dt)` (i)
For right loop: `varepsilon = i_(2)(2R) + L(di//dt)` (ii)
and `i_(1) + i_(2) = i` (iii)
Solve to get `(di)/(dt) = (2varepsilon)/(3L)`
p.d.across `R: V_(R) = i_(1)R = varepsilon - L(di//dt) = varepsilon - L ((2 E)/(3L)) = (varepsilon)/(3)`