An inductor of inductance L=400 mH and resistor of resistance `R_(1) = 2(Omega) and R_(2) = 2 (Omega)` are connected to a battery of emf E = 12 Vas shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at time t =0. What is the potential drop across L as a function of time? After the steady state is reached, the switch is opened. What is the direction and the magnitude of current through `R_(1)` as a function of time?

Explanation

`I_(1) = (E)/(R_(1)) = (12)/(2) = 6A, E = L(dl_(2))/(dt) +R_(2)xxl_(2)`
`I_(2) = I_(0)(1-e^(-t//t_(c ))), rArr I_(0) = (E)/(R_(2)) = (12)/(2) = 6A`
`t_(c ) = (L)/(R ) = (400 xx 10^(-3))/(2) = 0.2, I_(2) 6(1-e^(-t//0.2))`
Potential drop across `L`
`V_(L) = E-R_(2)I_(2) = 12-2 xx 6(1-e^(-bt)), = 12e^(-5t)`