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?

Answer

`12e^(-5t), 6e^(-10t)`

Explanation

`I_(1)=E/R_(1)`
`I_(2)=E/R_(2)(1-e^(-t//tau))`
`V_(L)=E e^(-t//tau)=Ee^(-Rt//L)=12 e^(-5t)`
when switched is opened current will decay exponatial with time constant.
`tau^(,)=(L/(R_(1)+R_(2)))=4/(2+2)=0.1`

`I^(,)=E/R_(2) e^(-10t)=6e^(-10t)`
`implies I_(0)=E/R_(2)=6 Amp`
due to inductance direction of current can not be changed instantaneously.