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?

Options

A. `6e^(-5t)V`
B. `(12)/(t)e^(3t)V`
C. `6(1-e^(-t)/(0.2))V`
D. `12e^(-5t)V`

Answer

D

Explanation

In the branch containing `L` and `R^(2)`
`i=(E)/(R_(2))(1-e^(-(R_(2)t)/(L)))`
`(di)/(dt)=(E)/(R_(2))e^((-R_(2)t)/(L)).(R_(2))/(L)=(E)/(L)e^(-(R_(2)t)/(L))`
`V_(L)=L(di)/(dt)=Ee^(-(R_(2)t)/(L))=12e^(-(2xxt)/(400xx10^(-3))`
`=12e^(5t)V`