An inductor of inductance L = 400 mH and resistors of resistance `R_(1) = 2Omega` and `R_(2) = 2Omega` are connected to a battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at `t = 0`. The potential drop across L as a function of time is

Options

A. `12/te^(-31) V`
B. `6(1-e^(-t//0.2))V`
C. `12 e^(-5t)V`
D. `5 e^(-5t)V`

Answer

C

Explanation

`V_(L)=epsilone-(R_(2)t)/L=12.e-(2t)/(400xx10^(-3))=12 e^(-5t)`