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

Answer

D

Explanation

`I_(1) = (12)/(2) = 6A " " E = L(dI_(2))/(dt) + R_(2) xx I_(2)`
`I_(2) = I_(0)(1-e^(-t//tc))`
`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` is