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

Answer

C

Explanation

In the R-L circuit, growth of current is given by
`I = (E)/(R ) [ - e^(-R_(2) t//L)]`
`:. (dI)/(dt) = - (E)/(R_(2)) (-(R_(2))/(L)) e^(R_(2) t //L)`
Potential drop across L is
`V_(L) = L (dI)/(dt) = L (- (E)/(R_(2))) e^(-R_(2)t//L) (- (R_(2))/(L))`
`V_(L) = E e^(-R_(2) t//L) = 12 e^(-2t//0.4) = 12 e^(-5 t)` volt