Question

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?

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

Answer 1

Correct Answer - C
(c)Growth in current in LR_(2) brach when switch is closed is given by
`i=(E)/(R^2)[1-e^(-R_(2)t//L)] implies (di)/(dt)=(E)/(R^2)*(R_2)/(L)e^(-R_(2)t//L)=(E)/(L)ee^(-R_(2)t/L)`
Hence, potential drop across
`L=(E/L e^(-R_(2)t//L))L=Ee^(-R_(2)t//L)=12e^(-(2t)/(400xx10^(-3)))`.