Question

Consider the circuit shown in figure. (a) Find the current through the battery a long time after the switch $S$ is closed. (b) Suppose the switch is again opened at $t=0$. What is the time constant of the discharging circuit?

Answer 1

(a) The inductor does not work in DC. When the switch is closed the current charges so at first inductor works. But after a long time the current flowing is constant
Thus effect of inductance vanishes
$i=\frac{E}{R_{net}}=\frac{E}{\frac{R_1{R}_2}{R_1+R_2}}=\frac{E(R_1+R_2)}{R_1{R}_2}$
(b) When the switch is opened the resistors are in series
$\tau =\frac{L}{R_{net}}=\frac{L}{R_1+R_2}$