Consider the circuit shown in .(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? (c ) Find t current through the inductor after one time constant.

Explanation

(a) The inductor does not work in DC (When switch is closed)
Thus effect of inductance vanishes.
`i= (E )/(R_(net))`
`==(E )/(((R_1)+(R_(2)))/((R_1)+(R_(2)))) (E(R_1+R_(2)))/(R_1R_(2)`
(b) When the switch is opened the resistance are in series.
(tau)=(L)/(R_(net))=(L)/(R_1+R_(2))`.