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 opened at $t=0$. What is the time constant of the decay circuit? (c) Find the current through the inductor after one time constant.

Answer 1

Current through the battery after a long time,

$I=\frac{E}{R_{eq}}=\frac{E(R_1+R_2)}{R_1{R}_2}$ as then inductor will not hinder.

When the circuit comes in steady state, that is after a long times, the current through the inductor is

$I_o=\frac{E}{R_1}$

And, the time constant is $\tau =\frac{L}{R_1}$ since only $R_1$ affects the current in the inductor, otherwise potential difference is always $E$ across that branch.

Hence,

$I=I_o(1-e^{-\frac{t}{\tau }})$

$⟹I=\frac{E}{R_1}(1-e^{-\frac{R_{1}t}{L}})$

At $t=\tau$

$I=\frac{E}{R_1}(1-\frac{1}{e})$