Question

In the circuit given, switch 'S' is open for long time and then it is closed at t$=0$. Then current through the inductor at time 't' is given by ______.

Answer 1

Steady state inductor behave short circuit

$S-open$

Current through inductor $i_1=\frac{E}{R}$

$S-closed$

Now steady state current through inductor is

$I_2=\frac{E}{(R\parallel R)}=\frac{E}{\left(\frac{R}{2}\right)}=\frac{2E}{R}$

At any time $t$ current through inductor is $I$

$I=I_2-(I_2-I_1)e^{-\frac{t}{\tau }}$

Where $T=Timeconstant=\frac{L}{\left(\frac{R}{2}\right)}=\frac{2L}{R}$

Now, $I=\frac{2E}{R}-[\frac{2E}{R}-\frac{E}{R}]e^{-Rt/2L}$

$=\frac{E}{R}[2-e^{-\frac{Rt}{2L}}]$