Question

In the circuit shown above, switch $S$ is closed at time $t=0$. Find the current through the inductor as a function of time $t$.

Answer 1

At $t=0$, Current through inductor will be zero.
At $t=\infty$, Net emf$=\frac{2/2+4/1}{1/2+1/1}=\frac{10}{3}V$
Net resistance $=\frac{2\times 1}{2+1}=\frac{2}{3}\Omega$
$\therefore$ $i=\frac{10/3}{2/3}=5A$

To find equivalent time constant short circuit both the batteries and find net resistance across inductor:
$R_{net}=\frac{2\times 1}{2+1}=\frac{2}{3}\Omega$
$\therefore$ ${\tau }_L=\frac{L}{R_{net}}=\frac{1\times {10}^{-3}}{2/3}=\frac{3}{2000}s$

Current through inductor will increase exponentially from $0$ to $5A$, $i=5(1-e^{-\frac{2000t}{3}})$