An inductor $(\mathrm{L}=0.03\mathrm{H})$ and a resistor  $(\mathrm{R}=0.15\mathrm{k}\mathrm{\Omega })$ are connected in series to a battery of 15V EMF in a circuit shown below. The key $K_1$ has been kept closed for a long time. Then at $t=0,$ $K_1$ is opened and key $K_2$  is closed simultaneously. At $\mathrm{t}=1\mathrm{ms}$, the current in the circuit will be $(e^5\cong 150)$

After long time Inductor behaves as short – circuit. At $t=0$, Inductor behaves as short – circuited 

$\therefore$Current$I_0=\frac{E_0}{R}=\frac{15\mathrm{V}}{0.15\mathrm{k\Omega }}=100\mathrm{mA}$

As ${\mathrm{K}}_2$  is closed , current through the Indicator starts decay , which is given at any time ‘t’ as

 $I=I_0{e}^{-\frac{Rt}{L}}=\left(100mA\right)e^{\frac{-t\times 15000}{3}}$

$\mathrm{I}=\left(100\mathrm{mA}\right){\mathrm{e}}^{-\frac{{10}^{-3}\times 15000}{3}}$

$=\left(100\mathrm{mA}\right){\mathrm{e}}^{-5}=0.67\mathrm{mA}$