Two resistors of $10\Omega and 20\Omega$ and an ideal inductor of 10 H are connected to  a 2 V battery as shown in figure. Key K is  inserted at time t = 0. The initial (t = 0) and final $\left(t\to \infty \right)$ currents through the battery are

At t = 0, i.e., when the key is just pressed, no current exists inside the inductor. So $10\Omega and 20\Omega$ resistors are in series and a net resistance of $\left(10+20\right)=30\Omega$ exists across the circuit. 

$Hence, I_1=\frac{2}{30}=\frac{1}{15}A$

As$t\to \infty$, the current in the inductor glows to attain a maximum valve. i,e., the entire current passes through the inductor and no current passes through $10\Omega$ resistor

$Hence, I_2=\frac{2}{20}=\frac{1}{10}A$