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 $\mathrm{\Omega }$ and 20 $\mathrm{\Omega }$ resistors are in series and a net resistance of (10 + 20) = 30 $\mathrm{\Omega }$ exists across the circuit.
$\text{ Hence, }{I}_1=\frac{2}{30}=\frac{1}{15}\mathrm{A}$
As $t\to \infty$, the current in the inductor grows to attain a maximum value, i.e., the entire current passes through the inductor and shortcircuits 10 $\mathrm{\Omega }$ resistor.
$\text{ Hence, }{I}_2=\frac{2}{20}=\frac{1}{10}\mathrm{A}$