In the circuit shown in figure $E=18V,L=2H,R_1=3\Omega ,R_2=6\Omega ,$ switch S is closed at  $t=0.$Match the following.

Column I		Column II
A	Current through  $R_1$at $t=0$	P	6A
B	Current through  $R_1$at $t=\infty$	Q	3A
C	Current through $R_2$ at $t=0$	R	Zero
D	Current through $R_2$ at $t=\infty$	S	Infinite

At $t=0$ inductor behaves like a zero-resistance element

 $\therefore$ current through $R_1$ is $i_1=\frac{E}{R_1}$

 Current through $R_2$ is $i_2=\frac{E}{R_2}$

At $t=\infty$ , inductor behaves like an infinite resistance element $(i.e.$breaks at the inductor)

$\therefore$ Current through the Resistor $R_1=Zero$

Current through the resister $R_2=\frac{E}{R_2}=3\mathrm{A}$