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In the circuit shown in figure $E = 18 V, L = 2 H, R_{1} = 3 Ω, R_{2} = 6 Ω,$ 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

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$A - r, B - p, C - q, D - q$

$A - r, B - q, C - s, D - p$

$A - p, B - s, C - q, D - r$

$A - q, B - p, C - r, D - s$

answer is A.

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Detailed Solution

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

$∴$ 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)

$∴$ Current through the Resistor $R_{1} = Z e r o$

Current through the resister $R_{2} = \frac{E}{R_{2}} = 3 A$

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