Interpret the circuit behavior at switching instants

This is a standard resistor-inductor switching circuit (the options indicate the battery current changes from one resistive value at $t=0$ to another at $t=\infty$).

We use the inductor properties:

• At $t=0^+$, current through an inductor cannot change instantaneously, so the inductor behaves like an open circuit if initial current is zero.
• At $t=\infty$, in steady-state DC, the inductor behaves like a short circuit.

At $t=0^+$

Since the inductor behaves as an open circuit, the branch containing the inductor (and hence the resistor associated with that branch) does not conduct initially.

Therefore the battery current flows only through $R_2$.

So, $I(0^+) = \frac{V}{R_2}$

At $t=\infty$

In steady state, the inductor acts as a short circuit. Then the circuit reduces to both resistors conducting in such a way that the battery sees the equivalent resistance of $R_1$ and $R_2$ in parallel.

Thus, $R_{\text{eq}} = \frac{R_1R_2}{R_1+R_2}$

Hence battery current is $I(\infty) = \frac{V}{R_{\text{eq}}} = \frac{V(R_1+R_2)}{R_1R_2}$

Match with options

We found:

• At $t=0$: $\frac{V}{R_2}$
• At $t=\infty$: $\frac{V(R_1+R_2)}{R_1R_2}$

This matches Option B.

Comparison with stored answer

Stored correct answer is B, which matches the derived result.