1. Understand the circuit at switching

Three branches are connected in parallel across a $5\text{ V}$ battery:

• Inductor $L_1 = 1\text{ mH}$ with internal resistance $3\,\Omega$
• Inductor $L_2 = 2\text{ mH}$ with internal resistance $4\,\Omega$
• Resistor $R = 12\,\Omega$

We need the ratio $\frac{I_{\max}}{I_{\min}}$ of the total current drawn from the battery.

Since inductors oppose sudden change in current:

• At $t=0^+$, both inductor branches behave like open circuits.
• At $t\to\infty$, both inductors behave like short circuits, but their internal resistances remain.

So the total current is minimum at $t=0^+$ and maximum at steady state.

2. Current at $t=0^+$

At the instant of switching on, no current flows through the inductor branches. Only the $12\,\Omega$ resistor draws current.

Thus, $I_{\min} = \frac{V}{R} = \frac{5}{12}\text{ A}$

3. Current at $t\to\infty$

In steady state, each inductor acts as a wire, so each inductor branch has only its internal resistance.

So the three parallel branches are now:

• $3\,\Omega$
• $4\,\Omega$
• $12\,\Omega$

Branch currents are: $I_1 = \frac{5}{3}\text{ A}$ $I_2 = \frac{5}{4}\text{ A}$ $I_R = \frac{5}{12}\text{ A}$

Therefore total current, $I_{\max} = \frac{5}{3} + \frac{5}{4} + \frac{5}{12}$

Taking LCM $=12$, $I_{\max} = \frac{20+15+5}{12} = \frac{40}{12} = \frac{10}{3}\text{ A}$

4. Compute the ratio

$\frac{I_{\max}}{I_{\min}} = \frac{\frac{10}{3}}{\frac{5}{12}}$

$= \frac{10}{3}\times \frac{12}{5} = 8$

5. Final answer

$\boxed{8}$

This matches the stored correct answer.