JEE Physics 2023 Question with Solution

Given: Three identical resistors each have resistance $R = 12 \, \Omega$ and two identical inductors each have self inductance $L = 5 \, \text{mH}$. The battery emf is $12 \, \text{V}$.

Find: The current through the battery long after the switch is closed.

After a long time in a DC circuit, inductors behave as short circuits because the current becomes steady.

So the circuit reduces to three identical resistors connected in parallel across the battery.

The equivalent resistance is:

$$\frac{1}{R_{\text{eq}}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R}$$ $$\frac{1}{R_{\text{eq}}} = \frac{1}{12} + \frac{1}{12} + \frac{1}{12} = \frac{3}{12}$$ $$R_{\text{eq}} = \frac{12}{3} = 4 \, \Omega$$

Using Ohm's law:

$$I = \frac{V}{R_{\text{eq}}} = \frac{12}{4} = 3 \, \text{A}$$

Therefore, the current through the battery is $3 \, \text{A}$.

Given: Long time after closing the switch, the circuit is in steady DC state.

Find: The battery current.

In steady state, each inductor acts as a wire. Hence the network becomes three equal resistors of $12 \, \Omega$ in parallel.

For three equal resistors in parallel, equivalent resistance is:

$$R_{\text{eq}} = \frac{R}{3} = \frac{12}{3} = 4 \, \Omega$$

So,

$$I = \frac{12}{4} = 3 \, \text{A}$$

Therefore, the correct numerical answer is 3.