JEE Physics 2023 Question with Solution

Given: In steady state, the capacitor is connected in a circuit with resistors $6 \, \Omega$ and $4 \, \Omega$ in series, a source of $3 \, \text{V}$, and capacitor $C = 4 \, \mu\text{F}$.

Find: The charge accumulated in the capacitor.

In steady state, no current passes through the capacitor. Hence, the capacitor acts as an open circuit.

The total resistance in the conducting branch is

$$R_{\text{total}} = 6 + 4 = 10 \, \Omega$$

So, the current in the circuit is

$$i_1 = \frac{3}{10} = 0.3 \, \text{A}$$

The potential difference across the $6 \, \Omega$ resistor is

$$V = i_1 \cdot R = 0.3 \cdot 6 = 1.8 \, \text{V}$$

Since the capacitor is in parallel with the $6 \, \Omega$ resistor, the potential difference across the capacitor is also $1.8 \, \text{V}$.

Now,

$$Q = C \cdot V$$

Substituting the values,

$$Q = 4 \cdot 1.8 = 7.2 \, \mu\text{C}$$

Therefore, the charge accumulated in the capacitor is $7.2 \, \mu\text{C}$. The correct option is C.

Given: The capacitor is in steady state.

Find: The voltage across the capacitor before calculating charge.

At steady state, the capacitor branch carries no current, so that branch is effectively open. The remaining circuit current flows only through the resistors.

Because the capacitor is in parallel with the $6 \, \Omega$ resistor, both elements have the same potential difference. After finding the circuit current as $0.3 \, \text{A}$, the voltage across that resistor becomes

$$V = IR = 0.3 \times 6 = 1.8 \, \text{V}$$

Hence the capacitor also has $1.8 \, \text{V}$ across it. Using

$$Q = CV$$

we get

$$Q = 4 \, \mu\text{F} \times 1.8 \, \text{V} = 7.2 \, \mu\text{C}$$

Therefore, the correct option is C.