Capacitors in DC Circuits & RC question from JEE Main 2024 (1 Feb, Shift 2)

Answer: (d)

1. Steady state: current flows left node $\to 4\ \Omega \to G\,(2\ \Omega) \to 6\ \Omega \to$ right node; $I = \dfrac{6}{4+2+6} = 0.5\text{ A}$.
2. Drops: $4\ \Omega$: 2 V, $G$: 1 V, $6\ \Omega$: 3 V.
3. $C_1$ is across $6\ \Omega$: $V_1 = 3\text{ V}$, $Q_1 = 4\times3 = 12\ \mu\text{C}$.
4. $C_2$ is across $G$ and $6\ \Omega$: $V_2 = 1+3 = 4\text{ V}$, $Q_2 = 6\times4 = 24\ \mu\text{C}$.
5. $\dfrac{Q_1}{Q_2} = \dfrac{12}{24}$, so the answer is (d) $\dfrac{1}{2}$.