Capacitors in DC Circuits & RC question from JEE Main 2023 (13 Apr, Shift 2)

Answer: (b)

1. Steady state: no current in the capacitor branch, so $I = \dfrac{3}{4+6} = 0.3\text{ A}$ through the bottom $6\ \Omega$.
2. Capacitor branch is across that $6\ \Omega$: $V_C = 0.3\times6 = 1.8\text{ V}$
3. $Q = CV = 4\times1.8 = 7.2\ \mu\text{C}$
4. So the answer is (b) $7.2\ \mu\text{C}$.