Capacitors in DC Circuits & RC question from MHT CET 2024 (9 May, Shift 1)

Answer: (c)

1. In steady state no current flows in the capacitor branch, so only $1\ \Omega$, $2\ \Omega$, $3\ \Omega$ (in parallel) carry current.
2. $\dfrac{1}{R} = 1 + \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{11}{6} \Rightarrow R = \dfrac{6}{11}\ \Omega$
3. $I = \dfrac{6}{6/11} = 11\text{ A}$
4. No current through $4\ \Omega$, so the capacitor gets the full 6 V: $Q = 0.5\times6 = 3\ \mu\text{C}$.
5. So the answer is (c) 11 A and $3\ \mu\text{C}$.