MHTCET 26 Apr 2025 Shift 1 Paper

Concept:The series combination of $C_1$ and $C_2$ is connected in parallel with $C_3$ across the battery.Charge on series capacitors is the same, while parallel capacitors have charge proportional to their capacitance.Explanation:From the circuit, $C_1 = C$, $C_2 = 2C$, and $C_3 = 3C$.$C_1$ and $C_2$ are in series, so their equivalent capacitance is:$$C_{12} = \frac{C_1 C_2}{C_1 + C_2} = \frac{C \cdot 2C}{C + 2C} = \frac{2}{3}C$$This series branch is in parallel with $C_3$, so the voltage across both is the battery voltage $V$.Charge on $C_3$ is:$$Q_3 = C_3 V = 3CV$$Charge on the series branch is:$$Q_{12} = C_{12} V = \frac{2}{3}CV$$In series, charge on $C_1$ equals the charge on the branch:$$Q_1 = Q_{12} = \frac{2}{3}CV$$Therefore, the required ratio is:$$\frac{Q_3}{Q_1} = \frac{3CV}{\frac{2}{3}CV} = \frac{9}{2} = 4.5$$Answer:The ratio of charge on $C_3$ and $C_1$ is $4.5$.Correct option: D.