Question

Two capacitors $C_{1}$ and $C_{2}=2 C_{1}$ are connected in a circuit with a switch between them as shown in the figure. Initially the switch is open and $C_{1}$ holds charge $Q$. The switch is closed. At steady state, the charge on each capacitors will be

• A. $Q, 2 Q$
• B. $\frac{Q}{3}, \frac{2 Q}{3}$
• C. $\frac{3 Q}{2}, 3 Q$
• D. $\frac{2 Q}{3}, \frac{4 Q}{3}$

Answer 1

Answer: (B)

Charge redistributes on the capacitors in the ratio of their capacitance.
$\frac{Q_{1}}{Q_{2}}=\frac{C_{1}}{C_{2}}=\frac{C_{1}}{2 C_{1}}=\frac{1}{2}$
$Q_{1}: Q_{2}=1: 2$
$\therefore Q_{1}=\frac{1}{1+2} \times Q=\frac{Q}{3}$
$Q_{2}=\frac{2}{1+2} Q=\frac{2}{3} Q$