Question

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

In steady state, both the capacitors are at the same potential. Therefore,
$\frac{Q_{1}}{C_{1}}=\frac{Q_{2}}{C_{2}} $
or $\frac{Q_{1}}{C}=\frac{Q_{2}}{2 C}$
or $ Q_{2}=2 Q_{1} \dots$(i)
Also $ Q_{1}+Q_{2}=Q \dots$(ii)
Solving (i) and (ii), we get
$\therefore Q_{1}=\frac{Q}{3}$ and $Q_{2}=\frac{2}{3} Q$