Question

2.30. In the circuit shown in the figure, initially, $\mathrm{K}_{1}$ is closed, and $\mathrm{K}_{2}$ is open. What are the charges on each capacitor? Then $\mathrm{K}_{1}$ was opened, and $K_{2}$ was closed. What will be the charge on each capacitor now?

• A. Charge on $C_{1}$gets redistributed such that $V_{1} = V_{2}$
• B. Charge on $C_{1}$ gets redistributed such that $q_{1}' = q_{2}'$
• C. Charge on $C_{1}$gets redistributed such that C_{1}V_{1} = C_{2}V_{2} = C_{1}V
• D. Charge on $C_{1}$ gets redistributed such that $q_{1}' + q_{2}' = 2q$

Answer 1

Answer: (A)

Charge on $C_{1}$gets redistributed such that $V_{1} = V_{2}$

Answer 2

: From the figure, when $K_{1}$is closed and $K_{2}$ is open, $C_{1}$is charged to potential V acquiring a total charge $q = C_{1}V$. When $K_{1}$is open and $K_{2}$is closed, battery is cut off $C_{1}$and $C_{2}$are in parallel. The charge on $C_{1}$is shared between $C_{1}$and $C_{2}$such that $V_{1} = V_{2}$. As after is no loss of charge, $q'_{1} + q'_{2} = q$