Question

In the given circuit, initially $K_{1}$ is closed and $K_{2}$ is open. Then $K_{1}$ is opened and $K_{2}$ is closed. If $q'_{1}$ and $q'_{2}$ are charges on $C_{1 }$ and $C_{2}$ and $V_{1}$ and $V_{2}$ are the voltages respectively, then

• 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)

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 there is no loss of charge, $q'_{1} +q'_{2} = q$.