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 $q_1^{\prime }$ and $q_2^{\prime }$ 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 $q_1^{\prime }=q_2^{\prime }$
• 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 $q_1^{\prime }+q_2^{\prime }=2q$

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, $q_1^{\prime }+q_2^{\prime }=q$.