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 $V_1$ are the voltage 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)

charge on $C_1$ gets redistributed such that $V_1$ = $V_2$

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 there is no loss of charge $q_1^{\prime }+q_2^{\prime }=q$