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)

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, $q'_{1} + q'_{2} = q$