Q: In the circuit shown in figure initially key K_{1} is closed and key K_{2} open. Then K_{1} is opened and K_{2} is closed (order is important). [Take Q_{1}‘ and Q_{2}‘ as charges on C_{1} and C_{2} and V_{1} and V_{2} as 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}‘ = Q_{2}‘

(c) Charge on C_{1} gets redistributed such that C_{1} V_{1} + C_{2} V_{2} = C_{1} E

(d) Charge on C_{1} gets redistributed such that Q_{1}‘ + Q_{2}‘ = Q

Ans: (a) & (d)

Solution : When key K_{1} is closed and key K_{2} open capacitor C_{1} is charged by the battery . When K_{1} is opened and K_{2} is closed the charged stored by the capacitor C_{1} will be redistributed between C_{1} and C_{2} . As C_{1} and C_{2} are in parallel hence V_{1} = V_{2}