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Question

# Consider the situation shown in figure. The switch S is open for a long time and then closed. Find the charge flown through the battery when the switch is closed.

A
CE/2
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B
CE
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C
CE/4
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D
0
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Solution

## The correct option is A CE/2When S is open. Both capacitors are in series and its equivalent capacitance will be Ceq=C2 Charge on the each plate of capacitor will be Q=CE2 So charge distribution on the plate of PPC will be as shown below When S is closed: ΔV across the capacitor on the right is 0. So, Qf=CE Thus, charge flown through battery will be ΔQ=CE−CE2=CE2 Hence, option (a) is correct.

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