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Q.

A capacitor of capacity C  is connected with a battery of potential V in parallel. The distance between its plates is reduced to half at once, assuming that the charge remains the same. Then to charge the capacitance upto the potential V again, the energy given by the battery will be

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a

CV2/4

b

CV2/2

c

3CV2/4

d

CV2

answer is D.

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Detailed Solution

Extra charge Q = (2CV – CV) = CV flows through potential V of the battery. Thus W = QV =  CV2
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