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