Questions
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
detailed solution
Correct option is D
Extra charge Q = (2CV – CV) = CV flows through potential V of the battery. Thus W = QV = CV2Talk to our academic expert!
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A parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved further apart by means of insulating handles, then
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