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Question

A capacitor of capacitance C is charged by connecting it to a battery of e.m.f. E volts. The capacitor is now disconnected and reconnected to the same battery with polarity reversed. The heat energy developed in the connecting wire is:

A
CE2
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B
2CE2
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C
12CE2
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D
Zero.
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Solution

The correct option is D 2CE2
ΔW=ΔU+ΔH
Work done by battery,
W=2CE×E
Change in energy stored in capacitor,
ΔU=0
Thus, Heat dissipated,
ΔH=2CE2

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