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