A 5μF capacitor is charged to 12 V. The positive plate of the capacitor is connected to the negative terminal of a 12 V battery and vice versa. Find the heat developed in the connecting wires :
A
72μJ
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B
720μJ
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C
1.44μJ
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D
144μJ
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Solution
The correct option is C1.44μJ The capacitor is connected to the battery so its potential remains constant. As they are connected in opposite polarity so potential across capacitor is V=12+12=24V Here the stored energy in capacitor will develop heat in the connecting wires. Thus, the heat, H=U=12CV2=0.5×5×10−6×242=1.44μJ