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+3 votes
58.8k views
in Physics by (60.8k points)

A 5.0μF capacitor is charged to 12V. The positive plate of this capacitor is now connected to the negative terminal of a 12V battery and vice versa. Calculate the heat developed in the connecting wires."

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1 Answer

+2 votes
by (60.8k points)

When the capacitor is connected to the battery, a charge Q = CE appears on one plate and –Q on the other. When the polarity is reversed, a charge –Q appears on the first plate and +Q on the second. A charge 2Q, therefore passes through the battery from the negative to the positive terminal. 

The battery does a work.

 W = Q × E = 2QE = 2CE2 

In this process. The energy stored in the capacitor is the same in the two cases. Thus the workdone by battery appears as heat in the connecting wires. The heat produced is therefore, 

2CE2 = 2 ×5 × 10–6 × 144 = 144 × 10–5J = 1.44mJ [have C = 5μf V = E = 12V]

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