A 5.0μF capacitor is charged to 12V. The positive plate of this capacitor is now connected
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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]
plzz marl as brainlieat answer
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]
plzz marl as brainlieat answer
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