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Question
A 5⋅0 µF capacitor is charged to 12 V. The positive plate of this capacitor is now connected to the negative terminal of a 12 V battery and vice versa. Calculate the heat developed in the connecting wires.
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Solution
As a capacitor of capacitance C is connected to the battery of voltage V, one of its plates gets charged to Q = CV and the other gets charged to −Q coulomb. When the polarity is reversed, charge −Q appears on the first plate and charge +Q appears on the second plate. So to maintain these charges, charge 2Q passes through the battery from the negative terminal to the positive terminal.
The work done by the battery is given by `W = 2QV = 2CE^2`
In the given process, the capacitor's energy remains the same in both cases. The work done by the battery appears as heat in the connecting wires.
Now,
Heat produced :
`H = 2CE^2`
`H = 2 xx (5 xx 10^-6) xx 144`
`H = 144 xx 10^-5 "J"= 1.44 "mJ"`
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