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.

Answer 1

The heat developed in the connecting wires is 1.44 mJ

Explanation:

The work done by the battery is given by = W = 2 Q V = 2 $CE^2$

Hence the capacitor's energy remains the same in both cases and the work done by the battery appears as heat in the connecting wires.

Now, Heat produced = H = 2 $CE^2$

$H=2 \times\left(5 \times 10^{-6}\right) \times 144$

$H=144 \times 10^{-5} \mathrm{J}=1.44 \mathrm{mJ}$