Question

A $5.0\mu F$ capacitor is charged to 12 V. The positive 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

Work done by battery$,w=Q_{total}\times E=2QE=2C{E}^2$

in this process, the energy stored in the capacitor is same in two cases.

Thus, work done by battery appears as heat is connecting wires.

$\therefore$ Heat produced $=$ work done by battery

$W=H=2C{E}^2$

$=2\times 5\times {10}^{-6}\times (11/2{)}^2$

$=1.44\times {10}^{-5}J$.