Subjective Type

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.

Solution

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

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

$$W=H=2CE^2$$

$$=2\times 5\times 10^{-6}\times (1 1/2)^2$$

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

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