Question

$A5.0\mu F$ capacitor is charged to 12 V. The positive plate of this capacitor is now connected to the negative terminal is now connected to the negative terminal of a 12 V batter and vice versa. Calculate the heat developed in the connecting wires.

Answer 1

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 1st plate and + Q on the 2nd. A charge 2Q, therefore passes through the battery from the -ve to the +ve terminal. The battery does a work.

$W=Q\times E=2QE=2C{E}^2$

In this process, the energy stored in the capacitor is the same in the capacitor is the same in the two cases. Thus the work done by the battery appears as heat in the connecting wires. Heat produced-is therefore,

$2C{E}^2=2\times 5\times {10}^{-6}\times 144$

$[haveC=5\mu F,v=E=12V]$

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