Question

A capacitor of capacitance C is charged by connecting it to a battery of e.m.f. E volts. The capacitor is now disconnected and reconnected to the same battery with polarity reversed. The heat energy developed in the connecting wire is:

• A. $C{E}^2$
• B. Zero.
• C. $\frac{1}{2}C{E}^2$
• D. $2C{E}^2$

Answer 1

Answer: (D)

$\Delta W=\Delta U+\Delta H$
Work done by battery,
$\therefore W=2CE\times E$
Change in energy stored in capacitor,
$\Delta U=0$
Thus, Heat dissipated,
$\Delta H=2C{E}^2$