Question

An initially discharged capacitor C is fully charged by a constant emf in series with a resistor R. (a) Show that the final energy stored in the capacitor is half of the energy supplied by the emf. (b) By direct integration of i2R in the charge time, show that the internal energy dissipated by the resistor is also half the energy supplied by the emf.

Answer 1

A.

The energy supplied by the emf we know - QV

and energy stored in the capacitor -

use Q=CV

B.

We know energy dissipated across the resistance

where

so total power dissipated across the resistance

Half of the power generated by EMF

Proved