An initially uncharged capacitor C is fully charged by a constant emf in series with a capacitor R. (a) We have to show that the final energy stored in the capacitor is half the energy supplied by the emf. (b) By direct integration of over the charging time, we have to show that the internal energy dissipated by the resistor is also half the energy supplied by the emf.

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(a)

The charging equation for a capacitor of capacitance C connected with a resistor or resistance R to a source of emf is

By differentiating with respect to the variable t, we will obtain the current as a function of time, , during the charging process. That is

The energy supplied by the emf in charging the capacitor fully will be given by the integral

The final energy stored in the capacitor is

Therefore, it is half the energy supplied by the source of emf.

(b)

We next calculate the internal energy dissipated in the resistor.

Joule heat per second is

.

Therefore, the total energy dissipated in the resistor during the charging process will be given by integrating over the charging time. That is

This is also equal to half the energy supplied by the emf in charging the capacitor.