Question: An initially uncharged capacitor C is fully charged by a device of constant emf connected in series with a resistor. R (a) Show that the final energy stored in the capacitor is half the energy supplied by the emf $\epsilon$device. (b) By direct integration of $i^{2}R$over the charging time, show that the thermal energy dissipated by the resistor is also half the energy supplied by the emf device.

An initially uncharged capacitor C is fully charged by a device of constant emf connected in series with a resistor R .

As the capacitor in an RC circuit is being charged, some energy supplied by the emf device remains within the plates of the capacitor C and also goes to the resistor as thermal energy.

Formulae:

The charge within the capacitor as a function of time,

$q=q_0\left(1-e^{-t/RC}\right)$ (i)

The charge stored within the plates of a capacitor,

q=CV (ii)

The rate at which the emf device suppliers energy,

$P_{\zeta }=i\epsilon$ (iii)

The current flowing in a device,

$i=\frac{dq}{dt}$ (iv)

The energy stored within the capacitor plates,
$U_c=\frac{1}{2}C{\epsilon }^2$ (v)

Now, substituting equation (ii) value in equation (i) for the charge as a function of time for a capacitor, the equation is given as:

$q=C\epsilon \left(1-e^{-t/RC}\right)$

Now, the charging current within the device can be given using equation (iv) and the above value as follows:

$$\begin{gathered} i=\frac{d}{dt}C\epsilon \left(1-e^{-t/RC}\right) \\ =\frac{\epsilon }{R}{e}^{-t/RC} \end{gathered}$$(a)

Thus, the energy supplied to the emf device is given using the above value (a) in equation (iii) and integrating the power value with time as follows:
$U={\int }_0^{\infty }{P}_{z}dt$

Substitute the values in the above expression, and we get,

$$\begin{gathered} U={\int }_0^{\infty }\epsilon \left(\frac{\epsilon }{R}{e}^{-t/RC}\right)dt \\ =\frac{{\epsilon }^2}{R}{\int }_0^{\infty }\epsilon \left(e^{-t/RC}\right)dt \\ =\frac{{\epsilon }^2}{R}{\left[\frac{e^{-t/RC}}{\left(-1/RC\right)}\right]}_0^{\infty } \\ =\frac{{\epsilon }^2}{R}\left(-RC\right)\left[0-1\right] \\ =C{\epsilon }^2 \end{gathered}$$

Substitute the values in the above expression, and we get,

$U=2U_c$

Hence, from the above-integrated value, it can be seen that the energy supplied by the emf device is equal to twice the capacitor store energy.

Now, simply integrating the given rate valueand $i^{2}R$using equation (a), the rate at which the thermal energy is transferred to the resistor is given as follows:

Substitute the values in the above expression, and we get,

$$\begin{gathered} U_R={\int }_0^{\infty }\epsilon \left(\frac{\epsilon }{R}{e}^{-t/RC}\right)Rdt \\ =\frac{{\epsilon }^2}{R}{\int }_0^{\infty }\left(e^{-2t/RC}\right)dt \\ =\frac{{\epsilon }^2}{R}{\left[\frac{e^{-2t/RC}}{\left(-2/RC\right)}\right]}_0^{\infty } \\ =\frac{{\epsilon }^2}{R}\left(\frac{RC}{2}\right)(0-\text{'}1) \\ =\frac{C{\epsilon }^2}{2} \end{gathered}$$

Substitute the values in the above expression, and we get,

$U_R=\frac{U_c}{2}$

Hence, in the above integration, it can be seen that the thermal energy by the resistor is also half the energy supplied by the emf device.