91Ó°ÊÓ

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,

$P_{\mathrm{Î\P }}=i\mathrm{Î\mu }$ (iii)

The current flowing in a device,

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

The energy stored within the capacitor plates,
$U_c=\frac{1}{2}C{\mathrm{Î\mu }}^2$ (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:

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

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

$$\begin{gathered} i=\frac{d}{dt}C\mathrm{Î\mu }\left(1-e^{-t/RC}\right) \\ =\frac{\mathrm{Î\mu }}{R}{e}^{-t/RC} \end{gathered}$$(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:
$U={∫}_0^{∞}{P}_{z}dt$

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

$$\begin{gathered} U={∫}_0^{∞}\mathrm{Î\mu }\left(\frac{\mathrm{Î\mu }}{R}{e}^{-t/RC}\right)dt \\ =\frac{{\mathrm{Î\mu }}^2}{R}{∫}_0^{∞}\mathrm{Î\mu }\left(e^{-t/RC}\right)dt \\ =\frac{{\mathrm{Î\mu }}^2}{R}{\left[\frac{e^{-t/RC}}{\left(-1/RC\right)}\right]}_0^{∞} \\ =\frac{{\mathrm{Î\mu }}^2}{R}\left(-RC\right)\left[0-1\right] \\ =C{\mathrm{Î\mu }}^2 \end{gathered}$$

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,

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

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.