ALTERNATING VOLTAGE APPLIED TO A CAPACITOR
Alternating source connected to a capacitor of capacitance C. Such a circuit is known as purely capacitive circuit. The capacitor is periodically charged and discharged when alternating voltage is applied to it.
The alternating voltage applied across the capacitor is given by
$V = V_0 \sin \omega t \quad 1$
Let q be the charge on the capacitor at any instant.
Then, potential difference across the capacitor,
$V_C = \frac{q}{C}$
But
$V_C = V \quad \text{or} \quad \frac{q}{C} = V = V_0 \sin \omega t$
$\therefore q = V_0 C \sin \omega t$
Now,
$I = \frac{dq}{dt} = \frac{d}{dt}(V_0 C \sin \omega t)$
$I= V_0 C (\cos \omega t) \omega$
$I= \frac{V_0}{(1 / C \omega)} \cos \omega t$
Since
$\cos \omega t = \sin \left(\omega t + \frac{\pi}{2}\right)$
$I= \frac{V_0}{(1 / C \omega)} \sin \left(\omega t + \frac{\pi}{2}\right)$
$\therefore I = I_0 \sin \left(\omega t + \frac{\pi}{2}\right) \quad ...(2)$
where
$I_0 = \frac{V_0}{(1 / C \omega)}$
is the peak value of a.c.
Comparison of eqns. (i) and (ii) shows that current leads the e.m.f. by an angle \( \pi/2 \) in a purely capacitive a.c. circuit as shown in Figure
The capacitive reactance is the effective opposition offered by a capacitor to the flow of current in the circuit.
Comparing
$I_0 = \frac{V0}{(1 / C \omega)}$
with
$I_0 = \frac{V_0}{R}$
we conclude that \(\left( \frac{1}{C \omega} \right)\) has the dimensions of resistance. The term \((1 / C \omega)\) is known as Capacitive reactance ($X_C$).
Thus, capacitive reactance,
$X_C = \frac{1}{C \omega}$
Power Supplied to a Capacitor :
The voltage and current in an a.c. circuit having capacitor only are given by
$V = V_0 \sin \omega t \quad ...(1)$
and
$I = I_0 \sin \left(\omega t + \frac{\pi}{2}\right) \quad ...(2)$
Instantaneous Power Supplied to Capacitor :
$P_C = VI = (V_0 \sin \omega t) \left[ I_0 \sin \left(\omega t + \frac{\pi}{2}\right) \right]$
$P_C= V_0 I_0 \sin \omega t \cdot \sin \left(\omega t + \frac{\pi}{2}\right)$
Since
$\sin \left(\omega t + \frac{\pi}{2}\right) = \cos \omega t$
$\therefore P_C = V_0 I_0 \cos (\omega t) \sin (\omega t)$
$P_C =\frac{2}{2}. V_0 I_0 \cos (\omega t) \sin (\omega t)$
$P_C= \frac{V_0 I_0}{2} \sin 2 \omega t$
$(\because 2 \cos \omega t \sin \omega t = \sin 2 \omega t)$
Average Power Supplied to Capacitor :
$\overline{P_C} = \text{average of } P_C \text{ for full cycle}$
$= \text{average of } \left( \frac{V_0 I_0}{2} \sin 2 \omega t \right)$
$= \frac{V_0 I_0}{2} \cdot \text{average of } (\sin 2 \omega t)$
Since average of \(\sin 2 \omega t\) for a full cycle is zero,
$\overline{P_C}= 0$ for full cycle