7.5 AC Voltage applied to a Capacitor

Figure 7.8 shows an ac source $\epsilon$ generating ac voltage $v=v_{m}sin\omega t$ connected to a capacitor only, a purely capacitive ac circuit.

When a capacitor is connected to a voltage source in a dc circuit, current will flow for the short time required to charge the capacitor. As charge accumulates on the capacitor plates, the voltage across them increases, opposing the current. That is, a capacitor in a dc circuit will limit or oppose the current as it charges. When the capacitor is fully charged, the current in the circuit falls to zero

When the capacitor is connected to an ac source, as in Fig. 7.8, it limits or regulates the current, but does not completely prevent the flow of charge. The capacitor is alternately charged and discharged as the current reverses each half cycle. Let q be the charge on the capacitor at any time t. The instantaneous voltage v across the capacitor is

$$v=\frac{q}{C}$$

From the Kirchhoff’s loop rule, the voltage across the source and the capacitor are equal,

$$v_{m}sin\omega t=\frac{q}{C}$$

To find the current, we use the relation $i=\frac{dq}{dt}$

$$i=\frac{d}{dt}(v_{m}Csin\omega t)=\omega C{v}_{m}cos(\omega t)$$

Using the relation, $cos(\omega t)=sin(\text{omegat}+\frac{pi}{2})$, we have

$$i=i_{m}sin(\omega t+\frac{\pi }{2})$$

where the amplitude of the oscillating current is $i_m=\omega C{v}_m$ . We can rewrite it as

$$i_m=\frac{v_m}{(1/\omega C)}$$

Comparing it to $i_m=v_m/R$ for a purely resistive circuit, we find that $(1/\omega C)$ plays the role of resistance. It is called capacitive reactance and is denoted by $X_c$ ,

$$X_c=1/\omega C$$

so that the amplitude of the current is

$$i_m=\frac{v_m}{X_c}$$

The dimension of capacitive reactance is the same as that of resistance and its SI unit is ohm ($\omega$). The capacitive reactance limits the amplitude of the current in a purely capacitive circuit in the same way as the resistance limits the current in a purely resistive circuit. But it is inversely proportional to the frequency and the capacitance.

A comparison of Eq. (7.16) with the equation of source voltage, Eq. (7.1) shows that the current is $\pi /2$ ahead of voltage. Figure 7.9(a) shows the phasor diagram at an instant $t_1$ . Here the current phasor I is $\pi /2$ ahead of the voltage phasor V as they rotate counterclockwise. Figure 7.9(b) shows the variation of voltage and current with time. We see that the current reaches its maximum value earlier than the voltage by one-fourth of a period.

The instantaneous power supplied to the capacitor is

$$p_c=iv=i_{m}cos(\omega t)v_{m}sin(\omega t)$$ $$=i_m{v}_{m}cos(\omega t)sin(\omega t)$$ $$=\frac{i_m{v}_m}{2}sin(2\omega t)$$

So, as in the case of an inductor, the average power

$$?$$

since $sin(2\omega t)=0$ over a complete cycle. Figure 7.10 explains it in detail. Thus, we see that in the case of an inductor, the current lags the voltage by $\pi /2$ and in the case of a capacitor, the current leads the voltage by $\pi /2$.