13.2: Alternating Voltage across a Capacitor

At any time, the charge $Q$ on the capacitor is related to the potential difference $V$ across it by $Q=CV$. If there is a current in the circuit, then $Q$ is changing, and $I=C\dot V$.

FIGURE $\text{XIII.3}$

Now suppose that an alternating voltage given by

$$\label{13.2.1}V=\hat{V}\sin \omega t$$

is applied across the capacitor. In that case the current is

$$I=C\omega \hat{V}\cos \omega t,\label{13.2.2}$$

which can be written

$$I=\hat{I}\cos \omega t,\label{13.2.3}$$

where the peak current is

$$\hat{I}=C\omega \hat{V}\label{13.2.4}$$

and, of course

$$I_{\text{RMS}}=C\omega V_{\text{RMS}}.$$

The quantity $1/(C\omega)$ is called the capacitive reactance $X_C$. It is expressed in ohms (check the dimensions), and, the higher the frequency, the smaller the reactance. (The frequency $\nu\text{ is }\omega /(2\pi)$.)

When we come to deal with complex numbers, in the next and future sections, we shall incorporate a sign into the reactance. We shall call the reactance of a capacitor $-1/(C\omega)$ rather than merely $1/(C\omega)$, and the minus sign will indicate to us that $V$ lags behind $I$. The reactance of an inductor will remain $L\omega$, since $V$ leads on $I$.

Comparison of equations $\ref{13.2.1}$ and $\ref{13.2.3}$ shows that the current and voltage are out of phase, and that $V$ lags behind $I$ by 90°, as shown in Figure XIII.4.

FIGURE $\text{XIII.4}$