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# 13.2: Alternating Voltage across a Capacitor

FIGURE $$\text{XIII.3}$$

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 $$I=C\dot V$$.

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 90o, as shown in figure XIII.4.

FIGURE $$\text{XIII.4}$$