# 13.2: Alternating Voltage across a Capacitor

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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 \(\text{XIII.4}\)