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 \(\text{XIII.4}\)