# 10.6: AC Power

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When a current I flows through a resistance \(R\), the rate of dissipation of electrical energy as heat is \(I R^2\). If an alternating potential difference \(V = \hat{V} \sin \omega t \) is applied across a resistance, then an alternating current \(I = \hat{I} \sin \omega t \) will flow through it, and the rate at which energy is dissipated as heat will also change periodically. Of interest is the *average* rate of dissipation of electrical energy as heat during a complete cycle of period \(P = 2\pi /\omega\).

Let \(W\) = instantaneous rate of dissipation of energy, and \(\overline W\) = average rate over a cycle of period \(P = 2π /\omega\). Then

\[\begin{align} \overline W P = \int_0^P W\, dt &= R\int_0^P I^2\,dt \\[5pt] &= R\hat{I}^2\int_0^P \sin^2 \omega t\,dt \\[5pt] &= \dfrac{1}{2}R\hat{I}^2\int_0^P (1-\cos 2\omega t)\,dt \\[5pt] &= \dfrac{1}{2}R\hat{I}^2 \left[t-\dfrac{1}{2\omega}\sin^2 \omega t \right]_0^{P=2\pi/\omega } \\[5pt] &= \dfrac{1}{2}R\hat{I}^2 P.\label{10.6.1} \end{align}\]

Thus

\[\overline W = \frac{1}{2}R\hat{I}^2\label{10.6.2}\]

The expression \(\frac{1}{2}\hat{I}^2\) is the mean value of \(I^2\) over a complete cycle. Its square root \(\hat{I}/\sqrt{2}=0.707\hat{I}\) is the root mean square value of the current, \(I_{RMS}\). Thus the average rate of dissipation of electrical energy is

\[\label{10.6.3}\overline W = RI_{RMS}^2.\]

Likewise, the RMS EMF (pardon all the abbreviations) over a complete cycle is \(\hat{V}/\sqrt{2}\).

Often when an AC current or voltage is quoted, it is the RMS value that is meant rather than the peak value. I recommend that in writing or conversation you always make it *explicitly clear* which you mean.

Also of interest is the *mean* induced voltage \(\overline V\) over half a cycle. (Over a full cycle, the mean voltage is, of course, zero.) We have

\[\label{10.6.4}\overline V P/2=\int_0^{P/2}V\,dt=\hat{V}\int_0^{P/2}\sin \omega t\,dt = \dfrac{\hat{V}}{\omega} \left[\cos \omega t\right]_{\frac{P}{2}=\frac{\pi}{2}}^0 = \dfrac{\hat{V}}{\omega}(1-\cos \pi)=\dfrac{2\hat{V}}{\omega}.\]

Remembering that \(P = 2π /\omega\), we see that

\[\label{10.6.5}\overline V = \dfrac{2\hat{V}}{\pi}=0.6366\hat{V} = \dfrac{2\sqrt{2}V_{RMS}}{\pi}=0.9003 V_{RMS}.\]