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# 10.6: AC Power

• • Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

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

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

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