A generator makes an emf that swings back and forth as $\sin\omega t$, and that is what comes out of a wall outlet. A voltage that spends half its time negative raises a question: how do you describe its "size", and how much power does it deliver? The answer is the rms value, the number printed on every appliance.
Key idea
$V = V_0\sin\omega t$. Its rms (root-mean-square) value is
$$V_{\text{rms}} = \frac{V_0}{\sqrt{2}},$$
and likewise $I_{\text{rms}} = I_0/\sqrt{2}$. A resistor's average power is $\bar{P} = I_{\text{rms}}V_{\text{rms}} = V_{\text{rms}}^2/R$, the same as a steady voltage of $V_{\text{rms}}$ would give.
Why it works
Power in a resistor is $V^2/R$, and $V^2 = V_0^2\sin^2\omega t$ is never negative, so the power is positive all the time even though $V$ averages to zero. Over a whole cycle, $\sin^2$ and $\cos^2$ average to the same value, and they add to 1 at every instant, so each averages to $\tfrac{1}{2}$. Then $\bar{P} = \dfrac{V_0^2}{2R} = \dfrac{(V_0/\sqrt{2})^2}{R}$. Squaring, averaging (mean) and taking the root is exactly how $V_{\text{rms}}$ is defined, which is where the name comes from.
Worked examples
Worked example: A household outlet
A North American outlet supplies $120\ \text{V}$ rms. Find its peak voltage, and the current and resistance of a $100\ \text{W}$ light bulb on it.
- $V_0 = \sqrt{2}(120) = 170\ \text{V}$: the voltage swings between about $\pm 170\ \text{V}$.
- $I_{\text{rms}} = \bar{P}/V_{\text{rms}} = 100/120 = 0.833\ \text{A}$, and $R = V_{\text{rms}}^2/\bar{P} = 144\ \text{Ω}$.
Answer: $170\ \text{V}$ peak; $0.833\ \text{A}$ rms
Worked example: Average and peak power
A source with peak voltage $50\ \text{V}$ drives a $20\ \Omega$ resistor. Find the average and the largest power.
- $\bar{P} = \dfrac{V_0^2}{2R} = \dfrac{50^2}{2(20)} = 62.5\ \text{W}$.
- At the peaks, $P = V_0^2/R = 125\ \text{W}$: twice the average.
Answer: $62.5\ \text{W}$ average, $125\ \text{W}$ peak
Worked example: A triangle wave
A triangle wave rises and falls in straight lines between $-V_0$ and $+V_0$. Find its rms value.
- Over the first quarter cycle $V = V_0\dfrac{4t}{T}$; the other quarters are mirror images with the same values of $V^2$.
- $$\overline{V^2} = \frac{4}{T}\int_0^{T/4}V_0^2\frac{16t^2}{T^2}\,dt = \frac{V_0^2}{3}$$
- $V_{\text{rms}} = V_0/\sqrt{3} \approx 0.577V_0$, a bit less than a sine wave's $0.707V_0$. The factor $1/\sqrt{2}$ belongs to sine waves only.
Answer: $V_{\text{rms}} = V_0/\sqrt{3}$
Common mistake
"AC averages to zero, so it delivers no power." The voltage averages to zero, but the power depends on $V^2$, which is positive whichever way the current flows: a toaster heats up on both halves of each cycle. Check: average $V^2$, not $V$, and you get $V_0^2/2$.