AC values

The idea

Mains electricity is a sine wave: it pushes current one way, then the other, 50 times a second in the UK. The “230 V” printed on everything isn’t the top of that wave. It’s the RMS value, the steady DC voltage that would heat a kettle by the same amount. The wave itself peaks at 325 V, and any part connected to mains has to survive that peak.

The three numbers

Symbol	Name	What it is	UK mains
$V_{pk}$	Peak	The top of the wave	325 V
$V_{RMS}$	RMS	The heating-equivalent value. What meters and labels quote	230 V
$V_{avg}$	Average	The mean height of one hump of the wave	207 V

Convert	Multiply by
RMS → peak	$\sqrt{2}\approx 1.41$
Peak → RMS	$1/\sqrt{2}\approx 0.707$
Peak → average	$2/\pi \approx 0.637$
RMS → average	$0.9$

Where the numbers come from

RMS means root of the mean of the square. We square because heating power goes as $V^2$.

1. The wave is $V=V_{pk}\sin \theta$, so $V^2=V_{pk}^2{\sin }^2\theta$.
2. Over a whole cycle, ${\sin }^2\theta$ averages to exactly $\frac{1}{2}$.
3. So the mean of the square is $\frac{1}{2}{V}_{pk}^2$, and its root is

$V_{RMS}=\frac{V_{pk}}{\sqrt{2}}\Rightarrow V_{pk}=\sqrt{2}\times 230=325.3\text{V}$

Average of one hump = area ÷ width. The area under $\sin \theta$ from $0$ to $\pi$ is exactly 2, and the width is $\pi$:

$V_{avg}=\frac{2}{\pi }{V}_{pk}=0.637V_{pk}\frac{V_{avg}}{V_{RMS}}=\frac{0.637}{0.707}=0.9$

The same ratios apply to current through a resistive load, so $I_{avg}=0.9I_{RMS}$. You’ll need that for bridge losses.

Worked example

A robot’s charger is plugged into a 110 V RMS (US) supply. What peak voltage must its rectifier survive, and what’s the average of each hump?

Answer

$V_{pk}=1.41\times 110=155.6$ V. $V_{avg}=0.637\times 155.6=99.1$ V (or $0.9\times 110=99$ V). A diode rated $V_{RRM}=400$ V would be fine. At 230 V you’d want 600–800 V.

Common mistakes

• Using 230 V where the question needs the peak. Capacitors charge to the peak, and diodes must block the peak.
• Using RMS current for a diode’s loss. Diodes use average current (see Bridge rectifiers).
• Trusting a cheap multimeter on PWM. Ordinary meters only read RMS correctly for a clean 50 Hz sine. Use a true-RMS meter or a scope.

Check yourself

1. A sine wave has a peak-to-peak voltage of 40 V. What's its RMS value?

Peak = 20 V, so RMS = 0.707 × 20 = 14.1 V.

2. Why is RMS defined with a square?

Because heating power in a resistor is $V^2/R$. Averaging $V^2$ gives a voltage that delivers the same power as DC.