AC Voltage: RMS, Peak and What Your Meter Reads

Pick a waveform, a peak voltage and a load. The page works out the RMS, the power it actually delivers, and what an averaging multimeter would say about it — which for two of the three shapes is not the same number.

Worked example — AC Voltage: RMS, Peak and What Your Meter Reads

Peak voltage	325.2691 V
RMS voltage	230.0000 V
Mean of the rectified wave	207.0727 V
Crest factor, peak ÷ RMS	1.41421
Form factor, RMS ÷ mean	1.11072
Power into the load	529.000 W
What an averaging meter would read	230.0000 V
How wrong that is	0.0000%

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — AC Voltage: RMS, Peak and What Your Meter Reads

An alternating voltage has no useful average — over a whole cycle a symmetric wave spends as long negative as positive and the mean is zero. What is useful is the RMS, the root of the mean of the square, because squaring removes the sign and power depends on the square. P = V²ᵣₘₛ/R, so the RMS is exactly the DC voltage that would heat the same resistor identically, and that is what makes it the number worth putting on a label. This page takes the definition as read: Wasteland Signal derives it from first principles and shows that neither phase nor frequency enters. What is here instead is what follows from it — the crest factor, peak over RMS, and the form factor, RMS over rectified mean, which is the number an instrument needs and the number it usually guesses.

What each symbol means

V_pk: the peak, in volts. What the insulation, the switch and the transistor have to survive
V_rms: the root-mean-square, in volts. The DC voltage that would heat the same resistor identically, and the one on the label
|v|: the rectified wave. Its plain average is what an averaging meter actually measures, before it multiplies
crest: peak ÷ RMS. 1.41421 for a sine, 1 for a square, 1.73205 for a triangle
form: RMS ÷ mean. 1.110721 for a sine, which is the number a cheap meter applies to everything

Assumes: Symmetric waveforms with no DC offset, a purely resistive load, and an ideal averaging meter — one that rectifies perfectly and has unlimited bandwidth. A real one has neither: the diodes in its rectifier are not ideal near zero, and its bandwidth is usually a few kilohertz, so the errors quoted here are the ones that remain after the instrument is otherwise perfect. The three shapes are ideal too. A real inverter's output is not a clean square but a modified one with a gap at the zero crossing, and its form factor is different again.
Breaks when: On any load that is not a resistor. Power in an inductive or capacitive load is not V²ᵣₘₛ/R and not VᵣₘₛIᵣₘₛ either — current and voltage peak at different moments, and the ratio of the real power to their product is the power factor, which this page does not model at all. It breaks on a DC offset: the RMS of a wave sitting on a DC level is the root of the sum of the squares of the two parts, which is why a half-wave rectified supply cannot be described by anything above. And it breaks where the arithmetic is fine but the reading is not: a true-RMS meter has a crest factor limit, typically around 3, and a waveform spikier than that overloads its input long before the number it prints looks wrong.

Your meter multiplies by 1.110721 and hopes 🖖

An averaging multimeter — which is most of the cheap ones — does not measure RMS at all. It rectifies the wave, measures the plain average of that, and multiplies by 1.110721, which is the ratio of RMS to rectified mean for a sine. Hand it a sine and it is exactly right. Press A cheap inverter and it reads 361.2831 V where the truth is 325.2691 — 11.0721% high. Press A triangle, same peak and it reads 3.8088% low. The error is not noise and it does not average out: it is a fixed property of the shape, and the instrument has no way to detect which shape it is holding. That is what the words true RMS on a more expensive meter are being sold for.

The same peak, three times the power 🖖

Press the first three presets in order. The peak voltage is 325.2691 V in all of them and the power into the same 100 Ω is 529.000 W, 1,058.000 W and 352.667 W. Square against triangle is a factor of exactly 3, at identical height. Heating does not care how tall a waveform gets; it cares about the average of the square of it, and a square wave spends all its time at full height while a triangle spends almost none. This is also why the peak is the wrong number to design insulation from and the wrong number to design a heater from — they are different questions and they have different answers.

The socket says 230 and the capacitor sees 325 🖖

Both numbers are real and they are for different readers. 230 V is what heats a kettle, and 325.2691 V is what the insulation, the switch contacts and the semiconductor have to survive, twice per cycle, fifty times a second. The gap matters most where a capacitor is involved, because a capacitor charges towards the peak and stays there: a transformer labelled 12 V hands a smoothing capacitor 16.9706 V before any load is connected. The worked problem below is a component chosen from the label that fails anyway, and it fails for exactly this reason.

Problem solved in full

The 16 V capacitor on a 12 V transformer that fails anyway 6 steps
A project needs 12 V DC. You buy a transformer labelled 12 V AC, rectify it with a bridge and smooth it with a capacitor rated at 16 V. It works on the bench and fails in the field. Work out why before reading on — press A 12 V transformer to see the first step on the tool.

Open this problem in the tool
1. The label is an RMS figure, which is what a transformer's AC rating always means. The peak is √2 times it.
  
  $V_{m a x} = 2 V_{rms} = 1.41421 \cdot 12 = 16.9706 V$
2. A smoothing capacitor charges towards the peak and stays near it, so 16.9706 V is what it sees, not 12.
  
  $V_{C} \to V_{m a x} = 16.9706 V$
3. A bridge rectifier costs two diode drops on the way, about 0.7 V each.
  
  $2 \times 0.7 = 1.4 V$
4. So the capacitor sits at 15.571 V unloaded. That is under its 16 V rating — by 2.68%, which is the entire margin the design has.
  
  $16.9706 - 1.4 = 15.571 V \frac{15 . 571}{16} = 0.9732$
5. Now the part that only shows up in the field. Mains is specified with a tolerance, commonly +10%, and the transformer's output follows it: 13.2 V RMS instead of 12.
  
  $12 \times 1.10 = 13.2 V$
6. Redo the same three steps at 13.2 V and the capacitor is over its rating, on a day nothing was wrong with it.
  
  $2 \cdot 13.2 - 1.4 \div 16 18.6676 V 17.268 V 1.0792$
Answer

17.268 V across a 16 V capacitor, which is 7.9% over its rating, on a perfectly ordinary high-mains day. Nothing in the design was miscalculated — the error was reading a label as if it described the voltage the part would see, when it describes the heating equivalent of a wave whose peak is 41.4% higher. The working rule that comes out of it is to pick the capacitor from the peak at the top of the supply tolerance and then leave margin on top of that, which is why a 12 V rail is usually smoothed by a 25 V part and never by a 16 V one.

$\frac{17 . 268}{16} = 1.0792 \Rightarrow + 7 . 9%$

Learning path

What Ohm’s law leaves out

References (2)

Insight 1 is about a measurement error, and the metrology literature treats the square wave as the standard case for exactly that reason — it is the waveform a true-RMS instrument is checked against: P. M. Tzvetkov and K. S. Galabov, "True RMS voltmeter calibration by reference square waveform signal and algorithm for measurement and processing of results." 2021 XXXI International Scientific Symposium Metrology and Metrology Assurance (MMA), 2021, pp. 1-4. DOI 10.1109/mma52675.2021.9610853
The derivation this page deliberately does not repeat lives on another tool in this repo rather than in a citation, and is worth naming as a source in its own right: Wasteland Signal works RMS out from the amplitudes and shows that phase and frequency never enter. The RMS definition, its cross-product argument and the independence from phase: see /tidal/wasteland-signal on this site, whose worked problem derives the whole of it.

By the 42Math team · Reviewed against published sources. Corrections welcome.