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In a AC sine wave, the RMS voltage value is \$\frac{V_{p}}{\sqrt{2}}\$ where \$V_{p}\$ is the peak voltage whereas the average voltage value over half the cycle is \$V_{p}\cdot \frac{2}{\pi}\$

If we take the average of the entire sine wave the result would yield 0, since both half's of the sine wave cancel each other out. In order to fix this issue we've come up with the RMS value.

What I don't get is why bother using RMS, can't we just use the average voltage from half the cycle to compute the power.

Use \$P = \frac{V_{avg}^{2}}{R}\$ instead of \$P = \frac{V_{RMS}^{2}}{R}\$

I've heard that the RMS voltage value is the same as the DC voltage for calculating power. But why RMS? And not the average voltage value for half the cycle.

Any help is grately appreciated.

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    \$\begingroup\$ Why not use the average voltage instead of RMS to compute the power (into a resistive load)? Because you get the wrong answer. \$\endgroup\$
    – John D
    Aug 31, 2018 at 0:48
  • \$\begingroup\$ Well that gives you a different amount of power, doesn't it? And if you measure the power, you will find that it gave you the wrong amount of power.... \$\endgroup\$
    – user253751
    Aug 31, 2018 at 1:04
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    \$\begingroup\$ Because \$\frac{V_m}{\sqrt 2}.\frac{I_m}{\sqrt 2}=\frac{V_mI_m}{2}=P=\$ average power. \$\endgroup\$
    – Chu
    Aug 31, 2018 at 6:53

2 Answers 2

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They're not the same, in general. For a sinusoidal waveform, the difference is about 11%. The square root of the average of the square of a function is not the same as the average of a function.


The average over a half cycle is 2Vp/pi = 0.6366 Vp

The RMS is Vp/sqrt(2) = 0.707107 Vp


The ratio is 1.11.

The ratio will be different for different waveforms, but in general the RMS will always be higher. RMS gives you the power transfer into a resistive load, average does not (except for the degenerate case of DC).

In some cases you may actually want the average (but not for power into a resistive load).

Cheap multimeters often just measure the average of the rectified waveform and read 11.1% high, assuming a sinusoidal waveform.

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  • \$\begingroup\$ Just to add a thing: the simple average does not account for those moments when the voltage is higher than the mean. But they are there, and they produce more current so, speaking about power, they count "twice", or "more". \$\endgroup\$ Jan 15 at 7:23
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It's all about power. We want 100 volts AC and 100 volts DC to heat a resistor to the same temperature. A worthy goal. The RMS formula gives us that "corrected" AC voltage. Average voltage and peak voltage are useless in terms of POWER. We are about to find out why.

Our analysis need only consider a quarter of a cycle. The other three-quarters yield the same area under the curve, the same squares, the same power.

Power is a function of the square of the voltage (Power=V²/R). Finding the square of a DC voltage is easy. How do we find the square of a voltage changing as a sine wave. You chop up the sine wave and square the voltage at each slice. Graphically, you can take each voltage slice and draw it as a square shooting out into the z-axis. Add up all the square areas, then divide by the number of squares. That gives you the average square area (mean of the squares) for that sine wave with a peak Vp. Now we take the square root of that average square area to get the voltage that produces it. And that is, in simplified graphical terms, the Vrms formula. We found the average sized square for the power formula and from that found the length of the edge of the square (Vrms).

enter image description here

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