# Why use RMS instead of the average voltage of half the cycle for power

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.

• Why not use the average voltage instead of RMS to compute the power (into a resistive load)? Because you get the wrong answer. – John D Aug 31 '18 at 0:48
• 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.... – user253751 Aug 31 '18 at 1:04
• Because $\frac{V_m}{\sqrt 2}.\frac{I_m}{\sqrt 2}=\frac{V_mI_m}{2}=P=$ average power. – Chu Aug 31 '18 at 6:53

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.