# Why V rms instead of V average?

I'm looking at an equation for average power in a signal

$$p_{avg} = \frac{1}{R} v_{rms}^2$$

and wondering why it isn't

$$p_{avg} = \frac{1}{R} |v|_{avg}^2$$

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Because the square of the average is not always the average of the squares, not even for positive numbers. 0 and 10 average to 5, square that to get 25. But the average of their squares (0 and 100) is 50. Not even close! Why the square in the first place? Power is Voltage * current, but the current is itself proportional to the Voltage, so the power is proportional to the Voltage squared. –  Wouter van Ooijen Sep 14 '12 at 20:14

Simple: the average of a sine is zero.

Power is proportional to voltage squared:

$P = \dfrac{V^2}{R}$

so to get average power you calculate average voltage squared. That's what the RMS refers to: Root Mean Square: take the square root of the average (mean) of the squared voltage. You have to take the square root to get the dimension of a voltage again, since you first squared it.

This graph shows the difference between the two. The purple curve is the sine squared, the yellowish line the absolute value. The RMS value is $\sqrt{2}/2$, or about 0.71, the average value is $2/\pi$, or about 0.64, a difference of 10 %.

RMS gives you the equivalent DC voltage for the same power. If you would measure the resistor's temperature as a measure of dissipated energy you'll see that it's the same as for a DC voltage of 0.71 V, not 0.64 V.

edit
Measuring average voltage is cheaper than measuring RMS voltage however, and that's what cheaper DMMs do. They presume the signal is a sine wave, measure the rectified average and multiply the result by 1.11 (0.71/0.64) to get the RMS value. But the factor 1.11 is only valid for sinewaves. For other signals the ratio will be different. That ratio got a name: it's called the signal's form factor. For a 10 % duty cycle PWM signal the form factor will be $1/\sqrt{10}$, or about 0.316. That's a lot less than the sine's 1.11. DMMs which are not "True RMS" will give large errors for non-sinusoidal waveforms.

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To your first point, I edited my second equation to use the average absolute value, which is what I meant. What I'm not seeing is why the order of the two operations (average and square) matters. Average voltage squared, vs average squared voltage. –  Rob N Sep 14 '12 at 17:13
Because of the square-law relationship the average of the power and the average of the voltage are two very different things. –  Dave Tweed Sep 14 '12 at 17:27
@RobN, the instantaneous power is $p(t) = v^2(t)/R$. The average power is the time average of $p(t)$. Thus, the average power is proportional to the time average of the squared voltage. Also, the order matters because the average of the squares is not equal to the square of the average. –  Alfred Centauri Sep 14 '12 at 19:33
Note that the average of the square of a sine is one half. The inverted and phase shifted curve fits exactly into the valleys in the original curve, a consequence of Pythagoras' law, and their sum is a constant 1. –  starblue Sep 14 '12 at 19:43

Now speaking in terms of equations:

$P_{avg}= avg(P_{inst})$

Now $P_{inst} = v(t) \cdot i(t)$ where $v(t)$ and $i(t)$ are instantaneous voltage and current resp. Hence

$P_{inst} = \dfrac{(v(t))^2}{R}$

$P_{avg} = avg(\dfrac{((v(t))^2}{R})$

$P_{avg}= \dfrac{V_{rms}^2}{R}$

As RMS = $\sqrt{\text{average of squares of inst.}}$

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The why is simple.

You want 1 W = 1 W.

Imagine a primitive heater, a 1 ohm resistor.

Consider 1 VDC into a 1 ohm resistor. Power consumption is obviously 1 W. Do that for one hour, and you burn one watt-hour, generating heat.

Now, instead of DC, you want to feed AC to the resistor, and produce the same heat. What AC voltage do you use?

It turns out that RMS voltage gives you the result you want.

THAT'S why RMS is defined the way it is, to make the power numbers come out right.

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