Why V rms instead of V average?

Question

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

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

Answer 1

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\$ (i.e. \$\frac{1}{\sqrt{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.

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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.

Answer 2

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 \left(\dfrac{((v(t))^2}{R} \right) \$

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

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

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

Answer 3

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.

Answer 4

Because the power equal to V^2/R so that you calculate the average of the squared voltages along the sinusoidal wave to get V^2avg. For simplicity we take the average of this mean then we can deal with it as we wish.

Answer 5

The answer is the reason given by John R. Strohm and the explanation is as follows: (requires a few additions to stevenvh's answer)

You see when you send a DC through a resistor and an AC wave through a resistor the resistor get heated up in both cases, but according to the equation for the average value the heating effect for ac should be 0 but its not why? This is because the when the electrons move in a conductor they strike atoms and this energy imparted to the atoms are consequently felt as heat, now AC does the same thing only the the electrons are moving in different directions but the energy transfer here is independent of the direction and so the conductor heats up all the same.

When we find the average value the ac components are cancelled out and hence fail to explain why the heat is generated but RMS equation rectifies that - as stevenvh says by taking the square and then the square root we are transposing the negative portion to the top of the axis such that the positive and negative portions don't cancel off.

This is why that we say that the average and the RMS values of a DC wave is the same.

The same applies to any real world signal (by this I mean imperfect - not pure AC) as Fourier series says that any wave can be replaced by a correct combination of sine and cosine waves and since the frequencies of the waves are higher (integer multiples of the base frequency) they too get cancelled out, isolating the DC component.

The above is the reason that we define RMS value as the equivalent value of DC that generates the same amount of heat as the AC wave.

Hope this helps.

PS: I know that the explanation for how heat is generated is quite ambiguous but I am at loss to find a better one, I went with it anyway because it helps convey the message

Answer 6

Where did the Vrms equation come from? Why does it work?

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.

By definition we are looking for equivalent AC and DC 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, 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 area (mean of the squares) for that sine wave with a peak Vp.

Now that we have the average square area, we take the square root of it to get the voltage that gives us that average square area and... oh my god we've derived the Vrms formula!

Answer 7

The same question was bugging me. Looking on the web turned up answers which didn't ease the existential angst. The most common one was that the average of a full cycle of a sine wave is zero, so we use RMS instead. The implication of this line of thought seems to be that the square root of the average of the squares (RMS) of a series is equal to the average of the absolute values of the series, which is not the case.

$$ \sqrt{\frac{1^2 + 2^2 + 3^2}{3}} = \sqrt{\frac{14}{3}} = 2.16... $$

$$ \frac{|1|+|2|+|3|}{3} = 2 $$

Close, but no cigar. Lets try to solve the problem using a vanilla average instead of thinking about this RMS stuff.

Ok, so we're trying to figure out how much AC voltage we need to dissipate the same amount of average power through a resistive load as DC voltage. I'll use the following equation to calculate the power:

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

The power through a resistive load with DC: $$ P_{DC} = \frac{V_{DC}^2}{R} $$

The instantaneous power through a resistive load with AC: $$ P_{AC} = \frac{V_{P}^2}{R}.sin^2\theta $$

The average power through a resistive load with AC: $$ \overline{P_{AC}} = \frac{1}{2\pi}\int_{0}^{2\pi} \frac{V_{P}^2}{R}sin^2\theta \,d\theta $$

$$ = \frac{V_{P}^2}{2 \pi R}\int_{0}^{2\pi}sin^2\theta \,d\theta $$

$$ = \frac{V_{P}^2}{2 \pi R}\int_{0}^{2\pi}\frac{1}{2}-\frac{cos2\theta}{2}\theta \,d\theta $$

$$ = \frac{V_{P}^2}{2 \pi R}\left[\frac{\theta}{2}-\frac{sin2\theta}{4}\right]_{0}^{2\pi} $$

$$ = \frac{V_{P}^2}{2 \pi R}\pi $$

$$ = \frac{V_{P}^2}{2R} $$

The whole point of this RMS stuff is to find how much AC voltage we need to dissipate the same amount of power through a resistive load as DC voltage:

$$ \overline{P_{AC}} = P_{DC} $$

$$ \frac{V_{P}^2}{2R} = \frac{V_{DC}^2}{R} $$

$$ V_{p}^2 = 2 V_{DC}^2 $$

$$ V_{p} = \sqrt{2}V_{DC} $$

We've arrived at the same conversion formula without any RMS stuff. So why bother with the RMS incantation? If you review the math you'll find that it follows the same pattern as the math for RMS. The square comes from the square in the power equation. I'm not a mathematician (or well versed in the history of maths), but I'm guessing this similarity is where the RMS stuff comes from (RMS is used in many fields). There is no need for any knowledge of RMS to derive the results.

Answer 8

I see all sorts of "proofs" for RMS calculations. RMS is not something that is inherently "correct", it is a useful measure for particular situations.

So let's get down to the details that make it useful for a particular situation: energy generally is interesting as a conserved measure and can often be represented as a time integral of power, and power as a product of two measures.

Electric power at a given moment can be defined as the product of current and voltage. There are lots of situations where current and voltage (or other power-defining measures) are in a proportional relation. When this is the case (and it often is, but one cannot depend on it similarly as with conservation of energy) and we are talking about stationary conditions (like AC circuits that repeat the same energy flow with each cycle), RMS values are convenient for simplifying calculations a whole lot.

This is primarily the case for resistive work loads, but reactive loads (capacities and inductivities) can also benefit from calculations based on RMS values, though in a less direct manner.

That doesn't give RMS values some physical meaning. They are a mathematical tool that is particularly useful in the context of AC calculations.

Answer 9

y(x) = |x| is not differentiable, because y'(0) is undefined.

y(x) = sqrt(x * x) is differentiable.

However they are otherwise equivalent.

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Vrms = average(abs(v(t))) = average(sqrt(v(t) * v(t)))

Why did they pick one definition over the other? Well, one is an average of a differentiatable function.