# Justification of root mean square [duplicate]

In the top answer to the question Why do we use Root Mean Square (RMS) values when talking about AC voltage, the following was stated:

This RMS is a mathematical quantity (used in many math fields) used to compare both alternating and direct currents (or voltage). In other words (as an example), the RMS value of AC (current) is the direct current which when passed through a resistor for a given period of time would produce the same heat as that produced by alternating current when passed through the same resistor for the same time.

The RMS value, specifically applied to a sinusoidal voltage source $V_\mathrm{p}$ is given by:

$$V_{\mathrm{RMS}} = {V_\mathrm{p} \over {\sqrt 2}}$$

Here is where my intuition conflicts (where it probably goes off).

I'd imagine the average voltage that would be "felt" (direction insignificant / i.e. absolute value) by the circuit would equate to the true average value of the voltage, which is given by integrating a half period and dividing by the length of the period (the mathematical procedure of finding the average height of a function over a given interval)

I.e. the avg. Voltage, it seems to me, should be given by the equation:

$$V_{\mathrm{avg.}} = {2V_\mathrm{p} \over {π}}$$

I know that the two conversion coefficients are close but I simply cannot see why the RMS value is the one that conforms to reality. Please enlighten me! :)

The instantaneous power expended in a resistor is proportional to $V^2$ (i.e. independent of its direction), so the average power expended is given by the mean square voltage!

The square of the average absolute voltage will not yield the power expended in the resistor.

Edit: And actually this close-to-duplicate question does have more extensive answers (though not the top one!) that essentially make the same point. Why do we use Root Mean Square (RMS) values when talking about AC voltage

i.e. for a given resistance in an AC circuit with a peak voltage $V_p$, the power expended is $P = V_{p}^{2}/2R = V_{\rm RMS}^{2}/R$, where $V_{\rm RMS}=V_{p}/\sqrt{2}$.

• Writing down the formula for power in terms of voltage (peak) amplitude and RMS would probably not be a bad idea here. – DanielSank Oct 13 '14 at 21:13
• Will a incandescent light bulb connected to a V_RMS DC voltage source, have the same (visible) bightness as one connected to a AC source? – Lyndon White Oct 13 '14 at 23:21
• @Oxinabox Yes, I think that must be the case, so long as no temperature changes take place on the length of an AC cycle. – Rob Jeffries Oct 13 '14 at 23:27
• Yes, an incandescent light bulb fed with a DC voltage will have the same brightness as when fed with the same RMS voltage of sinusoidal AC. This is one of the fundamental concepts relating to electric utilities and power distribution. – Hot Licks Oct 14 '14 at 0:20
• (As Rob suggests it depends on the AC cycle being fast enough that the temperature changes through one cycle can be ignored, and it also depends on the AC cycle being slow enough that inductive and capacitive effects can be ignored.) – Hot Licks Oct 14 '14 at 0:22

know that the two conversion coefficients are close but I simply cannot see why the RMS value is the one that conforms to reality.

It's true that we can ask for the average of the magnitude of a sinusoid over time and calculate it.

This is a useful quantity if, for example, we want the time average voltage at the output of a full wave rectifier which, ideally, outputs the absolute value of the input voltage.

However, the rms (root of the mean of the square) voltage is useful in AC circuits (which are linear as opposed to rectifier circuits) simply because this is what we need to find the time average power delivered to a resistive load.

Why? Because resistor power is proportional to the the square of the voltage across.

As I have written before, if one asks the question:

What is the DC voltage required to deliver the same time average power to a resistor as an AC voltage?

The answer is: the rms value of the AC voltage. This is mathematically easy to show (as I and others have done here) and it agrees with experiment.

Here, the average value of $V(t)$ clearly isn't $2V_p/\pi$ that you've obtained, it's zero. The half-cycle business doesn't make much sense given that if you chose $\pi/2$ to $\pi$, instead of 0 to $\pi/2$, you'd have a negative average, so which would be the true "average" $+2V_p/\pi$ or $-2V_p/\pi$? Instead, we can get the arithmetic mean of the positive values by squaring them (i.e., the root-mean-square).
Typically we get the RMS voltage for use to find the average power, $P_{avg}$. Since power is $P(t)=V^2(t)/R$ and, assuming $R$ is constant, then, $$P_{avg}=\left\langle P(t)\right\rangle=\left\langle\frac{V^2}{R}\right\rangle=\frac1R\langle V^2\rangle\equiv\frac{V_{rms}^2}{R}$$ which brings us directly to the root-mean-square. A similar method can be used for current, $P(t)=I(t)^2R$: $$P_{avg}=\langle I^2R\rangle=R\langle I^2\rangle\equiv RI_{rms}^2$$
The definition of RMS is Root Mean Square and means the square root of the avarage of the square of some quantity, that is $a_{RMS}=\sqrt{\langle a^2 \rangle}$. So for the case of a sinusoidal current, we would get $$V_{RMS}^2=\frac{1}{T}\int_0^T\left(V_p\sin\frac{2\pi t}{T}\right)^2dt=\frac{V_p^2}{2}\implies\\ V_{RMS}=\frac{V_p}{\sqrt2}$$