# Why RMS for the average voltage in AC current?

Efeguleroglu
We write rms formula as
$$v_{rms}=\sqrt\frac{\int_a^b[f(x)]^2dx}{|b-a|}$$
I know if we take arithmetic mean average voltage will be 0. So we want all voltage values to be positive. Why don't we do that:
$$V_{average}=\frac{\int_a^b\sqrt{[f(x)]^2}dx}{|b-a|}$$
That's first what I did:
$$\phi=ABcos(\omega t)$$
$$d\phi=-AB\omega sin(\omega t) dt$$
$$V=-\frac{d\phi}{dt}=AB\omega sin(\omega t)$$
If we take rms of $$AB\omega sin(\omega t)$$ we find $$\frac{AB\omega}{\sqrt{2}}$$
In my method it's this:
$$f(t)=AB\omega sin(\omega t)$$$$\frac{\int_0^{\frac{\pi}{2}}{f(t)}dt}{\frac{\pi}{2}}=\frac{2AB\omega}{\pi}$$
So please can someone explain why do I have such a confusion?

Last edited:

Dullard
RMS is not the same as 'average' voltage (as you note). Simply stated, the RMS value of a waveform is the DC voltage which would produce equivalent heating in a resistive load. The difference between the RMS and 'average' calculation is the result of the fact that power is proportional to the square of the voltage.

DaveE
Staff Emeritus
RMS is not the same as 'average' voltage (as you note). Simply stated, the RMS value of a waveform is the DC voltage which would produce equivalent heating in a resistive load.
Correct.

The difference between the RMS and 'average' calculation is the result of the fact that power is proportional to the square of the voltage.
Not correct. At least not without defining a specific circuit. You might say square of current also.

The universal statement that does not depend on a specific circuit is ##P(t)=V(t)I(t)## and ##P_{avg}=V_{RMS}I_{RMS}##

The important thing about RMS that it works for AC+DC signals, non-sinusoidal signals, intervals other than an integer number of cycles, aperiodic signals, and even signals that cannot be expressed as a function.

Asymptotic and vanhees71
Gold Member
2022 Award
Also it's important to remember the EMFs are not necessarily a voltage. The latter is a potential difference and thus it's only applicable if there's a potential, which is not the case if inductances are not negligible.

Efeguleroglu
Also it's important to remember the EMFs are not necessarily a voltage. The latter is a potential difference and thus it's only applicable if there's a potential, which is not the case if inductances are not negligible.
So the less the inductance is, the more precise the result of the rms emf is you say.

Gold Member
2022 Award
That's not what I say. It doesn't make any sense to me.

What you want to know for your electricity bill is how much power some appliance uses effectively, and that's given by the RMS of ##P(t)## given in #3.

Take as an example a circuit consisting of a real inductance, i.e., an ideal inductance ##L## in series with an Ohmic resistor ##R##. The equation reads
$$L \dot{i} + R i=U_0 \cos(\omega t)=U_0 \mathrm{Re} \exp(\mathrm{i} \omega t).$$
After some transient state the complex current (you can take the real part at the end of the calculation goes like
$$i(t)=i_0 \exp(\mathrm{i} \omega t).$$
Plugging this into the equation of motion you get
$$i_0 (\mathrm{i} L \omega + R)=U_0 \; \Rightarrow \; i_0=\frac{1}{Z} U_0$$
with
$$Z=R+\mathrm{i} \omega L.$$
Now the momentaneous power is
$$P(t)=i U =U_0^2 \cos(\omega t) \mathrm{Re} \left [\frac{1}{Z} \exp(\mathrm{i} \omega t) \right].$$
Now
$$\frac{1}{Z}=\frac{1}{R+\mathrm{i} \omega L} = \frac{R-\mathrm{i} \omega L}{R^2+\omega^2 L^2}.$$
And thus
$$P(t)=\frac{U_0^2}{R^2+\omega^2 L^2} \cos(\omega t) [R \cos (\omega t)+\omega L \sin(\omega t)].$$
Now you are not interested on the momenaneous power consumption but the time average. Now it's a periodic function with period ##T=2 \pi/\omega##, and thus you average as
$$\overline{P}=\frac{1}{T} \int_0^{T} P(t).$$
All you need are the integrals
$$\int_0^T \mathrm{d} t \cos^2(\omega t)=\frac{1}{2} \int_0^T \mathrm{d} t [\cos^2(\omega t)+\sin^2(\omega t)]=\frac{T}{2}, \quad \int_0^T \mathrm{d} t \cos(\omega t) \sin(\omega t)=\int_0^T \mathrm{d} t \frac{1}{2} \sin(2 \omega t)=0.$$
Thus your average power consumption is
$$\overline{P}=\frac{1}{2} U_0^2 \frac{R}{R^2+\omega^2 L^2}.$$

Efeguleroglu
I don't care my power consumption. But I got what I was chasing I think, thank you for that. It was just about definition. I constructed it on power.
$$ε(t)=ε_{max}sin(\omega t)$$
$$P(t)=\frac{{ε_{max}}^2 sin^2(\omega t)}{R}$$
$$P_{efficient}=\frac{\int_a^bP(t)dt}{b-a}=\frac{\int_a^b\frac{{ε_{max}}^2 sin^2(\omega t)}{R}dt}{b-a}=\frac{{ε_{efficient}}^2}{R}$$
$${ε_{rms}}={ε_{efficient}}$$