Why RMS for the average voltage in AC current?

In summary, the RMS (root mean square) formula is used to calculate the effective voltage in a circuit by taking the square root of the integral of the squared voltage divided by the length of the interval. This is different from the average voltage calculation, which would result in a value of 0. The difference between these calculations is due to the fact that power is proportional to the square of the voltage. The RMS calculation is important because it can be applied to various types of signals and circuits, and takes into account factors such as inductance. The key is to use the time-averaged power for the calculation, which can be found by taking the integral of the power over the interval and dividing by the length of the interval.
  • #1
Efeguleroglu
24
2
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?
 
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  • #2
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.
 
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  • #3
Dullard said:
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.

Dullard said:
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.
 
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  • #4
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.
 
  • #5
vanhees71 said:
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.
 
  • #6
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
$$\overline{P}=\frac{1}{2} U_0^2 \frac{R}{R^2+\omega^2 L^2}.$$
 
  • #7
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.
$$ε(t)=ε_{max}sin(\omega t)$$
$$P(t)=\frac{{ε_{max}}^2 sin^2(\omega t)}{R}$$
$$P_{efficient}=\frac{\int_a^bP(t)dt}{b-a}=\frac{\int_a^b\frac{{ε_{max}}^2 sin^2(\omega t)}{R}dt}{b-a}=\frac{{ε_{efficient}}^2}{R}$$
$${ε_{rms}}={ε_{efficient}}$$
 
  • #8
Yes, that's the point: You use the RMS of the "voltage" in this case of a pure Ohmic resistor, because it provides the time-averaged power. My example was just to demonstrate what happens if a inductivity is present. It's always the time-averaged power you are interested in, and that tells you, how this averaging has to be done.
 

1. What is RMS in AC current and why is it important?

RMS stands for "root mean square" and it is a way to measure the average voltage in an AC (alternating current) circuit. It is important because it takes into account both the positive and negative values of the voltage, giving a more accurate representation of the actual voltage in the circuit.

2. How is RMS calculated in AC current?

RMS is calculated by taking the square root of the average of the squared values of the voltage over a given period of time. This takes into account the fluctuations in voltage that occur in an AC circuit.

3. Why is RMS used instead of the peak voltage in AC current?

RMS is used instead of the peak voltage because it gives a more accurate representation of the actual voltage in the circuit. Peak voltage only measures the maximum value of the voltage, while RMS takes into account the entire waveform.

4. How does RMS affect the power output in AC circuits?

RMS is directly related to the power output in AC circuits. In fact, the power output is equal to the RMS voltage multiplied by the RMS current. This is because RMS takes into account both the voltage and current, giving a more accurate measure of the power being delivered.

5. Can RMS be used for any type of AC waveform?

Yes, RMS can be used for any type of AC waveform, including sine waves, square waves, and triangular waves. As long as the voltage is fluctuating in an alternating manner, RMS can be used to calculate the average voltage.

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