# V(rms) = V(ac) and V(ave) = V(dc), why?

Discussion in 'General Electronics Discussion' started by max_torch, Jul 27, 2014.

1. ### max_torch

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Feb 9, 2014
the root mean square of a voltage waveform corresponds to the AC voltage and the average corresponds to the DC voltage, but why is this so? In short, why does rms = ac and ave = dc?

here is a piece of the article on rms from wikipedia.org: http://en.wikipedia.org/wiki/Root_mean_square
I don't know how to connect this info to answer my question can anyone explain in a simpler way?...

5,164
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Dec 18, 2013
The RMS of an AC sine wave is the DC equivalent voltage. This the DC voltage required across say a resistor that will produced the same amount of power as your sine wave does. So for a 1 Volt DC source with a load of 1R the resistor will dissipate 1 Watt of power.

A sine wave is not DC so it can't produce the same power as the DC with the same peak voltage, because it is constantly changing. So to produce the same amount of power in the 1R load the sine wave peak voltage will have to be higher. Well it turns out it will have to be 1.414 Volts.

So if you had a sine wave with a peak value of 1.414 Volts then the DC equivalent voltage is going to be the RMS of this which is 1.414 Volts * 0.707. Where does the 0.707 come from?

This is how it is found. RMS = ROOT MEAN SQUARE.
You work backwards on this and do the SQUARE and then the MEAN and then the ROOT. So what does the SQAURE bit do. All it does is removes the negative part of the sine wave and puts it along side the positive part of the sine wave.

This makes it easier to work out the mean because the mean DC voltage of a sine wave with equal + and - voltages is equal to zero. So now the MEAN of this new waveform is going to be half, the same two waveforms / 2. Then the ROOT removes the original SQUARE and so you have Sqrt of 1/2 which is 0.707.

I think I got that right?

Hope this helps

3. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
That's a pretty good explanation Adam. I'm going to exercise my pedantry and clarify a few points though.

The RMS voltage of ANY AC WAVEFORM (regardless of its shape) is the voltage that, if it were DC, would produce the same MEAN POWER DISSIPATION, or the same HEATING EFFECT (which is the same thing), in the same RESISTIVE (OHMIC) load, when measured over a one cycle period (for a repetitive waveform). So for a given resistive load, 110V AC RMS will produce the same amount of heat as 110V DC.
More to the point, because the instantaneous voltage of the sinewave, measured at any instant during the cycle, is not always equal to the peak voltage.

Another way of saying this is that with any AC waveform, apart from a perfect rectangular wave with equal positive and negative voltages, if you draw a rectangle only just enclosing each half cycle, the rectangle is not fully filled with continuous voltage, so by definition, the mean voltage, or the mean power, in that half cycle cannot be as high as the peak voltage.
Which is 1.0V DC.
That's one part of it; the other part is that it takes into account the fact that for a resistive load, power is proportional to the SQUARE of the voltage. This is because current is proportional to voltage, and power is proportional to voltage × current. So power is proportional to voltage squared.

This is also shown by the formula P = V 2 / R, which is the result of combining Ohm's Law (I = V / R) with the Power Law (P = V × I). Substituting V/R for I in the second formula, you get P = V × (V / R), which is equivalent to P = (V × V) / R, which is equivalent to P = V2 / R. Therefore for a constant load resistance, power is proportional to the square of the voltage.

Here's another way to imagine it. For simplicity, let's use a 1Ω resistive load. Take any random point on any repetitive AC waveform that is being fed into that load. How much power will the load dissipate at that instant in time? If P = V2 / R, and R = 1Ω, then P = V2. So to calculate the power dissipation at that instant, you need to square the voltage at that point on the wave.

If you do this for every point in a cycle of the AC waveform, then take the mean of the results, you will get the mean power dissipated in the load during one cycle of the waveform. Then, if you take the square root of that mean power figure, you convert power back into voltage, and you get the equivalent DC (continuous) voltage that is required to produce that amount of power in the load.

For a sinewave, that voltage is equal to the peak voltage multiplied by 0.707. (0.707 is the square root of 0.5, or the reciprocal of the square root of 2.) For other waveforms, the relationship between the RMS and peak voltages is usually different.

As I mentioned, for a perfect symmetrical rectangular wave, the RMS voltage is equal to the peak voltage. For a triangle or sawtooth waveform, the RMS voltage is equal to the peak voltage multiplied by 0.577 (which is the square root of 1/3). It's possible to calculate these constants, and explain them, but my maths skills aren't up to it, and it might get boring. Have a look at https://en.wikipedia.org/wiki/Root_mean_square if you want a dry technical explanation.
I can't follow that explanation, but I'm pretty sure that bit is wrong. Steve or Harald can probably explain, if they have time.
All up to the last paragraph, AFAIK

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5,164
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Dec 18, 2013
Hi Kris what a great addition to my lame comments this is what I meant if I understand it correctly.

Thanks

5. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011

No, that's not right. Your second diagram shows the formula for the absolute value of the sinewave, i.e. what happens if you bridge-rectify it. It doesn't show the shape you get if you square the value of every point on the waveform.

The shape you get if you square half a cycle of a sinewave is actually the shape of a full cycle of a sinewave at twice the frequency! No doubt there's an explanation for this, but it's beyond me. It's another thing that Harald or Steve would be able to explain... possibly in a way I could grasp, possibly not! But it's true. I guess it ties in with the sqrt(2) ratio, and I guess a squaring circuit would work as a frequency doubler (assuming sinewave input). All very interesting, but beyond my rudimentary mathematical abilities.

Here's proof: a graph from a spreadsheet. The blue trace is a half cycle of sinewave, and the orange trace is the square of that half cycle.

Edit: Oh, it's obvious where the sqrt(2) ratio comes from. When you square a half cycle of sinewave, the resulting wave is a whole cycle of a sinewave at twice the frequency, with a peak-to-peak value equal to the peak value of the original sinewave. Since one cycle of a sinewave is symmetrical around the 0V line, and the centre line is half way up the half cycle, the mean value of the whole cycle is equal to half the amplitude of the half cycle.

So when you take the square root of the mean of the full cycle, you get the square root of half the amplitude of the half cycle, i.e. 0.707 × the amplitude of the half cycle.

Last edited: Jul 27, 2014

5,164
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Dec 18, 2013
Yeah ok you got me. I knew this just thought I would draw something to show what I meant. Should have guessed you would pick that up

5,164
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Dec 18, 2013
Yes correct as my diagram shows although wave shape could be better.

5,164
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Dec 18, 2013

9. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
No, what you thought were my knuckles was probably me snoring. I had to pull out of the Tour last night in the final stage. Didn't even make it to Paris.

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10. ### Laplace

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Apr 4, 2010

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