RMS versus AVERGAGE

In your first case, you are dealing with a square wave. The second is a sine wave. Sine waves fill up less of the curve under them because of the slopes of the sides.

mike

"panfilero"

** Cos they have different meanings in different situations.

Imagine a 12 volt battery and a load that is resistive and switched rapidly on and off with a 50% duty cycle.

The average current is 50% of the max value and the power in watts delivered into the load is that value times 12.

The current in the wires feeding the load has an RMS value of 70.7% of the max value - that number tell you what size conductors are needed and what size fuse to install.

RMS current: used to determine the heating effect on a resistance passing that current.

Average current: used when power is delivered from a fixed voltage or dissipated when the load has a fixed voltage - ie diodes or triacs etc.

.... Phil

It's only 0.707 for sine waves. That's because the average angle of a generator turning through 90 degrees (from 0 to peak) is 45 degrees. The average angle is 45, so the average voltage is the sine of 45, or

0.707 of peak.

For a triangle wave, the RMS is about 0.577 of peak. Square waves are

50% and complex waveforms can be anything.

The point about RMS is it's the same voltage that will generate the same heat as using a DC voltage (battery). So, a 10 volt peak (20 volt peak to peak) sine wave into 1 ohm will produce 7.07 ^2 / 1 =3D 50 watts. And a 7.07 volt battery connected to a 1 ohm load will produce the same thing (50 watts).

-Bill

"Bill Bowden is WRONG"

It's only 0.707 for sine waves.

** WRONG !!

That's because the average angle of a generator turning through 90 degrees (from 0 to peak) is 45 degrees. The average angle is 45, so the average voltage is the sine of 45, or

0.707 of peak.

** Total BOLLOCKS !!

The average value of a sine wave is 2/pi ( 0.6366) times the peak.

The ony way to derive the number is by using calculus.

.... Phil

Thanks Phil, your answer made some sense to me

Other answers that mentioned sine waves.... I never mentioned sine waves... I never even mentioned a square wave, I just said a wave that was high for 50% of the period... basically a pulse waveform with a duty cycle of 0.5, if you do the math on this, the rms value of a pulse is it's peak times the square root of the duty cycle, for Vin =3D

10V, and D =3D 0.5 you get 7.07V

Vin * sqrt(D) =3D 7.07V

much thanks!

No, that only applies to a sine wave. For pulses, it's a straight

1-to-1 because the sides are straight up and down. i.e., 50% duty cycle yields 50% power, and so on. But only for pulses and rectangular waves. For triangle waves, it's another equation, for sines you use RMS, and so on.

Maybe a good way to think of it would be the "effective" value.

Hope This Helps! Rich

Consider a case with a 10V supply feeding a 1 ohm load. The power delivered to the load is V^2/R = 100 watts. If you turn that off and on at a 50% duty cycle, the average power delivered will be 50 watts.

Now consider a 5V supply that is on 100% of the time. The power delivered to that same 1 ohm load is only 25 watts.

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Classic oops, the average value of a sine wave is zero.

NT

RMS = Root Mean Square, i.e. the square root of the mean of the square of the voltage.

Why use this measure? Because it provides an indication of equivalent power. The power dissipated by a resistance is equal to the product of the voltage and the current, P=VI. But the current is equal to the voltage divided by the resistance, I=V/R. Combining the two gives P=V^2/R, i.e. the power is proportional to the square of the voltage.

So the mean (average) power dissipated by a resistive load supplied with an AC voltage is proportional to the mean of the square of the voltage. For a DC voltage, the mean of the square of the voltage is just the square of the voltage. The two will provide the same power if:

(Vdc)^2 = mean(V(t)^2)

i.e. if: Vdc = sqrt(mean(V(t)^2))

E.g. the nominal UK mains supply is a 50Hz sine wave with a 340V peak. The RMS voltage is 340/sqrt(2) = 240V. The significance of this figure is that the 340V-peak sine wave will supply the same power into a resistive load as would a 240V DC supply.

Where does the sqrt(2) for sine waves come from? Consider the trigonometric identity:

sin(x)^2 = (1-cos(2x))/2

If the voltage is:

Vpk.sin(w.t)

the square of the voltage is:

Vpk^2.sin(w.t)^2 = Vpk^2.(1-cos(2.w.t))/2

This is a sine wave with twice the frequency, with a minimum of 0 and a maximum of Vpk^2, so the average value over a cycle is Vpk^2/2. Hence, the mean voltage-squared of a sine wave is half its peak voltage-squared, and thus the root-mean-square voltage is 1/sqrt(2) times the peak voltage.

--
http://en.wikipedia.org/wiki/Root_mean_square

You have to watch that sign change when it crosses the zero line. Basically you have to imagine the second half of the sine wave as being flipped upwards. Then integrate each and then add them together.

mike

That's a square wave with ground as the most-negative value (a combination of DC and AC).

And if the waveform was white noise, or a triangle wave, or a +V/-V square wave, there'd be a different number there. Check.

No, not generally. The thermal limit might depend only on RMS current, like in a resistor, but it ALSO depends on how the 'square wave' frequency compares to the thermal time constant. A two-day period square wave with 10A peak is gonna toast your 7.07A resistor. A two-microsecond period square wave might not.

More important, an iron-core inductor that takes 7A DC won't like the 10A peak AT ALL. The saturation point is a nonlinearity, and it's not responsive on any 'average' measure, you really need to look at the instantanenous values.

Yes, that should have been RMS voltage, not average.

I haven't seen many applications using average voltages other than meter movements and DMMs. I think if you calibrate a DC meter movement to read full scale at 10 volts DC and then apply a 10 volt peak sine wave (full wave rectified) you get the average of 6.366. So if the meter is calibrated to read RMS of a sine wave, or 7.07, there will be an error reading other waveforms. A 10 volt peak squarewave should read 5.55 instead of 5.00.

What other applications are there for average voltages?

-Bill

"Bill Bowden"

** It is the *average* value of the current charging a battery that is effective in producing charge - I have seen RMS responding meters ( ie moving iron types ) fitted to car battery chargers as a marketing scam.

The average value of the current flowing in a diode or triac is used to estimate power dissipation and hence compliance with maker's ratings.

Because of the scaling factor included in all non true rms AC volt and AC amp meters - one has to multiply the reading bu 0.9 to get the correct average value - PLUS be very aware of the particular meter's frequency response limitations.

... Phil

Are you sure it can't be derived from and the area of a circle (and limits)?

Of course, Archimedes determined the area of a circle long before Newton came along.

Best regards, Spehro Pefhany

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.... Phil

Ancient Egyptians calculated the area of a circle with (8d)/9 squared. So, a 1 foot diameter circle has 0.79 square feet of area compared to

0.7854 using a more precise value for pi. Not bad for old timers.

-Bill

"The Journey is the reward"

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"Bill Bowden"

Ancient Egyptians calculated the area of a circle with (8d)/9 squared. So, a 1 foot diameter circle has 0.79 square feet of area compared to

0.7854 using a more precise value for pi. Not bad for old timers.

** The approximation to " pi " taught to school children in my day was 22/7.

Get your pocket calc out and see how close it is.

.... Phil

Well the formula you wrote above is limited only to a half period of a wave... and it applies ONLY to square waves. If you had a Sine wave the Average would be Vin*D[(Cos(w*T/2))] The average value of a periodic function,any kind , square, sinusoidal, etc of voltage, current or whatever is zero in a time interval of 1 period.

Differently from the above formula the rms formula you have written applies to a time interval of 1 period and hence gets you a different result.

Both the rms and averages are a way of determing the mean value of a function.

If you were to find the rms of your periodic function within half a period you'd find it coincides with the "average value". Apart from what others mentioned we use rms because the doesn't give you an accurate mean value when the function becomes negative. Or more like with a zero average you don't go anywhere.