RMS vs Average voltage

I’m currently working on a project that deals with controlling an AC heater and it has led to a question that I really should be able to answer… but I can’t.

I understand how to get the values of the average voltage and the RMS voltage of an AC wave, but what I don’t understand is why RMS is used to calculate effective DC voltage. It seems, intuitively, that the area under an AC sine wave could be used to calculate the effective DC voltage. I get that the RMS equation deals nicely with the negative half of the wave, but so would taking the absolute value.

There is a lot of information on the internet about effective DC voltage, but most of it just resorts to the calculus behind the RMS equation; it does not explain why the RMS value is what you need to use instead of the average value. Is there a simple answer?

if the wave is a perfect sin then the rms value is Vpeak / 1.414 (root 2) easy . Using phase control of the heater element the resultant wave is not a perfect sinusoid or necessarily even symmetrical . in short the negative half is != top half so your calculation would be incorrect

yes but, since heater elements are almost 100% resistive, (no phase shifting between current and voltage) the output can be set with simple percentage type table values and no need to try to find rms values..... and that would not be true if the load had some appreciable inductive amount

Maybe this will answer your question. I got it from http://www.electronics-tutorials.ws/...s-voltage.html

The term “RMS” stands for “Root-Mean-Squared”. Most books define this as the “amount of AC power that produces the same heating effect as an equivalent DC power”, or something similar along these lines, but an RMS value is more than just that. The RMS value is the square root of the mean (average) value of the squared function of the instantaneous values. The symbols used for defining an RMS value are VRMS or IRMS.

The term RMS, ONLY refers to time-varying sinusoidal voltages, currents or complex waveforms were the magnitude of the waveform changes over time and is not used in DC circuit analysis or calculations were the magnitude is always constant. When used to compare the equivalent RMS voltage value of an alternating sinusoidal waveform that supplies the same electrical power to a given load as an equivalent DC circuit, the RMS value is called the “effective value” and is generally presented as: Veff or Ieff.

In other words, the effective value is an equivalent DC value which tells you how many volts or amps of DC that a time-varying sinusoidal waveform is equal to in terms of its ability to produce the same power. For example, the domestic mains supply in the United Kingdom is 240Vac. This value is assumed to indicate an effective value of “240 Volts RMS”. This means then that the sinusoidal RMS voltage from the wall sockets of a UK home is capable of producing the same average positive power as 240 volts of steady DC voltage as shown below.

If this misses the mark I apologize.

I have the software/ hardware part worked out. It's just that along the way I realized that I really don't know the answer to the basic question; why isn't the effective dc voltage equal to the area under ac curve, but instead the rms value.

because of the phase shift between voltage and current- for inductive and capacitive loads, but for pure resistive loads, the current and voltage are in sync

it has nothing to do with phase shift or inductive and capacitive loads its simple mathematics

power is whats being measured here
power = V squared / R
the effective power is the sum of the instantaneous powers measured in your wave form, try it for a triangular wave
R=1 ohm Vp=50 , Vave=25v Vrms=28.8

For starters, the average of a sine wave for a full cycle is 0.

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For starters, the average of a sine wave for a full cycle is 0.

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yes Charlie the average voltage = 0 and the average current =0 yet the load resistor gets hot . perhaps you have just discovered dark energy

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Originally Posted by richard

yes Charlie the average voltage = 0 and the average current =0 yet the load resistor gets hot . perhaps you have just discovered dark energy

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Well, Richard, you did bring up the math. And the mathematical reason we don't use average is that it's no where near correct, in fact the average IS 0. This is a sine function, so calculating the absolute value average is a bit more complicated. You need to integrate the area under the curve as was outlined above. The way you do that is to square all the instantaneous values, take the average, then take the square root of that. For a sine function, that is the peak to peak value * square root of 2 (0.707)