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## Re: Clarification on RMS and frequency

 Subject: Re: Clarification on RMS and frequency "John G" Tue, 6 Dec 2005 09:47:00 +1100 sci.electronics.basics
 ```"Bob Myers" wrote in message news:lZZkf.34\$Qy5.3@xxxxxxxxxxxxxxxxxxx > > "RR" wrote in message > news:GYrkf.10510\$ea6.9061@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx >> Hi, >> >> As I understand it, RMS is calculated on the area under the sine or >> cosine >> wave for the AC supply and the peak voltage. > > Sort of, but you'll get a better idea of what it's really all about > simply > by going through what "RMS" stands for in the first place. > > The problem is one of determining the "effective" voltage, current, or > whatever of an alternating source; in other words - and to use the > most > popular example - if I pass an AC current through a resistor, how much > power is dissipated in that resistor? How can I compare AC to DC in > this sense? > > You clearly can't use the peak voltage or current - the waveform isn't > at the peak but for an instant, so obviously doing a power or some > other such calculation based on that value would be wrong. The next > idea would probably be to try to find the "average" value of the > waveform, but that winds up even worse - if you average any "pure" > AC (meaning that it is symmetrical, regardless of the form of the > wave, > and spends as much time above zero as below), you get a result of > exactly zero. That's obviously not right, either, since the resistor > DOES > heat up. > > So instead, we start by squaring the function that describes the AC > waveform; if you square such a wave, everything winds up above > zero, right? Then find the average, or mean, of the squared waveform > (so now you have a constant value), and to correct for the squaring > operation you did in the first place, find the square root of that > average > value. > > In short, you are finding the (square) Root of the Mean of the Square. > "RMS," see? > > It happens to work out to 0.707 peak for a pure sinusoid; other > waveforms wind up with different fractions of the peak value, and > this is really only a shortcut which can be applied to those examples > where one of these "regular" waveforms is in question. But the > above process - square the wave, average it, and take the square > root of the result - applies to all. > > Obviously, frequency does not enter into this at all - RMS is 0.707 > of peak for ANY sinusoid, regardless of frequency. > >> >> I've attempted to follow the calculations here: >> http://www.alpharubicon.com/altenergy/understandingAC.htm >> and the frequency seems to always reduce to a factor of 0.5. > > The final result is just a numeric value - it has no frequency (that > is the result of averaging the squared waveform - an average has > no frequency, since it's a constant value, right?). You seem to get > into some unexpected frequencies during all of this because the > squared waveform is itself a periodic wave with a frequency 2X > the original (at least for anything but a 50% duty-cycle square > wave). > > >> So, why does an appliance designed for 240volts 50Hz care whether you > supply >> 240volts 60Hz or 240volts 100Hz, for that matter? > > When an appliance "cares" about the line frequency, one of two > things are generally at issue: > > 1. The appliance relies on the frequency of the line in order to > run at the proper speed - e.g., the motor in an electric clock. > > 2. The appliance contains components (typically, transformers or > similar magnetics) which are designed to operate at one frequency, > and will be less efficient at others. > > Bob M. > Bob, Where you a trained teacher? Thats one of the best replies I have seen here to a very basic question. Sometimes the posters get carried away with their own imagined expertise and launch off into long explanations that are technically way over the head of the original poster and sometimes not even accurate. -- John G Wot's Your Real Problem? ```
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