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realolman:
RMS measurements are very useful. In particular for an electrician, if you measure the RMS current in a wire and know the value of resistance in that wire, then you can calculate the power dissipated in the wire, which in turn is related to the temperature rise in the wire.
Consider a #12 copper wire. Its resistance is about 0.0016 ohms per foot. 20 A RMS thru this will produce about 0.64 W / ft. I have never measured a wire such as this to estimate the temperature rise.
Suppose you have a very peaked current waveform, such as from a switching power supply, then knowing the RMS current is important because peaked waveforms produce more heat than is predicted by the average current.
In the example I gave in a previous post of a rectangular pulse with a duty cycle of 1% the RMS current was 10% of the peak current. The average current is 1% of the peak value. So use of average current to calculate power would produce a gross understatement.
For a complex waveform represented by a mathematical equation one would use calculus to calculate the RMS value. For a waveform of arbitrary shape you can get a good estimate of the RMS value with a meter provided the peak to RMS ratio is not too large. This is called the crest factor. Another way is to --- determine the area under the curve that is the square of the actual current waveform, divide by the time period of the base of the area measured, then take the sq-root of this value. If you are interested I can describe how you might graphically do this. It is really the basis of integral calculus.
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