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Old October 19th 04, 10:39 AM
Roy Lewallen
 
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Let me add one more comment that will hopefully help in understanding this.

I think I might have identified another misconception that might be
contributing to the confusion. That misconception is that the RMS value
of a waveform is the "equivalent" or "heating" value. This leads to the
mistaken notion that the RMS value of the power must be the "equivalent"
or "heating" value.

Voltage and current, by themselves, don't do any heating. Heating takes
power. And the amount of heating is the determined by the average power.
If two different power waveforms have the same average value, they'll
create the same amount of heat and otherwise do the same amount of work,
even if their RMS values are different.

So what's the whole thing about RMS? The only importance of RMS is that
when you multiply the RMS values of two waveforms together (such as V *
V, I * I, or V * I), you get the average of the product of the two. That
is, RMS(i) * RMS(v) = Avg(i * v), where i and v are the instantaneous
values of the current and voltage, and i * v is the instantaneous power.
Likewise, RMS(v) * RMS(v) / R = Avg(v*v/R) and RMS(i) * RMS(i) * R =
Avg(i*i*R). In fact, RMS(x) * RMS(y) = Avg(x * y) where x and y can be
any periodic waveforms or quantities. (I derived these mathematically in
an earlier posting.) So we calculate the RMS values of voltage and
current only so we can use them to calculate the average power -- not
because the RMS value of any waveform is its "equivalent" or "heating"
value -- which it isn't.

Roy Lewallen, W7EL
 
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