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Re: AVERAGE APPARENT TRUE REACTIVE RMS ETC.



Original poster: "Charles Hobson by way of Terry Fritz <twftesla-at-uswest-dot-net>" <charles.a.hobson-at-btinternet-dot-com>

I think everyone by now probably has a handle on average power, apparent
power, true power, reactive power, rms volts and amps after seeing all the
excellent descriptions of them in the Tesla list, but I just couldn't resist
jumping in and putting forth my understanding of  where rms comes from.

In my example I should use a sine wave, but instead I shall use a cosine
wave form of a single cycle of voltage having a peak value of one Volt. The
peak value is usually called the amplitude value. Viewing this on an
oscilloscope one will see the cosine wave form starting at +1V (0 degrees),
decreases sinusoidal towards zero Volts (90 degrees), continue on to -1 Volt
(180 degrees), back up to zero (270 degrees) and return to +1 Volt (360
degrees)

Assume this voltage is connected to a one Ohm resistor.

Sketch this cosine waveform on some graph paper, x axis in degrees and y
axis in Volts.

Using a pocket calculator, find the cosine at convenient increments from 0
to 360 degrees, to check your sketch.
At each check, square the cosine value and plot a second curve directly
below the cosine curve.

You will get what looks like but not quite two cosine waves. They will all
be between zero and +1 (remember, a minus number times a minus number always
equals a plus number. So all squared values will be plus)

Since Watts = V^2/R  Then we can call the y axis on the second curve Watts.
(varying between 0 and +1)

If you draw a line in the x direction for 0 to 360 degrees at 1/2 Watt, you
will see that the two shapes above 1/2W can fit snugly in the two valleys
below 1/2W. Thus we have an average power of 1/2W dissipated in a one ohm
resistor when a voltage having a PEAK or amplitude value of 1 Volt is
applied. (Traditional integration methods will prove this)

Now, going backwards with a known average power of 1/2 W heating up a one
ohm resistor, i.e. 0.5W = V^2/1 ohm and find V we do  V=SQR 0.5 = 0.707!!!
which is the rms (effective) voltage causing the one ohm resistor to
dissipate 1/2W .

 "r" is for square root of  "m" mean or average  power  and "s" for squared
voltage.

I and I think most others capitalize Volts in honor of Alessandro Volta,
Watts in honor of James Watt, Andre Ampere in honor of Ampere and Ohms in
honor of Georg Ohm.

Thank you for your patience.

Chuck