🚀 go-pugleaf

>Your KWHr meter has always been RMS-responding.
>[...]
>Rich
RMS is not and should not be used in 'true' power meters.  They read just plain
average power ( I will use '<P>' for this).  The only reason people use RMS
to begin with is to be able to calculate <P> from V^2/R by using V=RMS(V(t)).
From: charless@cory.Berkeley.EDU (Charles R. Sullivan)
Path: cory.Berkeley.EDU!charless

Perhaps this example will help. To spare you the trigonometry, I 
will use a square wave voltage, 50% duty cycle, between 0 and 100V.
    RMS(V(t)) = sqrt( (100^2 + 0)/2 ) = 70.7 Volts.
With a 10 ohm load (resistive), I(t) is a sqare wave between 10A and 0, so
    <P(t)> = <V(t)*I(t)> =  (100*10 + 0)/2 = 500W
To calculate <P> = V^2/R, we use 
    RMS(V(t)) = sqrt( (100^2 + 0)/2 ) = 70.7 Volts,
so  <P> = 5000/10 = 500W, as before.

We needed the RMS trick, because the straight average voltage, 50V would
not give the right result; 50^2/10 = 250 W.

However, doing RMS(P(t)) yields a wierd, not-useful result:
RMS(P(t)) = sqrt( (1000^2 + 0)/2 ) = 707W.  I certainly hope the
power company doesn't use that to bill me.  (That would mean, for example,
that using a 1000W appliance for 1/2 hour would be billed the same as a
707W appliance for 1 hour!)
~
Perhaps the reason a 'true RMS power meter' is sometimes specified is that
if you made a power measurement by measuring voltage and current, with
separate meters, and then multiplied the 
measurements together *after* the meters averaged each independently, you would
prefer meters that used 'true RMS' measurements of the voltage and current.
This would result in correct average power in the case of a resistive
load.  However, it would not give correct results for a current waveform
that was not identical in shape and phase to the voltage wavform.  Hence
the need for power meters, which multiply the instantaneous voltage and 
current, and then average (or in the case of a kWhr. meter, integrate)
the result.

Charlie Sullivan                         charless@cory.berkeley.edu