There is a nice simple explanation in the ARRL handbook that I posted
earlier in the thread on "power output formula". Guess nobody read it.

2000 ARRL handbook 6.6 chapter 6, RMS VOLTAGES AND CURRENTS.

73
Gary K4FMX


On Tue, 19 Oct 2004 16:33:53 -0700, Roy Lewallen
wrote:

Paul, I apologize. My browser showed only the first of the two pages,
and I didn't realize that the second was there. While looking for the
quotation I found that the second page was simply scrolled off screen.

Yes, there is one thing (on that second page) I do disagree with the
author on, that the equivalent power, the product of Vrms and Irms, is
"RMS" power. I did a brief web search to find out who the author was so
I could contact him about that, and discovered that it's Joe Carr.
Unfortunately, he died a short time ago.

Maybe my suggestion about looking in non-mathematical texts for an
explanation wasn't such a good idea. It appears that some of the authors
of those texts don't fully understand the math either. I'll have to say
that you certainly have provided some evidence as to how widespread the
misconception is. Next time I'm downtown at Powell's Technical
Bookstore, I'll leaf through a few volumes oriented toward technicians
and see just how bad it is. All I have on my bookshelf in the way of
basic circuit analysis texts is two (Pearson and Maler, and Van
Valkenburg) which are intended for beginning engineering students, and
they of course both have it right. A popular elementary physics text
which I have, Weidner & Sells, _Elementary Classical Physics_, Vol. 2,
succinctly summarizes (p. 913): "Thus, the average rate at which thermal
energy is dissipated in the resistor is the product of the rms voltage
across it and the rms current through it." This follows immediately
below an equation showing the calculation of pav from the classical
definition of average which I posted some time ago.

In response to the question about Vrms, Irms, and equivalent power, I
said that when Vrms = 100 volts and Irms = 1 amp, Pavg = 100 watts and
Prms is about 122.5 watts. I haven't had any previous occasion to
calculate RMS power, so I might have made a mistake. According to my
calculation, for sinusoidal voltage and current, the RMS power equals
the average power (which is Vrms X Irms when the load is resistive)
times the square root of 1.5. Surely some of the readers of this group
can handle the calculus involved in the calculation -- it's at the level
taught to freshman engineering students, and now often taught in high
school. The calculation isn't hard, but it's a little tedious, so
there's ample opportunity to make a mistake. I'd very much appreciate if
one of you would take a few minutes and double-check my calculation. Or
check it with Mathcad or a similar program. I'll be glad to get you
started if you'll email me. And I'll be glad to post a correction if I
did make a mistake.

Roy Lewallen, W7EL

Paul Burridge wrote:

On Tue, 19 Oct 2004 12:20:29 -0700, Roy Lewallen
wrote:


I really appreciate the compliment, and will do my best to try and
deserve it.

Please look very carefully at the diagram at the URL you've posted, and
notice that it's a voltage waveform (see the labeling of the vertical
axis). Then read the text very carefully. Neither the diagram nor the
text contradict what I've said. If you think it does, post the reason
why, and I'll try to clear it up.


Okay, here's the bit that you seem to take exception to (it's spread
over both pages):

"We can define the real power in an AC circuit as the equivalent DC
power that would produce the same amout of heating in a resistive load
as the applied AC waveform. [In a purely resistive load] we can use
the root mean square (RMS) values (Vrms and Irms) to find this
equivalent or RMS power."

Then the equation "P = Vrms x Irms"

Where "P" here explicitly refers to RMS power.

This is something you have stated clearly that you disagree with.