Yes Chris, you are exactly right.

73
Gary K4FMX


On Wed, 20 Oct 2004 12:08:59 GMT, "Chris"
wrote:

Paul,

I've been following the thread still. Let me go back to the original
question for a moment. So is the 4W maximum power of a CB radio is actually
average power, not RMS, right? If it is modulated at 100% with a sine wave,
what wil the PEP be? Is 16W the correct answer?

Chris
"Roy Lewallen" wrote in message
...
| 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.