Bill Turner wrote:
On Mon, 18 Oct 2004 11:15:43 -0700, Roy Lewallen wrote:


The
problem with your approach is that there *is* such as thing as RMS power
(as I described in an earlier posting), and it is *not* the amount that
gives the same heating effect as the same amount of DC power. So while
you may think of RMS power in this way, your thinking is wrong.


__________________________________________________ _______

Let me ask one more question and then I'll shut up. Actually, three.

Suppose you have a 100 ohm resistor and you apply a 100 VRMS sine wave
to it. Will 1 amp of RMS current flow? I would think yes, but after
the previous discussions, I'm not taking anything for granted. :-)

Yes.

So if you have 100 VRMS and 1 ARMS, how much power is the resistor
dissipating? Just pure power in terms of heat generation, forget
whether it's RMS or anything else. How many watts?

100 watts. This is the average power. (The RMS power is about 122.5
watts, if I did the integration right -- corrections are welcome. RMS
power is calculated from the instantaneous power exactly like RMS
voltage or current is calculated from instantaneous voltage or current.)

And now the real question: How much DC voltage would you have to apply
to the resistor to get exactly the same power dissipation?

100 volts. The dissipation would be 100 watts.

Breathlessly awaiting,

I would hope that any Technician or higher class amateur could
immediately answer those questions, no waiting required.

The problem people seem to be having is the feeling (or, I'm afraid in
some cases, certainty) that an RMS value of a quantity times an RMS
value of another quantity HAS TO BE the RMS value of the product of the
quantities. But it isn't. Here's something you can check on your
calculator: Take sin(5 degrees) times sin(10 degrees). Is the product of
the two equal to sin(50 degrees)? While the product assumption works for
some functions, such as square and square root for example, it doesn't
work for others, like RMS and sine.

Unfortunately, if you (any reader, not Bill personally) aren't able to
follow the mathematics involved in calculating RMS and average values
(which I posted in brief form a short while ago), this is bound to
remain something of a mystery to you, and it looks like you're left with
four choices:

1. Hang tightly to a mistaken idea about the meaning of RMS and average,
and ignore or resist any explanation that contradicts them. This will
include explanations in any textbook.

2. Learn enough math to understand the definitions of average and RMS,
which should make the explanations understandable.

3. Assume that I and the textbook authors know what we're talking about
and believe us, without really understanding why.

4. Find a textbook oriented toward people without a very strong math
background, which hopefully can explain it in terms you can understand.

The fourth choice is probably the best for people who are really
interested in learning and can't devote the time necessary to learn the
math. The main problem with the first choice is that you'll be likely to
pass the misconceptions on to others. But the choice is yours.

Roy Lewallen, W7EL