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[Amps] RMS Power

To: <amps@contesting.com>
Subject: [Amps] RMS Power
From: wlfuqu00@uky.edu (Bill Fuqua)
Date: Mon, 11 Nov 2002 08:47:30 -0500
I do suggest that you find a text book and study it (make sure it is not 
either a audiophile or CB book).
An engineering text would be best.
Lets say you want to calculate the average power consumed by a
1 ohm resistor. For an example 10 seconds making a measurement every
second.

Each second you take a voltage measurement, square it and divide it by 1.
Add all those up and divide by 10 to get the average.

Now you want to determine a constant DC voltage required to produce the 
same power.
You take that number and take the square root of it (remember R=1).

What you have just done is calculated the RMS value of the voltage applied 
to the
resistor.

You have squared the voltages, calculated the mean(average) and then taken the
square root of it.  Root Mean Square

73
Bill wa4lav



At 11:17 PM 11/10/2002 +0000, you wrote:

>----- Original Message -----
>From: <MorgusMagnificen@aol.com>
>To: <amps@contesting.com>
>Sent: 10 November 2002 19:17
>Subject: [Amps] RMS Power
>
>
> > (This is a correction to my previous note).
> >
> > To those who say there is no such thing as RMS power, its time for you to
>go
> > back and re-learn your basic AC theory. RMS power is defined simply as the
> > time-average value of the energy in a circuit.
>Isn't that just the mean, or average power. I can see why you take the mean,
>or average of the voltage or current squared to get the average power but I
>can't get my head round how the average of (power squared) is meaningful.
>
>Steve
>
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