@almarg Hi Roger,
To use your example of 28.28 volts RMS into 8 ohms, corresponding to 100 watts, the corresponding RMS current is 28.28/8 = 3.54 amps.
Assuming a sine wave, the peak voltage is 28.28 x 1.414 = about 40 volts.
The corresponding peak current is 3.54 x 1.414 = 40/8 = about 5 amps.
The corresponding instantaneous peak power is 40 x 5 = 200 watts.
What Kijanki is saying is that the term "RMS power," if strictly interpreted, would imply 200 watts peak x 0.707 = 141.4 watts RMS. But of course what is really being referred to when that term is used is the product of RMS voltage and RMS current, which as you indicated is 100 watts in this example
So the widespread use of the term "RMS power" is, strictly speaking, a misnomer. That is Kijanki’s point, with which I agree.
Your first calculations in getting the peak voltage, current and power are indeed correct.
200 watts is the peak power however 100 watts is the RMS power. Thanks for not using the word average which does not apply. I will explain another way.
Kijanki needs a math lesson. 0.707 is the proper factor for the voltage and the current as you have demonstrated. One must however use that factor for both voltage and current. If you want to apply it to the already calculated peak power one must multiply that by 0.707 x 0.707=0.5 Correct?
Thus RMS power is 0.5 x peak power. It is incorrect to multiply power by 0.707 just once. One has to do it twice
If you read the OP’s last line he actually got it the 0.5 right but he called it average. I really dont see what he is going on about. A power amp behaves just like the power supplied to your home. You can use a power amps to run motors at various speeds and all sorts of things. This is basic electronics..