Ian Iveson wrote:

"Paul D. Spiegel" wrote


...Load = 4 ohms
Peak-to-Peak (P-P) output voltage before clipping is 8 volts.
Using the voltage power formula:
(8*8)/4=16 watts

However, it's customary to represent voltage with the root mean
square
(RMS) value. So 8 P-P becomes 5.66v rms when divided by 1.414
(sq. root
of 2). Now, (5.66*5.66)/4=8 watts...


RMS values of voltage *and* current *must* be used in the general
equation P=V.I, from which both of your variants can be derived
using ohms law.

That is because rms * rms = ms, the mean square. Since power is the
square, the result is mean power. Note that "rms power" is a common
misnomer. It's not a matter of custom, but of mathematics and
coherence.


...The constant current of the primary is 100ma. The output
transformer
has a winding ratio of 35:1 so this calculates to 3.5a at the
secondary.

Using the current power calculation:
(3.5*3.5)*4= 49 watts...


In your current calculation, you have calculated the dissipation due
to the bias, mostly that of the valves. For the dissipation by the
load, which is what you want, you need the current variation due to
the signal. You will find this will give the same answer as your
voltage calculation as long as you use rms current.

cheers, Ian



In addition to Ivan's information, in order to calculate true rms voltage to use
in your power calculation, the rms value of a SINE wave is .707 times the PEAK
voltage. The peak voltage is simply half of the peak-to-peak voltage.
(.707 is 1/1.414. this factor is only valid for sine waves.)

Paul

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