-- The RMS value of an AC waveform is (1/2)(√2) x (peak value)
So the peak value is (√2) x (RMS value)
-- The Average value of an AC waveform is (2/π) x (peak value)
So the peak value is (π/2) x (Average value).
-- So far, this is all very entertaining, but how does it help us answer the question.
Well, we found the peak value in terms of the RMS and in terms of the Average. So we can set these equal to each other, and solve for the Average in terms of the RMS. This sounds like such a good plan, I think I'll do it !
Peak = (√2) x (RMS value) and Peak also = (π/2) x (Average value).
So (√2) · RMS = (π/2) · Average .
Divide each side by π : (√2) · RMS / π = (1/2) · Average
Multiply each side by 2 : Average = (2/π) · (√2) · RMS .
You said that the RMS value is 80 V, so
Average = (2/π) · (√2) · (80)
Average = (2 · √2 / π) · (80)
Average = 160√2 / π
<em>Average = 72 volts </em>. (But be sure to read the 'gotcha' below.)
Now I'll go ahead and tell you the 'gotcha':
All of these numbers are true, as far as they go. But the 'average' is only true for 1/2 cycle of an AC wave. Picture an AC wave in your mind. You'll see that it spends just as much time being negative as it spends being positive. So the 'average' of any number of AC <em><u>whole cycles</u></em> is <em>zero.</em>
