# What gets wrong when I average AC voltage and current to get power?

Power is half the product of peak AC voltage and current (or product of its RMS voltage and current). But why is it wrong when I average first the voltage and current before getting the power?

I get that the average of AC voltage and current is zero. But what is then the use of the formula:

V(ave) = 2V/pi
AND
I(ave) = 2I/pi

This is the formula for averaging sinusoidal 360 degree voltage and current, right? Why can't I average the voltage and current to find power?

• Why can't I use concrete to make biscuits? Power = voltage x current. Try and fit that into your math. – Andy aka Apr 30 '20 at 14:16
• Power is half the product of peak AC voltage and current (or product of its RMS voltage and current So that would mean that 1/2 * AC voltage = RMS voltage? I do not agree with that. Go and look up what it really is. – Bimpelrekkie Apr 30 '20 at 14:26
• 1/sqrt2 * 1/sqrt2 =1/2 is what I mean by half – hontou_ Apr 30 '20 at 14:38
• You said that $V_{ave}=\frac{2V}{\pi}$ "is the formula for averaging sinusoidal 360 degree voltage." I am curious to know where you got that formula and, what its physical significance is supposed to be. – Solomon Slow Apr 30 '20 at 14:39
• Re, "...peak voltage by...0.637..." OK, right. That's one way to calculate the RMS voltage. RMS is useful to know because, for a pure sinewave power supply, and for a purely resistive load, The relationships between RMS voltage, power, and current are all the same formulas as the relationships between voltage, power, and current in a DC circuit. But basically, you should think of RMS calculations as a handy short-cut, and not as the physical explanation for anything. – Solomon Slow Apr 30 '20 at 15:57