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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?

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  • \$\begingroup\$ Why can't I use concrete to make biscuits? Power = voltage x current. Try and fit that into your math. \$\endgroup\$
    – Andy aka
    Commented Apr 30, 2020 at 14:16
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    \$\begingroup\$ 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. \$\endgroup\$
    – Bimpelrekkie
    Commented Apr 30, 2020 at 14:26
  • \$\begingroup\$ 1/sqrt2 * 1/sqrt2 =1/2 is what I mean by half \$\endgroup\$
    – hontou_
    Commented Apr 30, 2020 at 14:38
  • \$\begingroup\$ 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. \$\endgroup\$
    – Solomon Slow
    Commented Apr 30, 2020 at 14:39
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    \$\begingroup\$ 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. \$\endgroup\$
    – Solomon Slow
    Commented Apr 30, 2020 at 15:57

1 Answer 1

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You have made a false math assumption. You have caught from the wind "taking average and multiplication are distributive". That's not true as you have already found. You can check it with two voltage samples U1, U2 and two current samples I1, I2

The average power is (U1*I1 + U2*I2)/2. There's no way to reduce this to ((U1+U2)/2)*((I1+I2)/2)

You must calculate P(t)=U(t)*I(t). That's the momentary power. The average power is the average of U(t)*I(t) calculated over the period of interest. With sinusoidal current and voltage we calculate the average over one cycle.

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  • \$\begingroup\$ Ah, I get it, so my way of averaging is wrong. What is then the use of the formula for averaging V=2/PI? \$\endgroup\$
    – hontou_
    Commented Apr 30, 2020 at 14:44
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    \$\begingroup\$ Your 2V/Pi is the average of rectified sinusoidal voltage which has peak voltage =V. Rectifying = turning negative half cycles to positive. In electronics the formula is used tell the DC component of rectified sinusoidal voltage or current. \$\endgroup\$
    – user136077
    Commented Apr 30, 2020 at 14:51

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