Consider the power versus time graph of an AC signal:

The average power, P_{avg}, for one cycle is the area under the graph divided by 2π.
So if 0≤θ≤2π, p is the instantaneous power, i is the instantaneous current, and
I_{0} is the peak
current, then

P_{avg}= 1/2π×∫_{0}^{2π}p_{i}dθ = 1/2π×∫_{0}^{2π}i^{2}R dθ = R/2π×∫_{0}^{2π}i^{2}dθ = R/2π×∫_{0}^{2π}I_{0}^{2}sin^{2}θ dθ = I_{0}^{2}R/2π×∫_{0}^{2π}sin^{2}θ dθ = I_{0}^{2}R/2π×π = I_{0}^{2}R/2

In the case of sinusoidal current, the Root Mean Square (RMS) of an AC source is the DC source that gives the same average power.

I_{RMS}= I_{0}/√2

and if the peak
voltage of the AC signal is V_{0}, then

V_{RMS}= V_{0}/√2

which gives us

P_{avg}= I_{RMS}^{2}R P_{avg}= I_{RMS}V_{RMS}

Fischer-Cripps. A.C., *The Electronics Companion.* Institute of Physics, 2005.

Copyright © 2014 Barry Watson. All rights reserved.