Consider the power versus time graph of an AC signal:
The average power, Pavg, 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 I0 is the peak current, then
Pavg = 1/2π×∫02πpidθ = 1/2π×∫02πi2R dθ = R/2π×∫02πi2dθ = R/2π×∫02πI02sin2θ dθ = I02R/2π×∫02πsin2θ dθ = I02R/2π×π = I02R/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.
IRMS = I0/√2
and if the peak voltage of the AC signal is V0, then
VRMS = V0/√2
which gives us
Pavg = IRMS2R Pavg = IRMSVRMS
Fischer-Cripps. A.C., The Electronics Companion. Institute of Physics, 2005.
Copyright © 2014 Barry Watson. All rights reserved.