In Boylestad's Introductory Circuit Analysis 13th edition page 1176, there's an example about working out the total power dissipated by a circuit fed a nonsinusoidal signal. The signal is decomposed into:
- DC component: 63.6 V
- Fundamental: 70.71 V RMS
- Second harmonic: -29.98 V RMS
And the circuit is drawn like below in order to apply superposition.
(Not shown in the drawing above: the phase angle of the second harmonic is then changed to -90 so that all sources have the same polarity)
The current and average power for each component are found like this:
- DC component
I0 = 10.6 A
P0 = I02R = (10.6 A)2 (6 Ω) = 674.2 W - Fundamental
I1 = 1.85 A ∠-80.96°
P1 = I12R = (1.85 A)2 (6 Ω) = 20.54 W - Second harmonic
I2 = 0.396 A ∠-174.45°
P2 = I22R = (0.396 A)2 (6 Ω) = 0.941 W
And the total RMS current and total average power are found to be:
Irms = square root ((10.6 A)2 + (1.85 A)2 + (0.396 A)2) = 10.77 A
PT = Irms2R = (10.77 A)2 (6 Ω) = 695.96 W = P0 + P1 + P2
Why the total average power equals the sum of the powers from each component if the superposition theorem can't be applied to power? I understand PT = Irms2R, what I can't understand is why it's valid to sum P0 + P1 + P2.
(For context: I'm reviewing a few fine points I may have overlooked during EE undergraduation years ago)