AC Power Triangle

Figure 8.14 shows a simple AC circuit. Figure 8.14(a) is the time domain circuit while figure 8.14(b) is its power triangle. Given that v(t) = 16cosωt; i(t) = 4cos(ωt - π/6); and ω 377 rad/sec. Determine:
(a) The power factor, pf.
(b) The values of the real power P, the reactive power Q and the apparent power S of the power triangle.

The string:
S7P3A32 (Force - Push).
The math:

Pj Problem of interest is of type force. Power and energy problems are force problems. Electric power is a force-push.

(a) The power factor, pf = cosθ = cos30 = √3/2.
The power factor, pf = 0 for a purely inductive or capacitive load.
The power factor, pf = 1 for a purely resistive load. 0 < pf < 1, for every other case.

(b) Vrms = 0.707(16) = 11.312
Irms = 0.707(4) = 2.828
P = VrmsIrmscos30 = (√3/2)(11.312)(2.828) = 27.7 Watts.
Q = VrmsIrmssin30 = (1/2)(11.312)(2.828) = 15.995 VAR (volt-amperes-reactive)
S = VrmsIrms = (11.312)(2.828) = 31.990 VA (volt-amperes) = 27.7 + j15.995.

Blessed are they that have not seen, and yet have believed. John 20:29

TECTechnic Logo, Kimberlee J. Benart | © 2000-2021 | All rights reserved | Founder and Site Programmer, Peter O. Sagay.