Average Power Absorbed By Unbalanced Three Phase Wye-Connected Loads
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Average Power Absorbed By Unbalanced Three Phase Wye-Connected Loads The Three Phase AC circuit Of Figure 7.1 with unbalanced Wye-Connected loads has:
vs1 = 170cos(ωt) V
vs2 = 170cos(ωt + 2π/3) V
vs2 = 170cos(ωt - 2π/3) V
frequency, f = 60 Hz; Z1 = 0.5<20o Ω
Z2 = 0.35<0o Ω; Z3 = 1.7<-90o Ω
Determine the power absorbed by Z1, Z2 and Z3.

The strings: S7P3A32 (force-push).

The math:
Pj Problem of Interest is of type force (push). Equations: average power Pavg = VrmsIrmscosθ
Reactive power = Q = VrmsIrmssineθ
Ae = complex phasor notation for Acos(ωt + θ) = A<θ = Acosθ + jAsinθ

From Kirchoff's Current Law (KCL) at node O (figure 7.1):
(vog - vag)/Z1 + (vog - vbg)/Z2 + (vog - vcg)/Z3 = 0 -------(1)
So, vog[1/Z1 + 1/Z2 + 1/Z3] = (vag/Z1) + (vbg/Z2) + (vcg/Z3) ----------(2)

Left Side Of Equation (2) = vog[1/Z1 + 1/Z2 + 1/Z3]
So, vog[1/(0.5<20o) + 1/(0.35<0o) + 1/(1.7<-90o)] = vog[2<-20o + 2.857<0o + 0.588<90o]
= vog[2cos20o - j2sin20o + 2.857 + j0.588] = vog(4.74<0o).

Right Side Of Equation (2) = (vag/Z1) + (vbg/Z2) + (vcg/Z3)
So, 170<0o/0.5(<20o) + 170<120o/0.35(<0o) + 170<-120o/1.7(<-90o) = 340<-20o + 485.7<120o + 100<-30o
= 302.2<58.2o.

Left Side Of Equation (2) = Right Side Of Equation (2)
So, vog(4.74<0o) = 302.2<58.2o
So, vog = (302.2<58.2o)/(4.74<0o) = 63.82<58.2o.

Current in Z1 = Iz1 = voa/Z1 = (vog - vag)/Z1 = [(63.82<58.2o) - (170<0o)]/(0.5<20o) = 293<-41.8o.

Vrms = Vpeak/(2)1/2
Irms = Ipeak/(2)1/2
So, average power absorbed by Z1 = VrmsIrmscosθ = 146.8/(2)1/2(293)/(2)1/2(o.9397) = 21506 watts = 21.5 kW.

Average power in Z2 amd Z3 can be similarly determined.

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