Average AC Power And Reactive Power Absorbed By A Load
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Average AC Power And Reactive Power Absorbed By A Load

The AC circuit of figure 1.1 has:
effective voltage = RMS voltage = Vs = 320 V,
Internal impedance of Vs = Zs = 50 Ω + j100Ω
Load impedance, ZL = 200 Ω + j100Ω
Calculate:
(a) The average power and reactive power deliverd by Vs.
(b) The average power and reactive power absorbed by the load.
(c) A reactive element jX is added in parallel to ZL. Find the X such that power delivered to ZL is maximized.

The strings: S7P3A32 (force-push).

The math:
Pj Problem of Interest is of type force (push).

(a)Equations: average power Pavg = VrmsIrmscosθ
Reactive power = Q = VrmsIrmssineθ
Zeq = Zs + ZL
So, Zeq = 50 + j100 + 200 + j100 = 250 + j200
So, |Zeq| = [(250)2 + (200)2]1/2.
So, |Zeq| = [102500]1/2 = 320 Ω.
So, Irms = Vrms/Zeq = 320/320 = 1 A.
Cosθ = 250/320 = 0.78125.
So, Pavg = VrmsIrmscosθ = 320 x 1 x 0.78125 = 250 W.
Sineθ = 200/320 = 0.625
So, reactive power = Q = 320 x 1 x 0.625 = 200 VAR (volt-amperes-reactive).

(b) Pavg absorbed by Load = Pavg - Pavg dissipated by Zs.
Z = 50 + j100.
So, |Zs| = [(50)2 + (100)2)]1/2 = [12500]1/2 = 111.8.
So, VZs = IZs|Zs| = 1 x 111.8 =111.8
So, Pavg dissipated by Zs = 111.8 x 1 x (50/111.8) = 50 W
So, Pavg absorbed by Load = 250 - 50 = 200 W.
So, Qof Zs = 111.8 x 1 x (100/111.8) = 100 VAR
So, Qload = 200 - 100 = 100 VAR.

(c) Maximum power transfer theorem: to transfer maximum power to a load in an AC circuit, the equivalent source impedance and equivalent load impedance must be matched, that is, equivalent load impedance must be the conjugate of equivalent source impedance.
So, conjugate of (200 + j100 || jX) = 50 + j100.
X = -100 Ω satisfies the equation.

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