Problem 1For the system shown in Figure 12.9:a) Find PT, QT, and ST.b) Draw the power triangle.c) Find the power factorseen by the source E.d) Find Is.+E 120 V/50°Load 11100 VAR (C)750 WLoad 2950 VAR (L)100 WLoad 4700 VAR (C)200 WLoad 3Figure 12.91900 VAR (L)500 W

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Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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The circuit shows an unbalanced electrical installation powered by a positive sequence symmetrical three- phase network of 380 V compound voltage. The loads are single phase. Load 1 absorbs an active power of 950 W with f.d.p. 0.5 inductive. Load 2 absorbs an active power of 1,140 W with f.d.p. Unit. Load 3 absorbs from the network an active power of 760 W with f.d.p. 0.5 capacitive. Taking the network voltage URN as the phase reference, calculate: a) complex expressions of the line currents IR. Is. and IT RO- SO- To- : IR Is IT UL-380 V CARGA 1 P₁ = 950 W cos p1=0,5 ind. CARGA 2 P₂=1.140 W cos p₂=1 CARGA 3 P3=760 W cos p3=0,5 cap.
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The circuit shows an unbalanced electrical installation powered by a positive sequence symmetrical three- phase network of 380 V compound voltage. The loads are single phase. Load 1 absorbs an active power of 950 W with f.d.p. 0.5 inductive. Load 2 absorbs an active power of 1,140 W with f.d.p. Unit. Load 3 absorbs from the network an active power of 760 W with f.d.p. 0.5 capacitive. Taking the network voltage URN as the phase reference, calculate: a) complex expressions of the line currents IR. Is. and IT RO- SO TO IR Is IT U₁-380 V CARGA 1 P₁ =950 W cos p1=0,5 ind. CARGA 2 P₂=1.140 W cos p₂=1 CARGA 3 P3=760 W cos p3=0,5 cap.
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Problem 1 For the system shown in Figure 12.9: a) Find PT, QT, and ST. b) Draw the power triangle. c) Find the power factor seen by the source E. d) Find Is. + E 120 V/50° Load 1 1100 VAR (C) 750 W Load 2 950 VAR (L) 100 W Load 4 700 VAR (C) 200 W Load 3 Figure 12.9 1900 VAR (L) 500 W