Зф SourceZ LineЗф LoadSource:Zan = Zon = Zen = 0.1 + j1.5 NEan = 12020° vEon = 1202 - 120° v= 1202120° vEenLine:ZLinea= 1 + j2 0;ZLines= 1- j2 0;ZLinec= 0 + j2.2361 QLoad:Z,=Z=Ze = 40 + j30 Q*Fault located on the side of the 3-phase sourceFind the load line voltages V,, V,, Vc under unbalanced conditions1. Without the fault2. With faults

Given To find load line voltage 1) without fault 2) with fault here it is mentioned Va, Vb and Vc…

Source: Zan = Zon = Zen = 0.1 + j1.5 Q En = 12020° v En = 1202 - 120° v Een = 1202120° v %3D %3D %3D Line: ZLinea= 1 + j2 0; ZLines= 1 - j2 N; ZLinec= 0 + j2.2361 Q Load: Z=Zc=Ze = 40 + j30 N "Fault located on the side of the 3-phase source Find the load line voltages V,, V,, Vc under unbalanced conditions 1. Without the fault 2. With faults

Source: Zan = Zon = Zen = 0.1 + j1.5 Q En = 12020° v En = 1202 - 120° v Een = 1202120° v %3D %3D %3D Line: ZLinea= 1 + j2 0; ZLines= 1 - j2 N; ZLinec= 0 + j2.2361 Q Load: Z=Zc=Ze = 40 + j30 N "Fault located on the side of the 3-phase source Find the load line voltages V,, V,, Vc under unbalanced conditions 1. Without the fault 2. With faults

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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34 Source Zine 30 Load Source: Zan = Zbn = Zen = 0.1 + j1.5 Q Ean = 12020° v %3D %3D %3D Eon = 1502 - 143.13° v Ecn = 90490° v Line: ZLineA= 1 + j2 Q; ZLines= 1 - j2 Q; ZLinec= 0 + j2.2361 Q Load: Zan= 12 + j9 Q Zon= 12 - j9 N Zen= 15 + j0 Q *Fault located on the side of the 3-phase load Find the load line currents la , Ib , lc under unbalanced conditions 1. Without the fault 2. With fault
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