Зф SourceZLineЗф LoadSource:Zan = Zon = Zen = 0.1 + j1.5 QEan = 12020° vEon = 1202 - 120° vEon = 1202120° v%3DLine:ZLinea= 1 + j2 Q;ZLines= 1 - j2 0;ZLinec= 0 + j2.2361 0Load:Za=Z=Z = 40 + j30 Q*Fault located on the side of the 3-phase sourceFind the load line currents I, , I,, I, under unbalanced conditions1. Without the fault2. With faults

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

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

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter5: Transmission Lines: Steady-state Operation

Section: Chapter Questions

Problem 5.25MCQ

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