ZL = 2 + 4j N I Capacitors for pf correction Z, = 2 + 4j N VAB HE VBc VcA Load 1 Load 2 ZL = 2 + 4j N Load 1: Load 2: Total 3-phase S = 5 MVA at pf 0.8 lagging VIL = 25kV Total 3-phase P = 3 MW at pf 0.75 lagging VLL = 25kV

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
Chapter2: Fundamentals
Section: Chapter Questions
Problem 2.13P
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Note that the given 25 kV L-L voltage appears across the loads (right side of ZL) as denoted by Vab, Vbc, and Vca. Since there is no given voltage angle, you may assign angle zero to one of the line-to-neutral voltages.

1. Without the capacitors, what is the magnitude of the current passing through the line impedance ZL?

ZL = 2 + 4j N
I Capacitors for pf correction
ZL = 2 + 4j N
VAB
VBC
VCA
Load 1
Load 2
ZL = 2 + 4j N
Load 1:
Load 2:
Total 3-phase P = 3 MW at pf 0.75 lagging
VLL
Total 3-phase S = 5 MVA at pf 0.8 lagging
VLL
= 25kV
= 25kV
Transcribed Image Text:ZL = 2 + 4j N I Capacitors for pf correction ZL = 2 + 4j N VAB VBC VCA Load 1 Load 2 ZL = 2 + 4j N Load 1: Load 2: Total 3-phase P = 3 MW at pf 0.75 lagging VLL Total 3-phase S = 5 MVA at pf 0.8 lagging VLL = 25kV = 25kV
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