LESSON 8AC DISTRIBUTION SYSTEMExample 2.14: A Single-Phase AC Ring Main DistributionSystem, ABCA, turning in a clockwise direction, fed at Point Awith 240 V, supplies the following loads:Load PointLoad CurrentPower FactorB75 A0.875 lagC60 A0.9 lag• Calculate the Current in each section of the Distributor if the TotalImpedances of the various sections (go and return) are as follows:ZBC =(0.05 + j0. 1)ZCA = (0.025 + j0.5) N%3DZAB = (0.01 + j0.015) N

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LESSON 8 AC DISTRIBUTION SYSTEM Example 2.14: A Single-Phase AC Ring Main Distribution System, ABCA, turning in a clockwise direction, fed at Point A with 240 V, supplies the following loads: Load Point Load Current Power Factor 75 A 0.875 lag 60 A 0.9 lag Calculate the Current in each section of the Distributor if the Total Impedances of the various sections (go and return) are as follows: ZBC = (0.05 + j0. 1) N ZCA = (0.025 + j0.5) N %3D ZAB = (0.01 + j0.015) N %3D

LESSON 8 AC DISTRIBUTION SYSTEM Example 2.14: A Single-Phase AC Ring Main Distribution System, ABCA, turning in a clockwise direction, fed at Point A with 240 V, supplies the following loads: Load Point Load Current Power Factor 75 A 0.875 lag 60 A 0.9 lag Calculate the Current in each section of the Distributor if the Total Impedances of the various sections (go and return) are as follows: ZBC = (0.05 + j0. 1) N ZCA = (0.025 + j0.5) N %3D ZAB = (0.01 + j0.015) N %3D

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

Chapter6: Power Flows

Section: Chapter Questions

Problem 6.52P

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