Consider the system shown in below fig. phase 'b'is open due to conductor break.Calculate the sequence or symmetrical components of currents and also the neutral current. 1- 50L0° A 1,- 50120° A 5o6 50l120 la0 = 17.77 L60, la1 = 44.44 LO, la2 = 17.77 L-60 In = 60L60 None of the above la0 = 15.55 L60, la1 = 66.66 LO, la2 = 15.55 L-60 In = 30L60 la0 = 16.66 L60, la1 = 33.33 LO, la2 = 16.66 L-60 In = 50L60
Consider the system shown in below fig. phase 'b'is open due to conductor break.Calculate the sequence or symmetrical components of currents and also the neutral current. 1- 50L0° A 1,- 50120° A 5o6 50l120 la0 = 17.77 L60, la1 = 44.44 LO, la2 = 17.77 L-60 In = 60L60 None of the above la0 = 15.55 L60, la1 = 66.66 LO, la2 = 15.55 L-60 In = 30L60 la0 = 16.66 L60, la1 = 33.33 LO, la2 = 16.66 L-60 In = 50L60
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
Chapter4: Transmission Line Parameters
Section: Chapter Questions
Problem 4.31P: Figure 4.37 shows the conductor configuration of a three-phase transmission line and a telephone...
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