In Figure 1, the 3-phase source has a line-to-line voltage of 440 volts. Thedetails of the impedances are the following: Z₁ = 36-j15; Z₂ = 40 +130 0; Z₂ = 45+j600; Z4 = 75 + j0; Z5 = 50+j50 ; Z6 = 20+ j48.Determine the following:(a) Line currents: labi, and I(c) Line currents: lar Ibr, and ICT;(b) Line currents: Iaz, Ibz, and Iczi(d) PT, QT, ST, and pfr for the system;(e) W, and W, readings for 2-wattmeter method at lines a and b.Balanced3-Phaseab-bc-casourceCHFigure 1→laT-Ibr-ICT1lal²31Ibl1Ici→la2lb2→1c2251²₂OND

The line-to-line voltage of 3-phase source is VL=440 V.Two three-phase Y- and delta-connected loads…

In Figure 1, the 3-phase source has a line-to-line voltage of 440 volts. The details of the impedances are the following: Z₁ = 36-j15; Z₂ = 40 + /300 Z=45+j600; Z4 = 75+ j0; Z5 = 50+ j50 ; Z6 = 20 + j48 2. Determine the following: (a) Line currents: labi, and lei; (b) Line currents: Iaz, Ibz, and Iczi (c) Line currents: lar, Ibr. and ICT; (d) PT, QT, ST, and pfr for the system; (e) W, and W, readings for 2-wattmeter method at lines a and b. Balanced 3-Phase ab-bc-ca source b. CH Figure 1 -1₂T -ICT lal Z2 123 TI Ibi 1 Ici →la2 162 →lc2 26

In Figure 1, the 3-phase source has a line-to-line voltage of 440 volts. The details of the impedances are the following: Z₁ = 36-j15; Z₂ = 40 + /300 Z=45+j600; Z4 = 75+ j0; Z5 = 50+ j50 ; Z6 = 20 + j48 2. Determine the following: (a) Line currents: labi, and lei; (b) Line currents: Iaz, Ibz, and Iczi (c) Line currents: lar, Ibr. and ICT; (d) PT, QT, ST, and pfr for the system; (e) W, and W, readings for 2-wattmeter method at lines a and b. Balanced 3-Phase ab-bc-ca source b. CH Figure 1 -1₂T -ICT lal Z2 123 TI Ibi 1 Ici →la2 162 →lc2 26

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 2.23P: A single-phase source has a terminal voltage V=1200volts and a currentI=1530, which leaves the...

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