Anabc-sequence set ofvoltages feeds a balanced three phase wye-wye system The line and load mped nces...
A balanced positive-sequence wye-connected 60-Hz three-phase source has line-to-line voltages of VL = 440 V rms. This source is connected to a balanced wye-connected load. Each phase of the load consists of a 0.5-H inductance in series with a 50-Ω resistance. Assume that the phase of Van is zero. home / study / engineering / electrical engineering / electrical engineering questions and answers / A Balanced Positive-sequence Wye-connected 60-Hz Three-phase Source Has Line-to-line Voltages ... Question: A balanced positive-sequence wye-connected...
1. In a three-phase balanced wye-wye system, the source is an abc-positive sequence set of voltages with Van- 120 20 Vrms. The per phase impedance of the load is 5 +j62. If the line impedance per phase is 0 Ω, find the line currents, the line to line voltages at the source, and the line to line voltages at the load 120 20 Vms 2. In a three-phase balanced wye-wye system, the source is an abc-sequence set of voltages with...
In a three-phase balanced wye-wye system, the source is an abc-positive sequence set of voltages with Van = 120 000 Vrms. The per phase impedance of the load is 5 +J60. If the line impedance per phase is 0 0, find the line currents, the line to line voltages at the source, and the line to line voltages at the load. 1. Van 120 0" Vrms ZY = 5+j Va VA neutral
Q1.A balanced 3-Phase Y-connected generator with positive sequence has an impedance of 0.2 + 0.5j ф and an intemal voltage of 120 V/d. The generator feeds a balanced 3- Phase Y-connected load having an impedance of 39 + 28j Ωφ. The impedance of the line connecting the generator to the load is 0.8 + 1Sj Ωφ. The a-phase internal voltage of the generator is specified as the reference phasor a) Construct the a-phase equivalent circuit of the system b) Calculate...
Questions cover CLOS Question 3 In a balanced three-phase wye-wye system, the source is an abe-sequence set of voltages and Van-120/40 V rms. If the a-phase line current and line impedance are known to be air Arms and 1 +jl Ω, respectively, find the load impedance Ans)
Questions cover CLOS Question 3 In a balanced three-phase wye-wye system, the source is an abe-sequence set of voltages and Van-120/40 V rms. If the a-phase line current and line impedance are known...
I. A balanced three-phase positive sequence wye-connected 60Hz source voltage has line-neutral voltage of V, 1000 V. It is connected to a three- phase wye connected load whose impedance is z 50+j37.7 2 a. Find line currents ia ib, ic and line-line voltages Vab, Vbo Vac b. Find the total complex power delivered to the load. c. Find delta-connected capacitor bank needed to bring the PF to 0.95 (lagging) of system. d. Re-calculate currents la, ib, io with Case c....
I. A balanced three-phase positive sequence wye-connected 60Hz source voltage has line-neutral voltage of V, 1000 V. It is connected to a three- phase wye connected load whose impedance is z 50+j37.7 2 a. Find line currents ia ib, ic and line-line voltages Vab, Vbo Vac b. Find the total complex power delivered to the load. c. Find delta-connected capacitor bank needed to bring the PF to 0.95 (lagging) of system. d. Re-calculate currents la, ib, io with Case c....
1. A balanced three-phase Y load has one phase voltage of Vcn = 277∠45◦ V . If the phase sequence is negative sequence i.e. acb, calculate the line voltages Vca, Vab, and Vbc . 2. What are the phase voltages for a balanced three-phase Y load, if Vba = 12.47∠−35◦kV ? The phase sequence is positive sequence i.e. abc. Please help me solve this with full working out step by step.
1. A balanced three-phase wye-delta system has Van 12020 V rms and Za-(51+j45)2 Ifthe line impedance per phase is (0.4 +j 1.2 Ω, find the total complex power delivered to the load.
A balanced Δ-connected load has an impedance of 216−j288Ω/ϕ. The load is fed through a line having an impedance of 3+j5Ω/ϕ. The phase voltage at the terminals of the load is 7.0 kV . The phase sequence is negative. Use VAB as the reference. Calculate the line voltage Vab, Vbc, and Vca at the sending end of the line. Please show how you put it in the calculator.. I think that is where I keep getting errors.