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Homework 6

# Homework 6 - Homework 06 Randall Robert 9771 Problem 1...

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Homework 06 Randall Robert November 21, 2004 9771 Problem 1. Calculate RMS values of the line currents as well as the active an reactive powers in the circuit shown in Fig. 1, assuming that the load is supplied with a symmetrical voltage such that u R = 120 √(2) cos ωt V Load parameters R RS =3.5Ω, ωL ST =5.5Ω, R TR =4.5Ω, ωL TR =4.5Ω Assume V R =120√2e j90° V To find the line to line voltages V RS = V R – V S = √(3)120√(2)e j120° V V ST = V S – V T = √(3)120√(2)e j0° V V TR = V T – V R = √(3)120√(2)e j-120° V To find the line to line currents I RS = V RS /R RS = √(6)(120)(120°)[V]/(3.5Ω) = 83.983(120°)[A] I ST = V ST /(ωL ST = √(6)(120)(0°)[V]/(j5.5Ω) = 53.443(30°)[A] I TR = V TR /(R TR + ωL TR ) = √(6)(120)(-120°)[V]/(4.5 + j4.5Ω) = 46.188(-165°)[A] To find the line currents I R = I RS – I TR = 84.727(88.226°)[A] I S = I ST – I RS = 99.545(-27.529°)[A] I T = I TR – I ST = 98.783(156.951°)[A] Verification I R + I S + I T = 84.727(88.226°)[A] + 99.545(-27.529°)[A] + 98.783(156.951°)[A] = 0 Active Power P = V R I R cosΦ R +V S I S cosΦ S +V T I T cosΦ T = (120(√(2)))(84.727[A])cos(90°-88.226°)+(120(√(2)))(99.545[A])* cos(-130°+27.529°)+(120(√(2)))(98.783[A]cos(-150°+156.951°) = 27.365W Reactive Power Q = V R I R sinΦ R +V S I S

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