# Soln5 - 1 EE 212 Assignment 5 Solutions 1 A balanced 3-phase-connected load is energized by a 3-phase source that has line-to-line voltage of 208 V

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EE 212 - Assignment 5 - Solutions 1. A balanced 3-phase -connected load is energized by a 3-phase source that has line-to-line voltage of 208 V rms. and phase sequence abc. Per phase load impedance Z L = 12 + j16 ohms. Find the line and phase voltages and currents, and show them in a phasor diagram. Determine the real and reactive power delivered to the load. Per phase equivalent circuit : Taking phase voltage V ab as the reference, V ab  = 208 /0     0    V A j Z V I ab ab 0 0 13 . 53 4 . 10 16 12 0 208 - = + = = Figure 2 Phase currents: I ab = 10.4 -53.13 0 A I bc = 10.4 (-53.13 – 120) 0 = 10.4 -173.13 0 A I ca = 10.4 (-53.13 + 120) 0 = 10.4 66.87 0 A Line currents: I a = I ab - I ca = 10.4 -53.13 0 - 10.4 66.87 0 = 18.02 -83.1 0 A I b = I bc - I ab = 10.4 -173.13 0 - 10.4 -53.13 0 = 18.02 156.9 0 A I c = I ca - I bc = 10.4 66.87 0 - 10.4 -173.13 0 = 18.02 36.9 0 A Phase Voltages (Line Voltages are the same) : V ab = 208 0 0 Volts V bc = 208 -120 0 Volts V ca = 208 120 0 Volts 1 a b c a b 12+ j16 c 12+ j16 12+ j16 V L = 208 V 12+ j16 a b I ab V ab = 208 /0 0 V

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Phasor Diagram : Power per phase: P ph = I 2 R = (10.4) 2 * 12 = 1297.92 W Q ph = I 2 X = (10.4) 2 * 16 = 1730.56 VAr OR P ph = V ab * I ab Cos (53.13 ) = (208) * (10.4) Cos(53.13 ) = 1297.92 W Q ph = V ab * I ab Sin (53.13 ) = (208) * (10.4) Sin(53.13 ) = 1730.56 VAr Total Power: P = 3 * 1297.92 = 3893.76 W Q = 3 * 1730.56 = 5191.67 VAr 2 I ca I ab I bc I a V ab V ca V bc 30 0 53.13 0 -I ca I b I c
2.

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## This note was uploaded on 09/25/2010 for the course EE 212 taught by Professor Gander during the Spring '10 term at University of Saskatchewan- Management Area.

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Soln5 - 1 EE 212 Assignment 5 Solutions 1 A balanced 3-phase-connected load is energized by a 3-phase source that has line-to-line voltage of 208 V

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