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# HW 03 - Spring2011 Homework 3 Solutions 3.1(25 pt The turns...

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ELEN 3441 – Fundamentals of Power Engineering Spring 2011 Page | 1 Homework 3 – Solutions 3.1. (25 pt.) The turns ratio of this transformer is a = 8000/277 = 28.89. Therefore, the secondary impedances referred to the primary side are c) To simplify the calculations, we use the simplified equivalent circuit referred to the primary side of the transformer:

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ELEN 3441 – Fundamentals of Power Engineering Spring 2011 Page | 2 3.2. (20 pt.) We refer the circuit to the secondary (low-voltage) side. The feeder’s impedance referred to the secondary side is The secondary current I s is given by
ELEN 3441 – Fundamentals of Power Engineering Spring 2011 Page | 3 90 000 46.04 ; 46.04 31.79 2300 0.85 s IA == = ° s I a) The voltage at the power source of this system (referred to the secondary side) is () ( ) ( ) ' 2 300 0 46.04 31.79 1.12 4.11 46.04 31.79 0.12 0.5 2461.17 3.6 sl i n e E Q ZZ j jV ++= ° + ° + + ° + = source s s V' = V I I Therefore, the voltage at the power source is 14 000 2461.17 3.6 14.36 3.6 2 400 kV =∠ ° = ° sourse V b)

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HW 03 - Spring2011 Homework 3 Solutions 3.1(25 pt The turns...

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