Lecture 19

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Unformatted text preview: 0.952∠ − 18.2° g Solving for the (2) network we get (2) Ig (2) Im I (2) f Ef 1.05∠0° = = = − j7 j0.15 j0.15 E f 1.05∠0° = = = − j 2.1 j0.5 j0.5 = − j 7 − j 2.1 = − j 9.1 I g = 0.952∠ − 18.2° − j 7 = 7.35∠ − 82.9° This matches what we calculated earlier 11 ! Extension to Larger Systems The superposition approach can be easily extended to larger systems. Using the Ybus we have Ybus V = I For the second (2) system there is only one voltage source so I is all zeros except at the fault location ! M" #0$ # $ I = #− I f $ # $ #0$ # M$ & ' However to use this approach we need to first determine If 12 ! Determination of Fault Current Define the bus impedance matrix Z bus as −1 Z bus @ Ybus " Z11 L Then $ M O $ $ Z n1 L & V = Z busI "V1(2) # " M# $ 0 % $V (2) % Z1n # $ % $2 % M % $− I f % = $ M % %$ % $ (2) % Z nn % $ 0 % $V % ' n −1 $ M % $ (2) % & ' $Vn % & ' For a fault a b...
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This note was uploaded on 02/09/2014 for the course ECE 5042 taught by Professor Degroat during the Fall '13 term at Ohio State.

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