Fault Analysis using Symmetrical Components.

Discussion in 'General Electronics Chat' started by lam58, Aug 7, 2015.

  1. lam58

    Thread Starter Member

    Jan 3, 2014
    69
    0
    Hi could someone check my answer to the question below. THIS IS NOT HOMEWORK.

    Question:
    qasa.png

    My Answer:

     I_{fA} = 3I_{1A} = 3I_{2A} = 3I_{0A} \Rightarrow I_{fA} = \frac{3E}{Z_1 + Z_2 + Z_0} = \frac{33kV}{20+20+30} = 471.43A

    Then using the same method I got  I_{fB} = 942.86A
     
    Last edited: Aug 7, 2015
  2. lam58

    Thread Starter Member

    Jan 3, 2014
    69
    0
    Anyone?
     
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