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GATE 2014 EE (EE1) – Question 41

Power Systems · Symmetrical components, Symmetrical and unsymmetrical fault analysis · 2 marks · Multiple choice

An unbalanced three-phase system has phase I_a = 1∠−90° pu, negative-sequence I_b2 = 4∠−150° pu, and zero-sequence I_c0 = 3∠90° pu. Find |I_b| (pu).

  1. 1.00
  2. 7.81
  3. 11.53
  4. 13.00

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Show answer and explanation

Correct answer: (C) 11.53

Explanation

Let a=e^(j120°). The zero-sequence currents all equal3j. Negative-sequence Ib2=aIa2, so Ia2=Ib2/a=4j. Since Ia=−j, Ia1=Ia−Ia0−Ia2=−8j. Then Ib=3j+a²(−8j)+4∠−150°=−6√3+5j. Its magnitude is√(108+25)=√133=**11.5326pu,C**.