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An alternator has positive, negative and zero sequence impedances as j0.25, j0.25 and j0.5 respectively, is grounded with an impedance of j0.2. For a LLG fault on phase ‘b’ and ’c* the positive sequence current ofthe network is____________
  • a)
        -j2.204 p.u.            
  • b)
        j2.204p.u.
  • c)
      -j2.3 p.u.   
  • d)
      j2.3 p.u.
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
An alternator has positive, negative and zero sequence impedances as j...
The given problem involves determining the positive sequence current of the network during a LLG fault on phase b and c* of an alternator. To solve this, we need to consider the sequence impedance of the alternator and the grounding impedance.

Given data:
Positive sequence impedance (Z1) = j0.25
Negative sequence impedance (Z2) = j0.25
Zero sequence impedance (Z0) = j0.5
Grounding impedance (Zg) = j0.2

To find the positive sequence current, we can use the positive sequence equivalent circuit of the alternator during the fault condition.

1. Find the fault current:
During a LLG fault, the fault current will flow through phases b and c*. Since the fault is LLG, the zero sequence current will also flow through these phases. Therefore, the fault current can be calculated using the following equation:

Ifault = (V1 - V2) / (Z1 + Z2 + 2*Z0 + 2*Zg)

2. Calculate the positive sequence current:
The positive sequence current (I1) can be calculated using the following equation:

I1 = Ifault * (Z1 + Zg) / (Z1 + Z2 + 2*Z0 + 2*Zg)

Substituting the given values, we have:

Ifault = (V1 - V2) / (j0.25 + j0.25 + 2*j0.5 + 2*j0.2)
= (V1 - V2) / (j1.7)

I1 = Ifault * (j0.25 + j0.2) / (j0.25 + j0.25 + 2*j0.5 + 2*j0.2)
= Ifault * (j0.45) / (j1.7)
= Ifault * (j0.45/j1.7)

So, the positive sequence current of the network is -j2.204 p.u. (per unit).

Therefore, the correct answer is option 'A' (-j2.204 p.u.).
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An alternator has positive, negative and zero sequence impedances as j0.25, j0.25 and j0.5 respectively, is grounded with an impedance of j0.2. For a LLG fault on phase b and c* the positive sequence current ofthe network is____________a) -j2.204 p.u.b) j2.204p.u.c) -j2.3 p.u.d) j2.3 p.u.Correct answer is option 'A'. Can you explain this answer?
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An alternator has positive, negative and zero sequence impedances as j0.25, j0.25 and j0.5 respectively, is grounded with an impedance of j0.2. For a LLG fault on phase b and c* the positive sequence current ofthe network is____________a) -j2.204 p.u.b) j2.204p.u.c) -j2.3 p.u.d) j2.3 p.u.Correct answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An alternator has positive, negative and zero sequence impedances as j0.25, j0.25 and j0.5 respectively, is grounded with an impedance of j0.2. For a LLG fault on phase b and c* the positive sequence current ofthe network is____________a) -j2.204 p.u.b) j2.204p.u.c) -j2.3 p.u.d) j2.3 p.u.Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An alternator has positive, negative and zero sequence impedances as j0.25, j0.25 and j0.5 respectively, is grounded with an impedance of j0.2. For a LLG fault on phase b and c* the positive sequence current ofthe network is____________a) -j2.204 p.u.b) j2.204p.u.c) -j2.3 p.u.d) j2.3 p.u.Correct answer is option 'A'. Can you explain this answer?.
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