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At the terminal of a 3ϕ, 6.6 kV, 10 MVA alternator, a load R = 200 Ω is connected between two phases, and other phase is kept open. The sequence impedance of the alternator is Z1 = Z2 = j 5Ω and Z0 = j2Ω.
What is the current through the load resistance?
  • a)
    50 A
  • b)
    60 A
  • c)
    75 A
  • d)
    32 A
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
At the terminal of a 3, 6.6 kV, 10 MVA alternator, a load R = 200 &Ome...
Concept:
In a double line fault, the fault current is given by

Where Z1 is the positive sequence impedance
Z2 is the negative sequence impedance
Calculation:
V = 6.6 kV
Z1 = Z2 = j5Ω
Zf = 200 Ω
Z0 = j2Ω
This is the case of the line to line fault.

Ia = 0 (∵ open circuited)
From the formula of a double line fault or L-L fault,
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Community Answer
At the terminal of a 3, 6.6 kV, 10 MVA alternator, a load R = 200 &Ome...
Concept:
In a double line fault, the fault current is given by

Where Z1 is the positive sequence impedance
Z2 is the negative sequence impedance
Calculation:
V = 6.6 kV
Z1 = Z2 = j5Ω
Zf = 200 Ω
Z0 = j2Ω
This is the case of the line to line fault.

Ia = 0 (∵ open circuited)
From the formula of a double line fault or L-L fault,
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At the terminal of a 3, 6.6 kV, 10 MVA alternator, a load R = 200 Ω is connected between two phases, and other phase is kept open. The sequence impedance of the alternator is Z1 = Z2 = j 5Ω and Z0 = j2Ω.What is the current through the load resistance?a)50 Ab)60 Ac)75 Ad)32 ACorrect answer is option 'D'. Can you explain this answer?
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At the terminal of a 3, 6.6 kV, 10 MVA alternator, a load R = 200 Ω is connected between two phases, and other phase is kept open. The sequence impedance of the alternator is Z1 = Z2 = j 5Ω and Z0 = j2Ω.What is the current through the load resistance?a)50 Ab)60 Ac)75 Ad)32 ACorrect answer is option 'D'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about At the terminal of a 3, 6.6 kV, 10 MVA alternator, a load R = 200 Ω is connected between two phases, and other phase is kept open. The sequence impedance of the alternator is Z1 = Z2 = j 5Ω and Z0 = j2Ω.What is the current through the load resistance?a)50 Ab)60 Ac)75 Ad)32 ACorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At the terminal of a 3, 6.6 kV, 10 MVA alternator, a load R = 200 Ω is connected between two phases, and other phase is kept open. The sequence impedance of the alternator is Z1 = Z2 = j 5Ω and Z0 = j2Ω.What is the current through the load resistance?a)50 Ab)60 Ac)75 Ad)32 ACorrect answer is option 'D'. Can you explain this answer?.
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