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A three phase 100 kVA, 3000 V star connected alternator has an effective armature resistance is 0.2 Ω. A field current of 40 A produces a short circuit current of 200 A and open circuit emf of 1040 V line to line. The full load regulation of 0.8 leading power factor is – (in %)
Correct answer is between '-2,-1.5'. Can you explain this answer?
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Verified Answer
A three phase 100 kVA, 3000 V star connected alternator has an effecti...
EOC = 1040 V
E = 1701.23 V
% Regulation = 
=−1.776%Most Upvoted Answer
A three phase 100 kVA, 3000 V star connected alternator has an effecti...
Ohms per phase. The line voltage is:
The line voltage can be calculated using the formula:
VL = √3 × EL
Where VL is the line voltage, EL is the phase voltage.
Since the alternator is star connected, the phase voltage is equal to the line voltage divided by √3.
EL = VL / √3
Substituting the given values, we get:
EL = 3000 / √3 = 1732.05 V
The effective armature resistance is given as 0.2 ohms per phase. Therefore, the total armature resistance is:
RT = 0.2 × 3 = 0.6 ohms
The power factor of the alternator can be calculated using the formula:
PF = cos φ = P / S
Where P is the real power, S is the apparent power, and φ is the phase angle between the voltage and current.
Since the alternator is purely resistive, the phase angle is zero, and the power factor is 1.
The real power can be calculated using the formula:
P = VL × IL × PF
Where IL is the line current.
Substituting the given values, we get:
P = 3000 × IL × 1
Therefore, IL = P / (3000 × 1) = P / 3000
The apparent power is given by:
S = √3 × VL × IL
Substituting the values, we get:
S = √3 × 3000 × (P / 3000) = √3 × P
Since the alternator is rated for 100 kVA, we have:
S = 100000 VA
Therefore, √3 × P = 100000 VA
P = 100000 / √3 = 57735.03 W
The line current is:
IL = P / (VL × PF) = 57735.03 / (3000 × 1) = 19.245 A
Therefore, the line voltage is 3000 V, and the line current is 19.245 A.
The line voltage can be calculated using the formula:
VL = √3 × EL
Where VL is the line voltage, EL is the phase voltage.
Since the alternator is star connected, the phase voltage is equal to the line voltage divided by √3.
EL = VL / √3
Substituting the given values, we get:
EL = 3000 / √3 = 1732.05 V
The effective armature resistance is given as 0.2 ohms per phase. Therefore, the total armature resistance is:
RT = 0.2 × 3 = 0.6 ohms
The power factor of the alternator can be calculated using the formula:
PF = cos φ = P / S
Where P is the real power, S is the apparent power, and φ is the phase angle between the voltage and current.
Since the alternator is purely resistive, the phase angle is zero, and the power factor is 1.
The real power can be calculated using the formula:
P = VL × IL × PF
Where IL is the line current.
Substituting the given values, we get:
P = 3000 × IL × 1
Therefore, IL = P / (3000 × 1) = P / 3000
The apparent power is given by:
S = √3 × VL × IL
Substituting the values, we get:
S = √3 × 3000 × (P / 3000) = √3 × P
Since the alternator is rated for 100 kVA, we have:
S = 100000 VA
Therefore, √3 × P = 100000 VA
P = 100000 / √3 = 57735.03 W
The line current is:
IL = P / (VL × PF) = 57735.03 / (3000 × 1) = 19.245 A
Therefore, the line voltage is 3000 V, and the line current is 19.245 A.
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A three phase 100 kVA, 3000 V star connected alternator has an effective armature resistance is 0.2 Ω. A field current of 40 A produces a short circuit current of 200 A and open circuit emf of 1040 V line to line. The full load regulation of 0.8 leading power factor is – (in %)Correct answer is between '-2,-1.5'. Can you explain this answer?
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A three phase 100 kVA, 3000 V star connected alternator has an effective armature resistance is 0.2 Ω. A field current of 40 A produces a short circuit current of 200 A and open circuit emf of 1040 V line to line. The full load regulation of 0.8 leading power factor is – (in %)Correct answer is between '-2,-1.5'. 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 A three phase 100 kVA, 3000 V star connected alternator has an effective armature resistance is 0.2 Ω. A field current of 40 A produces a short circuit current of 200 A and open circuit emf of 1040 V line to line. The full load regulation of 0.8 leading power factor is – (in %)Correct answer is between '-2,-1.5'. 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 A three phase 100 kVA, 3000 V star connected alternator has an effective armature resistance is 0.2 Ω. A field current of 40 A produces a short circuit current of 200 A and open circuit emf of 1040 V line to line. The full load regulation of 0.8 leading power factor is – (in %)Correct answer is between '-2,-1.5'. Can you explain this answer?.
A three phase 100 kVA, 3000 V star connected alternator has an effective armature resistance is 0.2 Ω. A field current of 40 A produces a short circuit current of 200 A and open circuit emf of 1040 V line to line. The full load regulation of 0.8 leading power factor is – (in %)Correct answer is between '-2,-1.5'. 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 A three phase 100 kVA, 3000 V star connected alternator has an effective armature resistance is 0.2 Ω. A field current of 40 A produces a short circuit current of 200 A and open circuit emf of 1040 V line to line. The full load regulation of 0.8 leading power factor is – (in %)Correct answer is between '-2,-1.5'. 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 A three phase 100 kVA, 3000 V star connected alternator has an effective armature resistance is 0.2 Ω. A field current of 40 A produces a short circuit current of 200 A and open circuit emf of 1040 V line to line. The full load regulation of 0.8 leading power factor is – (in %)Correct answer is between '-2,-1.5'. Can you explain this answer?.
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