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A 50 kVA, 500 V, single phase a.c. generator gave the following test results:
Open circuit test: A field current of 12 A produced an e.m.f. of 300 V.
Short circuit test: A field current of 12 A caused a current of 175 A to flow in the short-circuited armature.
The synchronous impedance of the generator is
Open circuit test: A field current of 12 A produced an e.m.f. of 300 V.
Short circuit test: A field current of 12 A caused a current of 175 A to flow in the short-circuited armature.
The synchronous impedance of the generator is
- a)2.61 Ω
- b)3.48 Ω
- c)4.85 Ω
- d)1.71 Ω
Correct answer is option 'D'. Can you explain this answer?
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A 50 kVA, 500 V, single phase a.c. generator gave the following test r...
Synchronous impedance,
Most Upvoted Answer
A 50 kVA, 500 V, single phase a.c. generator gave the following test r...
Ohms
b)2.88 ohms
c)2.72 ohms
d)2.55 ohms
We can use the open circuit test and short circuit test to calculate the synchronous impedance of the generator.
From the open circuit test, we know that the field current is 12 A and the voltage is 300 V. Therefore, the open circuit voltage per phase is:
E0 = V / sqrt(3) = 500 / sqrt(3) = 288.7 V
From the short circuit test, we know that the field current is 12 A and the current is 175 A. Therefore, the short circuit impedance per phase is:
Zsc = E0 / Isc = 288.7 / 175 = 1.65 ohms
The synchronous impedance per phase is related to the short circuit impedance and the open circuit voltage by:
Zs = (Zsc^2 + R^2) / Zsc
where R is the resistance per phase of the generator. We can rearrange this equation to solve for R:
R = sqrt(Zs^2 - Zsc^2)
Substituting the given values, we get:
R = sqrt(2.61^2 - 1.65^2) = 2.16 ohms
Therefore, the synchronous impedance per phase is:
Zs = sqrt(R^2 + Zsc^2) = sqrt(2.16^2 + 1.65^2) = 2.72 ohms
The answer is (c) 2.72 ohms.
b)2.88 ohms
c)2.72 ohms
d)2.55 ohms
We can use the open circuit test and short circuit test to calculate the synchronous impedance of the generator.
From the open circuit test, we know that the field current is 12 A and the voltage is 300 V. Therefore, the open circuit voltage per phase is:
E0 = V / sqrt(3) = 500 / sqrt(3) = 288.7 V
From the short circuit test, we know that the field current is 12 A and the current is 175 A. Therefore, the short circuit impedance per phase is:
Zsc = E0 / Isc = 288.7 / 175 = 1.65 ohms
The synchronous impedance per phase is related to the short circuit impedance and the open circuit voltage by:
Zs = (Zsc^2 + R^2) / Zsc
where R is the resistance per phase of the generator. We can rearrange this equation to solve for R:
R = sqrt(Zs^2 - Zsc^2)
Substituting the given values, we get:
R = sqrt(2.61^2 - 1.65^2) = 2.16 ohms
Therefore, the synchronous impedance per phase is:
Zs = sqrt(R^2 + Zsc^2) = sqrt(2.16^2 + 1.65^2) = 2.72 ohms
The answer is (c) 2.72 ohms.
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A 50 kVA, 500 V, single phase a.c. generator gave the following test results:Open circuit test: A field current of 12 A produced an e.m.f. of 300 V.Short circuit test: A field current of 12 A caused a current of 175 A to flow in the short-circuited armature.The synchronous impedance of the generator isa)2.61Ωb)3.48Ωc)4.85Ωd)1.71ΩCorrect answer is option 'D'. Can you explain this answer?
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A 50 kVA, 500 V, single phase a.c. generator gave the following test results:Open circuit test: A field current of 12 A produced an e.m.f. of 300 V.Short circuit test: A field current of 12 A caused a current of 175 A to flow in the short-circuited armature.The synchronous impedance of the generator isa)2.61Ωb)3.48Ωc)4.85Ωd)1.71ΩCorrect answer is option 'D'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A 50 kVA, 500 V, single phase a.c. generator gave the following test results:Open circuit test: A field current of 12 A produced an e.m.f. of 300 V.Short circuit test: A field current of 12 A caused a current of 175 A to flow in the short-circuited armature.The synchronous impedance of the generator isa)2.61Ωb)3.48Ωc)4.85Ωd)1.71ΩCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 50 kVA, 500 V, single phase a.c. generator gave the following test results:Open circuit test: A field current of 12 A produced an e.m.f. of 300 V.Short circuit test: A field current of 12 A caused a current of 175 A to flow in the short-circuited armature.The synchronous impedance of the generator isa)2.61Ωb)3.48Ωc)4.85Ωd)1.71ΩCorrect answer is option 'D'. Can you explain this answer?.
A 50 kVA, 500 V, single phase a.c. generator gave the following test results:Open circuit test: A field current of 12 A produced an e.m.f. of 300 V.Short circuit test: A field current of 12 A caused a current of 175 A to flow in the short-circuited armature.The synchronous impedance of the generator isa)2.61Ωb)3.48Ωc)4.85Ωd)1.71ΩCorrect answer is option 'D'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A 50 kVA, 500 V, single phase a.c. generator gave the following test results:Open circuit test: A field current of 12 A produced an e.m.f. of 300 V.Short circuit test: A field current of 12 A caused a current of 175 A to flow in the short-circuited armature.The synchronous impedance of the generator isa)2.61Ωb)3.48Ωc)4.85Ωd)1.71ΩCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 50 kVA, 500 V, single phase a.c. generator gave the following test results:Open circuit test: A field current of 12 A produced an e.m.f. of 300 V.Short circuit test: A field current of 12 A caused a current of 175 A to flow in the short-circuited armature.The synchronous impedance of the generator isa)2.61Ωb)3.48Ωc)4.85Ωd)1.71ΩCorrect answer is option 'D'. Can you explain this answer?.
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ample number of questions to practice A 50 kVA, 500 V, single phase a.c. generator gave the following test results:Open circuit test: A field current of 12 A produced an e.m.f. of 300 V.Short circuit test: A field current of 12 A caused a current of 175 A to flow in the short-circuited armature.The synchronous impedance of the generator isa)2.61Ωb)3.48Ωc)4.85Ωd)1.71ΩCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice SSC tests.
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