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Two shunt generators A and B rated as 50 kW, 500 V and 100 kW, 500 V respectively operating in parallel deliver a total current of 250 A. The regulation of generators A and B are 6 % and 4 % respectively. The currents IA and IB delivered by the generators are
- a)IA = 125 A, IB = 125 A
- b)IA = 120 A, IB = 130 A
- c)IA = 62.5 A, IB = 187.5 A
- d)IA = 97.5 A, IB = 152.5 A
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Two shunt generators A and B rated as 50 kW, 500 V and 100 kW, 500 V r...
For generator A,
Full load current,
Full load voltage drop,
Voltage of A at load current
For generator B,
Voltage of A at load current IB,
Full load current,
Full load voltage drop,
Voltage of A at load current
For generator B,
Voltage of A at load current IB,
Most Upvoted Answer
Two shunt generators A and B rated as 50 kW, 500 V and 100 kW, 500 V r...
Given Data:
Generator A rating, P1 = 50 kW
Generator B rating, P2 = 100 kW
Generator A voltage, V = 500 V
Generator B voltage, V = 500 V
Total current, I = 250 A
Regulation of generator A = 6%
Regulation of generator B = 4%
To find:
Current delivered by generator A, IA and current delivered by generator B, IB.
Solution:
1. Power output of generator A and B
The power output of generator A and B can be calculated as:
P1 = V × IA
P2 = V × IB
Substituting the given values, we get:
P1 = 500 × IA
P2 = 500 × IB
2. Currents delivered by the generators
The total current is given as 250 A, which can be written as:
I = IA + IB
Substituting the given values, we get:
250 = IA + IB
3. Calculation of IA and IB
To calculate IA and IB, we need to solve the above two equations simultaneously.
Dividing equation (1) by equation (2), we get:
P1/P2 = IA/IB
Substituting the given values and simplifying, we get:
IA/IB = 1/2
Multiplying equation (2) by IA/IB, we get:
IA^2/2 + IA/2 = 250
Substituting IA/IB = 1/2, we get:
IA^2/2 + 1/2 × IA = 250
Multiplying both sides by 2, we get:
IA^2 + IA = 500
Solving the above quadratic equation, we get:
IA = 62.5 A or IA = - 63.5 A
Since IA cannot be negative, we discard the negative value and obtain:
IA = 62.5 A
Substituting IA = 62.5 A in equation (2), we get:
IB = 250 - IA
IB = 250 - 62.5
IB = 187.5 A
Therefore, the currents delivered by generator A and B are IA = 62.5 A and IB = 187.5 A respectively.
Hence, option (C) is the correct answer.
Generator A rating, P1 = 50 kW
Generator B rating, P2 = 100 kW
Generator A voltage, V = 500 V
Generator B voltage, V = 500 V
Total current, I = 250 A
Regulation of generator A = 6%
Regulation of generator B = 4%
To find:
Current delivered by generator A, IA and current delivered by generator B, IB.
Solution:
1. Power output of generator A and B
The power output of generator A and B can be calculated as:
P1 = V × IA
P2 = V × IB
Substituting the given values, we get:
P1 = 500 × IA
P2 = 500 × IB
2. Currents delivered by the generators
The total current is given as 250 A, which can be written as:
I = IA + IB
Substituting the given values, we get:
250 = IA + IB
3. Calculation of IA and IB
To calculate IA and IB, we need to solve the above two equations simultaneously.
Dividing equation (1) by equation (2), we get:
P1/P2 = IA/IB
Substituting the given values and simplifying, we get:
IA/IB = 1/2
Multiplying equation (2) by IA/IB, we get:
IA^2/2 + IA/2 = 250
Substituting IA/IB = 1/2, we get:
IA^2/2 + 1/2 × IA = 250
Multiplying both sides by 2, we get:
IA^2 + IA = 500
Solving the above quadratic equation, we get:
IA = 62.5 A or IA = - 63.5 A
Since IA cannot be negative, we discard the negative value and obtain:
IA = 62.5 A
Substituting IA = 62.5 A in equation (2), we get:
IB = 250 - IA
IB = 250 - 62.5
IB = 187.5 A
Therefore, the currents delivered by generator A and B are IA = 62.5 A and IB = 187.5 A respectively.
Hence, option (C) is the correct answer.
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Two shunt generators A and B rated as 50 kW, 500 V and 100 kW, 500 V respectively operating in parallel deliver a total current of 250 A. The regulation of generators A and B are 6 % and 4 % respectively. The currents IAand IBdelivered by the generators area)IA= 125 A, IB= 125 Ab)IA= 120 A, IB= 130 Ac)IA= 62.5 A, IB= 187.5 Ad)IA= 97.5 A, IB= 152.5 ACorrect answer is option 'C'. Can you explain this answer?
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Two shunt generators A and B rated as 50 kW, 500 V and 100 kW, 500 V respectively operating in parallel deliver a total current of 250 A. The regulation of generators A and B are 6 % and 4 % respectively. The currents IAand IBdelivered by the generators area)IA= 125 A, IB= 125 Ab)IA= 120 A, IB= 130 Ac)IA= 62.5 A, IB= 187.5 Ad)IA= 97.5 A, IB= 152.5 ACorrect answer is option 'C'. 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 Two shunt generators A and B rated as 50 kW, 500 V and 100 kW, 500 V respectively operating in parallel deliver a total current of 250 A. The regulation of generators A and B are 6 % and 4 % respectively. The currents IAand IBdelivered by the generators area)IA= 125 A, IB= 125 Ab)IA= 120 A, IB= 130 Ac)IA= 62.5 A, IB= 187.5 Ad)IA= 97.5 A, IB= 152.5 ACorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two shunt generators A and B rated as 50 kW, 500 V and 100 kW, 500 V respectively operating in parallel deliver a total current of 250 A. The regulation of generators A and B are 6 % and 4 % respectively. The currents IAand IBdelivered by the generators area)IA= 125 A, IB= 125 Ab)IA= 120 A, IB= 130 Ac)IA= 62.5 A, IB= 187.5 Ad)IA= 97.5 A, IB= 152.5 ACorrect answer is option 'C'. Can you explain this answer?.
Two shunt generators A and B rated as 50 kW, 500 V and 100 kW, 500 V respectively operating in parallel deliver a total current of 250 A. The regulation of generators A and B are 6 % and 4 % respectively. The currents IAand IBdelivered by the generators area)IA= 125 A, IB= 125 Ab)IA= 120 A, IB= 130 Ac)IA= 62.5 A, IB= 187.5 Ad)IA= 97.5 A, IB= 152.5 ACorrect answer is option 'C'. 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 Two shunt generators A and B rated as 50 kW, 500 V and 100 kW, 500 V respectively operating in parallel deliver a total current of 250 A. The regulation of generators A and B are 6 % and 4 % respectively. The currents IAand IBdelivered by the generators area)IA= 125 A, IB= 125 Ab)IA= 120 A, IB= 130 Ac)IA= 62.5 A, IB= 187.5 Ad)IA= 97.5 A, IB= 152.5 ACorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two shunt generators A and B rated as 50 kW, 500 V and 100 kW, 500 V respectively operating in parallel deliver a total current of 250 A. The regulation of generators A and B are 6 % and 4 % respectively. The currents IAand IBdelivered by the generators area)IA= 125 A, IB= 125 Ab)IA= 120 A, IB= 130 Ac)IA= 62.5 A, IB= 187.5 Ad)IA= 97.5 A, IB= 152.5 ACorrect answer is option 'C'. Can you explain this answer?.
Solutions for Two shunt generators A and B rated as 50 kW, 500 V and 100 kW, 500 V respectively operating in parallel deliver a total current of 250 A. The regulation of generators A and B are 6 % and 4 % respectively. The currents IAand IBdelivered by the generators area)IA= 125 A, IB= 125 Ab)IA= 120 A, IB= 130 Ac)IA= 62.5 A, IB= 187.5 Ad)IA= 97.5 A, IB= 152.5 ACorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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