SSC Exam > SSC Questions > A 100 kVA, 2200/440 V transformer has primary...
Start Learning for Free
A 100 kVA, 2200/440 V transformer has primary values R1=0.3 Ω, X1=1,1 Ω and secondary
R2=0.01Ω and X2=0.035 Ω.values and What are the total copper losses?
R2=0.01Ω and X2=0.035 Ω.values and What are the total copper losses?
- a)1240 W
- b)976 W
- c)1136 W
- d)2124 W
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
A 100 kVA, 2200/440 V transformer has primary values R1=0.3 Ω, X...
Full-load primary current,
Most Upvoted Answer
A 100 kVA, 2200/440 V transformer has primary values R1=0.3 Ω, X...
Ohms and X1=0.6 ohms, and secondary values R2=0.015 ohms and X2=0.03 ohms. Calculate:
a) The equivalent circuit referred to the primary side.
b) The voltage regulation of the transformer when supplying a full load of 100 kVA at 0.8 power factor lagging.
c) The efficiency of the transformer at full load and 0.8 power factor lagging.
Solution:
a) The equivalent circuit referred to the primary side can be obtained by multiplying the impedance values by the turns ratio squared, and keeping the same polarity:
R'_eq = (N2/N1)^2 R2 = (440/2200)^2 0.015 = 0.0006 ohms
X'_eq = (N2/N1)^2 X2 = (440/2200)^2 0.03 = 0.0012 ohms
The primary leakage reactance can be obtained by subtracting the referred secondary reactance from the original primary reactance:
X_1p = X1 - X'_eq = 0.6 - 0.0012 = 0.5988 ohms
The primary resistance is the same as the original primary resistance:
R_1p = R1 = 0.3 ohms
Therefore, the equivalent circuit referred to the primary side is:
b) The voltage regulation can be calculated as follows:
%VR = (VNL - VFL) / VFL x 100%
where VNL is the no-load voltage and VFL is the full-load voltage.
The no-load voltage can be calculated from the turns ratio:
VNL = V2 / (N2/N1) = 440 / (2200/440) = 88 V
The full-load voltage can be obtained from the power equation:
S = VFL x IFL
where S is the apparent power (100 kVA) and IFL is the full-load current. The full-load current can be calculated from the power factor:
cos(φ) = 0.8 lagging
sin(φ) = 0.6
PF = cos(φ) / (cos(φ)^2 + sin(φ)^2)^0.5 = 0.8 / 1 = 0.8
IFL = S / (VFL x PF) = 100,000 / (440 x 0.8) = 284.09 A
Therefore, VFL = S / IFL = 100,000 / 284.09 = 351.4 V
%VR = (88 - 351.4) / 351.4 x 100% = -74.98%
The negative sign indicates that the voltage regulation is a voltage drop, as expected for a transformer.
c) The efficiency can be calculated as follows:
η = output power / input power
where output power is the real power delivered to the load (100 kW) and input power is the sum of the real and reactive powers absorbed by the transformer. At full load and 0.8 power factor lagging, the reactive power absorbed by the transformer is:
Q = S x sin(φ) = 100,000 x 0.6 = 60 kVAR
The input power is therefore:
P_in = P_out + P_loss + Q^2 /
a) The equivalent circuit referred to the primary side.
b) The voltage regulation of the transformer when supplying a full load of 100 kVA at 0.8 power factor lagging.
c) The efficiency of the transformer at full load and 0.8 power factor lagging.
Solution:
a) The equivalent circuit referred to the primary side can be obtained by multiplying the impedance values by the turns ratio squared, and keeping the same polarity:
R'_eq = (N2/N1)^2 R2 = (440/2200)^2 0.015 = 0.0006 ohms
X'_eq = (N2/N1)^2 X2 = (440/2200)^2 0.03 = 0.0012 ohms
The primary leakage reactance can be obtained by subtracting the referred secondary reactance from the original primary reactance:
X_1p = X1 - X'_eq = 0.6 - 0.0012 = 0.5988 ohms
The primary resistance is the same as the original primary resistance:
R_1p = R1 = 0.3 ohms
Therefore, the equivalent circuit referred to the primary side is:
b) The voltage regulation can be calculated as follows:
%VR = (VNL - VFL) / VFL x 100%
where VNL is the no-load voltage and VFL is the full-load voltage.
The no-load voltage can be calculated from the turns ratio:
VNL = V2 / (N2/N1) = 440 / (2200/440) = 88 V
The full-load voltage can be obtained from the power equation:
S = VFL x IFL
where S is the apparent power (100 kVA) and IFL is the full-load current. The full-load current can be calculated from the power factor:
cos(φ) = 0.8 lagging
sin(φ) = 0.6
PF = cos(φ) / (cos(φ)^2 + sin(φ)^2)^0.5 = 0.8 / 1 = 0.8
IFL = S / (VFL x PF) = 100,000 / (440 x 0.8) = 284.09 A
Therefore, VFL = S / IFL = 100,000 / 284.09 = 351.4 V
%VR = (88 - 351.4) / 351.4 x 100% = -74.98%
The negative sign indicates that the voltage regulation is a voltage drop, as expected for a transformer.
c) The efficiency can be calculated as follows:
η = output power / input power
where output power is the real power delivered to the load (100 kW) and input power is the sum of the real and reactive powers absorbed by the transformer. At full load and 0.8 power factor lagging, the reactive power absorbed by the transformer is:
Q = S x sin(φ) = 100,000 x 0.6 = 60 kVAR
The input power is therefore:
P_in = P_out + P_loss + Q^2 /
|
Explore Courses for SSC exam
|
|
Similar SSC Doubts
A 100 kVA, 2200/440 V transformer has primary values R1=0.3 Ω, X1=1,1 Ω and secondary R2=0.01Ω and X2=0.035 Ω.values and What are the total copper losses?a)1240 Wb)976 Wc)1136 Wd)2124 WCorrect answer is option 'C'. Can you explain this answer?
Question Description
A 100 kVA, 2200/440 V transformer has primary values R1=0.3 Ω, X1=1,1 Ω and secondary R2=0.01Ω and X2=0.035 Ω.values and What are the total copper losses?a)1240 Wb)976 Wc)1136 Wd)2124 WCorrect answer is option 'C'. 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 100 kVA, 2200/440 V transformer has primary values R1=0.3 Ω, X1=1,1 Ω and secondary R2=0.01Ω and X2=0.035 Ω.values and What are the total copper losses?a)1240 Wb)976 Wc)1136 Wd)2124 WCorrect answer is option 'C'. 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 100 kVA, 2200/440 V transformer has primary values R1=0.3 Ω, X1=1,1 Ω and secondary R2=0.01Ω and X2=0.035 Ω.values and What are the total copper losses?a)1240 Wb)976 Wc)1136 Wd)2124 WCorrect answer is option 'C'. Can you explain this answer?.
A 100 kVA, 2200/440 V transformer has primary values R1=0.3 Ω, X1=1,1 Ω and secondary R2=0.01Ω and X2=0.035 Ω.values and What are the total copper losses?a)1240 Wb)976 Wc)1136 Wd)2124 WCorrect answer is option 'C'. 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 100 kVA, 2200/440 V transformer has primary values R1=0.3 Ω, X1=1,1 Ω and secondary R2=0.01Ω and X2=0.035 Ω.values and What are the total copper losses?a)1240 Wb)976 Wc)1136 Wd)2124 WCorrect answer is option 'C'. 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 100 kVA, 2200/440 V transformer has primary values R1=0.3 Ω, X1=1,1 Ω and secondary R2=0.01Ω and X2=0.035 Ω.values and What are the total copper losses?a)1240 Wb)976 Wc)1136 Wd)2124 WCorrect answer is option 'C'. Can you explain this answer?.
Solutions for A 100 kVA, 2200/440 V transformer has primary values R1=0.3 Ω, X1=1,1 Ω and secondary R2=0.01Ω and X2=0.035 Ω.values and What are the total copper losses?a)1240 Wb)976 Wc)1136 Wd)2124 WCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for SSC.
Download more important topics, notes, lectures and mock test series for SSC Exam by signing up for free.
Here you can find the meaning of A 100 kVA, 2200/440 V transformer has primary values R1=0.3 Ω, X1=1,1 Ω and secondary R2=0.01Ω and X2=0.035 Ω.values and What are the total copper losses?a)1240 Wb)976 Wc)1136 Wd)2124 WCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A 100 kVA, 2200/440 V transformer has primary values R1=0.3 Ω, X1=1,1 Ω and secondary R2=0.01Ω and X2=0.035 Ω.values and What are the total copper losses?a)1240 Wb)976 Wc)1136 Wd)2124 WCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A 100 kVA, 2200/440 V transformer has primary values R1=0.3 Ω, X1=1,1 Ω and secondary R2=0.01Ω and X2=0.035 Ω.values and What are the total copper losses?a)1240 Wb)976 Wc)1136 Wd)2124 WCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A 100 kVA, 2200/440 V transformer has primary values R1=0.3 Ω, X1=1,1 Ω and secondary R2=0.01Ω and X2=0.035 Ω.values and What are the total copper losses?a)1240 Wb)976 Wc)1136 Wd)2124 WCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A 100 kVA, 2200/440 V transformer has primary values R1=0.3 Ω, X1=1,1 Ω and secondary R2=0.01Ω and X2=0.035 Ω.values and What are the total copper losses?a)1240 Wb)976 Wc)1136 Wd)2124 WCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice SSC tests.
|
|
Explore Courses for SSC exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup