Electrical Engineering (EE) Exam > Electrical Engineering (EE) Questions > The efficiency of a 20 KVA, 2000/200 V, singl... Start Learning for Free
The efficiency of a 20 KVA, 2000/200 V, single phase transformer at unity pf is 98%. The given total losses at full load is 200 W. The pu resistance is?
- a)0.01
- b)0.1
- c)1.0
- d)0.0196
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The efficiency of a 20 KVA, 2000/200 V, single phase transformer at un...
pu resistance = Ohmic losses/KVA = 200/20000 = 0.01 pu.
Most Upvoted Answer
The efficiency of a 20 KVA, 2000/200 V, single phase transformer at un...
To find the per unit (pu) resistance of the transformer, we need to use the given information about the efficiency, total losses, and the transformer's power rating.
First, let's calculate the input power (Pin) and the output power (Pout) of the transformer.
Given:
Power rating (S) = 20 KVA
Input voltage (Vi) = 2000 V
Output voltage (Vo) = 200 V
Total losses (Pl) = 200 W
Efficiency (η) = 98% = 0.98
We know that,
Pout = Pin - Pl
Since the transformer is at unity power factor, the power is equal to the apparent power. Therefore,
Pout = S × η
Now, substitute the given values into the equations:
Pout = 20 KVA × 0.98
Pout = 19.6 KVA
Next, calculate the input power:
Pin = Pout + Pl
Pin = 19.6 KVA + 200 W
Pin = 19.6 KVA + 0.2 KVA
Pin = 19.8 KVA
Now, calculate the input current (Iin) using the formula:
Pin = Vi × Iin
Iin = Pin / Vi
Iin = 19.8 KVA / 2000 V
Iin = 9.9 A
The primary winding resistance (Rp) can be calculated using the formula:
Rp = (Pin / Iin^2) × η
Substituting the values:
Rp = (19.8 KVA / (9.9 A)^2) × 0.98
Rp = (19.8 KVA / 97.92 A^2) × 0.98
Rp = 0.202 KVA / 97.92 A^2
Rp = 0.00206 Ω
Finally, convert the resistance to per unit (pu) using the equation:
pu resistance = Rp / (Vi / Iin)^2
pu resistance = 0.00206 Ω / (2000 V / 9.9 A)^2
pu resistance = 0.00206 Ω / (202.02 Ω)^2
pu resistance = 0.00206 Ω / 40,804.08 Ω
pu resistance = 0.0000000505
Rounded to two decimal places, the per unit (pu) resistance is approximately 0.01 (Option A).
First, let's calculate the input power (Pin) and the output power (Pout) of the transformer.
Given:
Power rating (S) = 20 KVA
Input voltage (Vi) = 2000 V
Output voltage (Vo) = 200 V
Total losses (Pl) = 200 W
Efficiency (η) = 98% = 0.98
We know that,
Pout = Pin - Pl
Since the transformer is at unity power factor, the power is equal to the apparent power. Therefore,
Pout = S × η
Now, substitute the given values into the equations:
Pout = 20 KVA × 0.98
Pout = 19.6 KVA
Next, calculate the input power:
Pin = Pout + Pl
Pin = 19.6 KVA + 200 W
Pin = 19.6 KVA + 0.2 KVA
Pin = 19.8 KVA
Now, calculate the input current (Iin) using the formula:
Pin = Vi × Iin
Iin = Pin / Vi
Iin = 19.8 KVA / 2000 V
Iin = 9.9 A
The primary winding resistance (Rp) can be calculated using the formula:
Rp = (Pin / Iin^2) × η
Substituting the values:
Rp = (19.8 KVA / (9.9 A)^2) × 0.98
Rp = (19.8 KVA / 97.92 A^2) × 0.98
Rp = 0.202 KVA / 97.92 A^2
Rp = 0.00206 Ω
Finally, convert the resistance to per unit (pu) using the equation:
pu resistance = Rp / (Vi / Iin)^2
pu resistance = 0.00206 Ω / (2000 V / 9.9 A)^2
pu resistance = 0.00206 Ω / (202.02 Ω)^2
pu resistance = 0.00206 Ω / 40,804.08 Ω
pu resistance = 0.0000000505
Rounded to two decimal places, the per unit (pu) resistance is approximately 0.01 (Option A).
 | Explore Courses for Electrical Engineering (EE) exam |  |
Top Courses for Electrical Engineering (EE)View all
Top Courses for Electrical Engineering (EE)
Question Description
The efficiency of a 20 KVA, 2000/200 V, single phase transformer at unity pf is 98%. The given total losses at full load is 200 W. The pu resistance is?a)0.01b)0.1c)1.0d)0.0196Correct answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2026 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about The efficiency of a 20 KVA, 2000/200 V, single phase transformer at unity pf is 98%. The given total losses at full load is 200 W. The pu resistance is?a)0.01b)0.1c)1.0d)0.0196Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The efficiency of a 20 KVA, 2000/200 V, single phase transformer at unity pf is 98%. The given total losses at full load is 200 W. The pu resistance is?a)0.01b)0.1c)1.0d)0.0196Correct answer is option 'A'. Can you explain this answer?.
The efficiency of a 20 KVA, 2000/200 V, single phase transformer at unity pf is 98%. The given total losses at full load is 200 W. The pu resistance is?a)0.01b)0.1c)1.0d)0.0196Correct answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2026 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about The efficiency of a 20 KVA, 2000/200 V, single phase transformer at unity pf is 98%. The given total losses at full load is 200 W. The pu resistance is?a)0.01b)0.1c)1.0d)0.0196Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The efficiency of a 20 KVA, 2000/200 V, single phase transformer at unity pf is 98%. The given total losses at full load is 200 W. The pu resistance is?a)0.01b)0.1c)1.0d)0.0196Correct answer is option 'A'. Can you explain this answer?.
Solutions for The efficiency of a 20 KVA, 2000/200 V, single phase transformer at unity pf is 98%. The given total losses at full load is 200 W. The pu resistance is?a)0.01b)0.1c)1.0d)0.0196Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE). Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
Here you can find the meaning of The efficiency of a 20 KVA, 2000/200 V, single phase transformer at unity pf is 98%. The given total losses at full load is 200 W. The pu resistance is?a)0.01b)0.1c)1.0d)0.0196Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of The efficiency of a 20 KVA, 2000/200 V, single phase transformer at unity pf is 98%. The given total losses at full load is 200 W. The pu resistance is?a)0.01b)0.1c)1.0d)0.0196Correct answer is option 'A'. Can you explain this answer?, a detailed solution for The efficiency of a 20 KVA, 2000/200 V, single phase transformer at unity pf is 98%. The given total losses at full load is 200 W. The pu resistance is?a)0.01b)0.1c)1.0d)0.0196Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of The efficiency of a 20 KVA, 2000/200 V, single phase transformer at unity pf is 98%. The given total losses at full load is 200 W. The pu resistance is?a)0.01b)0.1c)1.0d)0.0196Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice The efficiency of a 20 KVA, 2000/200 V, single phase transformer at unity pf is 98%. The given total losses at full load is 200 W. The pu resistance is?a)0.01b)0.1c)1.0d)0.0196Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Electrical Engineering (EE) tests.
| Explore Courses for Electrical Engineering (EE) exam | |
Top Courses for Electrical Engineering (EE)
Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
