Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Questions  >  In a 25 KVA 3300/230V single phase transforme... Start Learning for Free
In a 25 KVA 3300/230V single phase transformer the iron and full load copper losses are 350 and 400 W respectively. Calculate the efficiency of the transformer at 1/2 full load, 0.8 power factor?
Most Upvoted Answer
In a 25 KVA 3300/230V single phase transformer the iron and full load ...
Problem:

Calculate the efficiency of a 25 KVA 3300/230V single-phase transformer at 1/2 full load, 0.8 power factor when the iron and full load copper losses are 350 and 400 W, respectively.


Solution:

Given:


  • Transformer rating (S) = 25 KVA

  • Primary voltage (V1) = 3300 V

  • Secondary voltage (V2) = 230 V

  • Iron losses (Pcore) = 350 W

  • Copper losses at full load (Pcu) = 400 W

  • Load at which efficiency is to be calculated (kVA) = 1/2 of full load = 12.5 KVA

  • Power factor (pf) = 0.8



Step 1: Calculate the primary and secondary currents at 1/2 full load

The transformer rating is given in KVA, and the load is given in kVA. Therefore, the load current can be calculated as:

Load current (I2) = Load (kVA) / Secondary voltage (V2) = 12.5 / 230 = 0.05435 A

The primary current can be calculated as:

Primary current (I1) = S / V1 = 25,000 / 3300 = 7.5758 A

The primary current at 1/2 full load can be calculated as:

I1' = I1 * (I2 / pf) = 7.5758 * (0.05435 / 0.8) = 0.5147 A


Step 2: Calculate the copper losses at 1/2 full load

The copper losses in a transformer are proportional to the square of the current. Therefore, the copper losses at 1/2 full load can be calculated as:

Pcu' = (I1')^2 * R = (0.5147)^2 * R

where R is the resistance of the transformer winding. As the resistance is not given, we assume that it is proportional to the full load copper losses. Therefore, we can write:

R' / R = (I1' / I1)^2 = (0.5147 / 7.5758)^2

R' = R * (0.0036)

Substituting this value of R' in the expression for Pcu', we get:

Pcu' = (0.5147)^2 * R' = (0.5147)^2 * R * (0.0036) = 0.0107 * Pcu = 4.28 W


Step 3: Calculate the total losses at 1/2 full load

The total losses in a transformer are the sum of the iron losses and the copper losses
Community Answer
In a 25 KVA 3300/230V single phase transformer the iron and full load ...
Explore Courses for Electrical Engineering (EE) exam

Top Courses for Electrical Engineering (EE)

In a 25 KVA 3300/230V single phase transformer the iron and full load copper losses are 350 and 400 W respectively. Calculate the efficiency of the transformer at 1/2 full load, 0.8 power factor?
Question Description
In a 25 KVA 3300/230V single phase transformer the iron and full load copper losses are 350 and 400 W respectively. Calculate the efficiency of the transformer at 1/2 full load, 0.8 power factor? 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 In a 25 KVA 3300/230V single phase transformer the iron and full load copper losses are 350 and 400 W respectively. Calculate the efficiency of the transformer at 1/2 full load, 0.8 power factor? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a 25 KVA 3300/230V single phase transformer the iron and full load copper losses are 350 and 400 W respectively. Calculate the efficiency of the transformer at 1/2 full load, 0.8 power factor?.
Solutions for In a 25 KVA 3300/230V single phase transformer the iron and full load copper losses are 350 and 400 W respectively. Calculate the efficiency of the transformer at 1/2 full load, 0.8 power factor? 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 In a 25 KVA 3300/230V single phase transformer the iron and full load copper losses are 350 and 400 W respectively. Calculate the efficiency of the transformer at 1/2 full load, 0.8 power factor? defined & explained in the simplest way possible. Besides giving the explanation of In a 25 KVA 3300/230V single phase transformer the iron and full load copper losses are 350 and 400 W respectively. Calculate the efficiency of the transformer at 1/2 full load, 0.8 power factor?, a detailed solution for In a 25 KVA 3300/230V single phase transformer the iron and full load copper losses are 350 and 400 W respectively. Calculate the efficiency of the transformer at 1/2 full load, 0.8 power factor? has been provided alongside types of In a 25 KVA 3300/230V single phase transformer the iron and full load copper losses are 350 and 400 W respectively. Calculate the efficiency of the transformer at 1/2 full load, 0.8 power factor? theory, EduRev gives you an ample number of questions to practice In a 25 KVA 3300/230V single phase transformer the iron and full load copper losses are 350 and 400 W respectively. Calculate the efficiency of the transformer at 1/2 full load, 0.8 power factor? 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.
10M+ students study on EduRev