---
title: "300 kVA transformer efficiency calculation"
subject: "Basic Electrical and Electronics Engineering"
chapter: 2
key: "semester_1/basic_electrical_and_electronics_engineering:transformer-300kva-efficiency"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/transformer-300kva-efficiency
site: BCT Engineering
---

# 300 kVA transformer efficiency calculation

Basic Electrical and Electronics Engineering · Chapter 2: Electrical Machines

- Asked in: 2081 Kartik (new course), 6 marks
- Video: [Transformer  | Calculation of  Efficiency at full load | UPF | 0.8 Lagging pf](https://www.youtube.com/watch?v=FXJm5sg9tIM) — FOR STUDENTS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/transformer-300kva-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_and_electronics_engineering/chapter-2

## Question

A 300 kVA single phase transformer has iron losses of 1.8 kW. The full load copper losses is 2500 watts. Calculate, (i) Efficiency at full load, 0.8 lagging pf (ii) kVA supplied at maximum Efficiency (iii) Maximum Efficiency at 0.8 lagging pf.

## Short answer

Full-load efficiency at 0.8 lagging pf is98.24%. Maximum efficiency occurs at254.56 kVA, yielding98.26% at0.8 pf.

## Answer

**Given:** $S=300$ kVA, $P_i=1.8$ kW, full-load $P_{cu}=2.5$ kW.

**(i) Full-load efficiency at 0.8 lagging pf.**

$$
\text{Output}=300\times0.8=240\ \text{kW}
$$

$$
\text{Total loss}=1.8+2.5=4.3\ \text{kW}
$$

$$
\eta=\frac{240}{240+4.3}\times100=98.24\%
$$

**(ii) kVA at maximum efficiency.** Efficiency is maximum when copper loss = iron loss:

$$
x=\sqrt{\frac{P_i}{P_{cu}}}=\sqrt{\frac{1.8}{2.5}}=0.8485
$$

$$
\text{kVA}=300\times0.8485=254.56\ \text{kVA}
$$

**(iii) Maximum efficiency at 0.8 pf.** Output $=254.56\times0.8=203.65$ kW. At this load, the copper loss equals the iron loss $=1.8$ kW, so the total loss is $2\times1.8=3.6$ kW:

$$
\eta_{max}=\frac{203.65}{203.65+3.6}\times100=98.26\%
$$

**Answer: (i) 98.24%, (ii) 254.56 kVA, (iii) 98.26%.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is full-load efficiency at 0.8 pf obtained?
   A. Output 300 kW; losses 1.8 kW only
   B. Output 240 kW; losses 4.3 kW; eta = 240/(240 + 4.3) = 98.24%
   C. Output 240 kW; losses 2.5 kW; eta = 98.97%
   D. Output 300 kVA; losses 4.3 kW; eta = 98.6%
   Answer: B. Output = 300 x 0.8 = 240 kW and total loss = iron + full-load copper = 4.3 kW. (Key point: Full-load efficiency)

2. At what kVA is efficiency maximum?
   A. 300 x 1.8/2.5 = 216 kVA
   B. 300 x 2.5/1.8 = 416.7 kVA
   C. 300 x sqrt(1.8/2.5) = 254.56 kVA
   D. 300 kVA
   Answer: C. Maximum efficiency occurs at x = sqrt(Pi/Pcu) = 0.8485 of full load. (Key point: Load for max efficiency)

3. What is the total loss at maximum efficiency?
   A. 4.3 kW
   B. 1.8 kW
   C. 2.5 kW
   D. 3.6 kW (2 x iron loss)
   Answer: D. Copper loss equals iron loss (1.8 kW) there, so the total is 3.6 kW. (Key point: Losses at maximum efficiency)

4. What are the output and maximum efficiency at 0.8 pf?
   A. 203.65 kW and 98.26%
   B. 240 kW and 98.24%
   C. 254.56 kW and 98.6%
   D. 203.65 kW and 97.2%
   Answer: A. Output = 254.56 x 0.8 = 203.65 kW; eta = 203.65/(203.65 + 3.6) = 98.26%. (Key point: Maximum efficiency)

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