---
title: "Capacity of transformer; efficiency at 25% overload"
subject: "Electrical Machines"
chapter: 2
key: "semester_4/electrical_machines:tf-150kva-overload-efficiency"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-150kva-overload-efficiency
site: BCT Engineering
---

# Capacity of transformer; efficiency at 25% overload

Electrical Machines · Chapter 2: Transformer

- Asked in: 2079 Jestha, 8 marks
- Video: [#How to calculate efficiency of a transformer](https://www.youtube.com/watch?v=Ys_TVn3IBrc) — Excellent school of Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-150kva-overload-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2

## Question

What is capacity of transformer? A 150 kVA single phase transformer has an iron loss of 700 W and a full load copper loss of 1800 W. Calculate the copper loss, iron loss, output power and efficiency of transformer at 0.8 power factor lagging when secondary is 25% overloaded.

## Short answer

At 25% overload (x=1.25): copper loss = 1.25² × 1800 = 2812.5 W, iron loss = 700 W, output power = 1.25 × 150 × 0.8 = 150 kW, efficiency = 150000/(150000+3512.5) × 100 = 97.71%.

## Answer

**Capacity of a transformer** is its rated apparent power output (in kVA) at rated voltage and frequency. It is given in kVA, not kW, because the copper loss depends on the current and the core loss on the voltage, both independent of the power factor of the load.

**Given:** 150 kVA, $P_i = 700$ W, $P_{cu,fl} = 1800$ W, pf $= 0.8$ lagging, load $= 125\%$ of full load, so $x = 1.25$.

**Copper loss at 25% overload**

$$
P_{cu} = x^2P_{cu,fl} = 1.25^2\times1800 = 2812.5\ \text{W}
$$

**Iron loss** is independent of load: $P_i = 700$ W.

**Output power**

$$
P_{out} = x\times S\times\cos\phi = 1.25\times150\times0.8 = 150\ \text{kW}
$$

**Efficiency**

$$
\begin{aligned}
\text{Total loss} &= 2812.5 + 700 = 3512.5\ \text{W} \\
\eta &= \frac{150\,000}{150\,000 + 3512.5} \times100 = 97.71\%
\end{aligned}
$$

**Answer: copper loss = 2812.5 W; iron loss = 700 W; output = 150 kW; efficiency = 97.71%**

## Check yourself

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

1. Why is transformer capacity quoted in kVA rather than kW?
   A. Because the power factor is always unity
   B. Copper loss depends on current and core loss on voltage, independent of power factor
   C. Because kW depends on frequency
   D. Because the core loss depends on the load pf
   Answer: B. Losses depend on V and I, not on pf, so the rating is apparent power. (Key point: Why kVA rating)

2. What is the copper loss at 25% overload?
   A. 2812.5 W
   B. 2250 W
   C. 1800 W
   D. 3600 W
   Answer: A. Pcu = x^2 x 1800 = 1.5625 x 1800 = 2812.5 W. (Key point: Copper loss at overload)

3. What is the efficiency at 125% load, 0.8 pf?
   A. 98.3%
   B. 96.5%
   C. 97.71%
   D. 98.9%
   Answer: C. Output is 150 kW and losses are 2812.5 + 700 = 3512.5 W, so eta = 150/153.5125 = 97.71%. (Key point: Efficiency at overload)

## Related in Electrical Machines

- [Efficiency of 25 kVA transformer at half load and overload](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A25kva-2200-220-efficiency-half-overload) — Electrical Machines I, chapter 2
- [150 kVA transformer: max efficiency load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-11000-230-max-efficiency-kva) — Electrical Machines, chapter 2
- [Full-load efficiency of 500 kVA transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A500kva-90pc-half-load-full-load) — Electrical Machines I, chapter 2
- [500 kVA transformer efficiency, full and half load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-500kva-6600-400-efficiency) — Electrical Machines, chapter 2
- [Full-load and half-load efficiency of 100 kVA transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A100kva-1200w-960w-efficiency) — Electrical Machines I, chapter 2
- [100 kVA transformer efficiency at full and half load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-100kva-efficiency-full-half-load) — Electrical Machines, chapter 2

## Related in other subjects

- [20 kVA transformer iron loss, efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Atransformer-20kva-iron-loss-efficiency) — Electrical Machine, chapter 2
- [Design: 500 kVA single phase transformer with tubes](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-500kva-single-phase-tubes) — Electric Machine Design, chapter 3, asked 2×
- [500 kVA transformer efficiency full/half load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Atransformer-500kva-efficiency-full-half) — Electrical Machine, chapter 2
- [Design: 150 kVA single phase transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-150kva-single-phase-windings) — Electric Machine Design, chapter 3, asked 4×