---
title: "Full-load efficiency of 500 kVA transformer"
subject: "Electrical Machines I"
chapter: 2
key: "semester_4/electrical_machines_i:500kva-90pc-half-load-full-load"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/500kva-90pc-half-load-full-load
site: BCT Engineering
---

# Full-load efficiency of 500 kVA transformer

Electrical Machines I · Chapter 2: Transformer

- Asked in: 2073 Magh, 8 marks
- Video: [Transformer efficiency: at Full and Half Load problem.](https://www.youtube.com/watch?v=ek2AQfUmAKo) — Dr Bharat Singh Choudhary
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/500kva-90pc-half-load-full-load
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2

## Question

A single phase 500 kVA transformer working at unity power factor has an efficiency of 90% at half load and iron loss of 2000 watt. Determine efficiency at full load.

## Short answer

Half-load output 250 kW, input 277.78 kW, so copper loss 25.78 kW. Full-load copper loss = 4×25.78 = 103.11 kW. Iron loss 2 kW. Full-load output 500 kW. Efficiency = 500/(500+2+103.11) = 82.63%.

## Answer

Efficiency $\eta = \frac{\text{output}}{\text{output} + P_i + P_{cu}}$; copper loss varies with the square of load.

#### Step 1: Losses at half load (upf)
$$
\begin{aligned}
\text{Output}_{1/2} &= 0.5 \times 500 \times 1 = 250\ \text{kW} \\
\text{Input}_{1/2} &= \frac{250}{0.90} = 277.78\ \text{kW} \\
\text{Total loss} &= 277.78 - 250 = 27.78\ \text{kW}
\end{aligned}
$$

#### Step 2: Copper loss at half and full load
$$
\begin{aligned}
P_{cu,1/2} &= 27778 - 2000 = 25778\ \text{W} \\
P_{cu,FL} &= \frac{P_{cu,1/2}}{(0.5)^2} = 4 \times 25778 = 103111\ \text{W} = 103.11\ \text{kW}
\end{aligned}
$$

#### Step 3: Efficiency at full load (upf)
$$
\begin{aligned}
\text{Output}_{FL} &= 500\ \text{kW} \\
\text{Losses} &= 2 + 103.11 = 105.11\ \text{kW} \\
\eta_{FL} &= \frac{500}{500 + 105.11}\times 100 = 82.63\%
\end{aligned}
$$

| Load | Output | Iron loss | Cu loss | Efficiency |
|---|---|---|---|---|
| Half | 250 kW | 2 kW | 25.78 kW | 90.00% |
| Full | 500 kW | 2 kW | 103.11 kW | 82.63% |

**Answer:** Full-load efficiency at unity pf = **82.63%**.

(The efficiency falls at full load because copper loss, which grows as the square of load, is much larger than iron loss here. The data are as given in the question; a real 500 kVA transformer would have far higher efficiency.)

## Check yourself

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

1. At half load upf, output 250 kW and efficiency 90%. Total loss is:
   A. 25 kW
   B. 27.78 kW
   C. 250 kW
   D. 22.2 kW
   Answer: B. Input = 250/0.9 = 277.78 kW, so loss = 27.78 kW. (Key point: Half-load losses)

2. Copper loss at half load is found by:
   A. Total loss minus iron loss (27.78 - 2 kW)
   B. Total loss plus iron loss
   C. Total loss divided by 4
   D. Equal to iron loss
   Answer: A. Iron loss is constant 2 kW, so Pcu,1/2 = 25.78 kW. (Key point: Copper loss at half)

3. Full-load copper loss is obtained from the half-load value by:
   A. Multiplying by 2
   B. Multiplying by 4
   C. Dividing by 4
   D. Multiplying by 8
   Answer: B. Copper loss varies as load squared, so FL loss = 25.78/(0.5)^2 = 4 x 25.78 = 103.11 kW. (Key point: Scaling to full load)

4. Full-load efficiency at upf is about:
   A. 90%
   B. 98.0%
   C. 82.63%
   D. 95.2%
   Answer: C. 500/(500 + 2 + 103.11) = 82.63%, lower than at half load because copper loss grows as the square of load. (Key point: Full-load efficiency)

## Related in Electrical Machines I

- [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
- [Iron loss and maximum efficiency of 20 kVA 220/2200 V transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A20kva-iron-loss-max-efficiency-current) — Electrical Machines I, chapter 2, asked 2×
- [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
- [Iron loss and maximum efficiency of 20 kVA 250/2500 V transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A20kva-250-2500-iron-loss-max) — Electrical Machines I, chapter 2, asked 2×
- [Equivalent circuit and half-load efficiency of 50 kVA transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A50kva-2500-250-test-secondary-half) — Electrical Machines I, chapter 2
- [Half-load efficiency of 200 kVA transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aefficiency-half-load-200kva-98) — Electrical Machines I, chapter 2

## Related in other subjects

- [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
- [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
- [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
- [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