---
title: "500 kVA transformer efficiency full/half load"
subject: "Electrical Machine"
chapter: 2
key: "semester_4/electrical_machine:transformer-500kva-efficiency-full-half"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/transformer-500kva-efficiency-full-half
site: BCT Engineering
---

# 500 kVA transformer efficiency full/half load

Electrical Machine · Chapter 2: Transformer

- Asked in: 2072 Asoj
- 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_machine/transformer-500kva-efficiency-full-half
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-2

## Question

A 500 KVA, 50 Hz, 6600 V/400 V, 1-phase transformer have primary and secondary winding resistances are 0.4 Ω and 0.001 Ω respectively. If the iron loss is 3.0 kW, calculate the efficiency at (a) full load (b) half full load.

## Short answer

Full-load copper loss 3858 W, iron loss 3000 W. At unity pf: η full load=500000/(500000+6858)=98.65%; η half load=250000/(250000+3965)=98.44%.

## Answer

The efficiency is found from output ÷ (output + iron loss + copper loss). The power factor is not given, so **unity power factor is assumed** (the result at 0.8 lagging is also shown).

#### Rated currents
- Primary: I₁ = 500 000 / 6600 = 75.76 A
- Secondary: I₂ = 500 000 / 400 = 1250 A

#### Copper loss at full load
- Primary: I₁²R₁ = (75.76)² × 0.4 = 2295.7 W
- Secondary: I₂²R₂ = (1250)² × 0.001 = 1562.5 W
- Total P_cu(fl) = 3858.2 W ≈ 3.858 kW
- Iron loss P_i = 3000 W (constant, same at all loads)

#### (a) Full load (unity pf)
```text
Output   = 500 kVA x 1.0           = 500 000 W
Losses   = 3000 + 3858.2           = 6858.2 W
Input    = 500 000 + 6858.2        = 506 858.2 W
η        = 500 000 / 506 858.2     = 0.9865
```
**Efficiency at full load = 98.65 %**

#### (b) Half load (unity pf)
Copper loss varies with (load)², so P_cu = (0.5)² × 3858.2 = 964.5 W.
```text
Output   = 250 000 W
Losses   = 3000 + 964.5            = 3964.5 W
η        = 250 000 / 253 964.5     = 0.9844
```
**Efficiency at half load = 98.44 %**

| Load | Power factor | Output (kW) | Cu loss (W) | Iron loss (W) | η (%) |
|---|---|---|---|---|---|
| Full | 1.0 | 500 | 3858.2 | 3000 | 98.65 |
| Half | 1.0 | 250 | 964.5 | 3000 | 98.44 |
| Full | 0.8 lag | 400 | 3858.2 | 3000 | 98.31 |
| Half | 0.8 lag | 200 | 964.5 | 3000 | 98.06 |

**Answer: η(full load) ≈ 98.65 %, η(half load) ≈ 98.44 % at unity pf.**

## Check yourself

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

1. Why is the total full-load copper loss 3858 W?
   A. Only primary loss is counted
   B. Both primary I1²R1 (2296 W) and secondary I2²R2 (1562 W) are added
   C. It is the iron loss
   D. It is the average of the two
   Answer: B. Both winding losses are added at rated currents. (Key point: Total copper loss)

2. What is the copper loss at half load?
   A. 1929 W
   B. 3858 W
   C. 964.5 W
   D. 482 W
   Answer: C. 0.5² × 3858.2 = 964.5 W. (Key point: Half-load Cu loss)

3. What is assumed since the power factor is not given?
   A. 0.8 lagging
   B. Unity power factor
   C. Zero power factor
   D. 0.8 leading
   Answer: B. Unity pf is assumed. (Key point: Pf assumption)

4. What is the half-load efficiency at unity pf?
   A. 98.44 %
   B. 98.65 %
   C. 97.0 %
   D. 99.1 %
   Answer: A. 250000/(250000 + 3000 + 964.5) = 98.44 %. (Key point: Half-load efficiency)

## Related in Electrical Machine

- [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
- [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
- [500 kVA 3-phase transformer efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-500kva-33-11kv-efficiency) — 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
- [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
- [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

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