---
title: "20 kVA transformer iron loss, efficiency"
subject: "Electrical Machine"
chapter: 2
key: "semester_4/electrical_machine:transformer-20kva-iron-loss-efficiency"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/transformer-20kva-iron-loss-efficiency
site: BCT Engineering
---

# 20 kVA transformer iron loss, efficiency

Electrical Machine · Chapter 2: Transformer

- Asked in: 2072 Magh
- Video: [Efficiency of Transformers | Example |Transformers | Lec - 28](https://www.youtube.com/watch?v=HQ_-jctQGIU) — Education 4u
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/transformer-20kva-iron-loss-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-2

## Question

A 20 kVA, 250/2500 V, 50 Hz, single phase transformer has Ro = 595.2 Ω, Xo = 187.26 Ω, Ro1 = 0.05 Ω and Xo1 = 0.14 Ω. Calculate (i) iron loss of the transformer (ii) efficiency of the transformer at half full load with 0.8 pf lagging.

## Short answer

Iron loss Pi=250²/595.2=105 W. At half load and 0.8 pf: output 8000 W, copper loss 80 W, efficiency=8000/(8000+105+80)=97.7%.

## Answer

**Given:** 20 kVA, 250/2500 V, 50 Hz. Ro = 595.2 Ω, Xo = 187.26 Ω (the no-load branch, referred to the 250 V primary); Ro1 = 0.05 Ω, Xo1 = 0.14 Ω (series, referred to primary).

#### (i) Iron loss
The iron loss is the power consumed in Ro at the rated primary voltage 250 V:
```text
Pi = V1²/Ro = 250²/595.2 = 105.0 W
```

#### (ii) Efficiency at half load, 0.8 pf lagging
```text
Rated primary current  I1 = 20 000/250 = 80 A
Half-load current      = 40 A
Cu loss = I² Ro1 = 40² × 0.05 = 80 W
Output  = 0.5 × 20 000 × 0.8 = 8000 W
η = 8000/(8000 + 105 + 80) = 8000/8185 = 97.74 %
```
Check: full-load Cu loss = 80² × 0.05 = 320 W, and at half load it is 0.25 × 320 = 80 W.

**Answer: (i) iron loss = 105 W; (ii) η ≈ 97.7 %.**

## Check yourself

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

1. How is the iron loss found from the equivalent circuit?
   A. V1²/Ro = 250²/595.2 = 105 W
   B. V1²/Xo
   C. I²Ro1
   D. Rated kVA × 0.005
   Answer: A. Iron loss is power dissipated in Ro at rated voltage. (Key point: Iron loss)

2. What is the copper loss at half load?
   A. 320 W
   B. 40 W
   C. 160 W
   D. 80 W
   Answer: D. Half-load current 40 A: 40² × 0.05 = 80 W. (Key point: Half-load Cu loss)

3. What is the efficiency at half load, 0.8 pf?
   A. 97.74 %
   B. 95.0 %
   C. 98.9 %
   D. 96.2 %
   Answer: A. 8000/(8000 + 105 + 80) = 97.74 %. (Key point: Efficiency)

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