---
title: "Full-load and half-load efficiency of 100 kVA transformer"
subject: "Electrical Machines I"
chapter: 2
key: "semester_4/electrical_machines_i:100kva-1200w-960w-efficiency"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/100kva-1200w-960w-efficiency
site: BCT Engineering
---

# Full-load and half-load efficiency of 100 kVA transformer

Electrical Machines I · Chapter 2: Transformer

- Asked in: 2073 Bhadra, 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/100kva-1200w-960w-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2

## Question

A transformer is rated at 100 kVA. At full load its copper loss is 1200 W and its iron loss is 960 W. Calculate: i) Efficiency at full load, unity power factor ii) Efficiency at half load, 0.8 power factor

## Short answer

Use efficiency = output/(output + iron loss + x^2 copper loss). At full load unity pf output is 100 kW, losses 2160 W, efficiency 97.89%. At half load 0.8 pf output is 40 kW, losses 1260 W, efficiency 96.95%.

## Answer

$\eta = \frac{xS\cos\phi}{xS\cos\phi + P_i + x^2P_{cu,FL}}$, with $S = 100$ kVA, $P_i = 960$ W (constant), $P_{cu,FL} = 1200$ W.

#### (i) Full load, unity pf ($x = 1$)
$$
\begin{aligned}
\text{Output} &= 1 \times 100 \times 1 = 100\ \text{kW} = 100000\ \text{W} \\
\text{Losses} &= 960 + 1^2 \times 1200 = 2160\ \text{W} \\
\eta &= \frac{100000}{100000 + 2160}\times 100 = 97.89\%
\end{aligned}
$$

#### (ii) Half load, 0.8 pf ($x = 0.5$)
$$
\begin{aligned}
\text{Output} &= 0.5 \times 100 \times 0.8 = 40\ \text{kW} = 40000\ \text{W} \\
P_{cu} &= 0.5^2 \times 1200 = 300\ \text{W} \\
\text{Losses} &= 960 + 300 = 1260\ \text{W} \\
\eta &= \frac{40000}{40000 + 1260}\times 100 = 96.95\%
\end{aligned}
$$

| Case | Output | Iron loss | Cu loss | Efficiency |
|---|---|---|---|---|
| FL, upf | 100 kW | 960 W | 1200 W | 97.89% |
| ½ FL, 0.8 pf | 40 kW | 960 W | 300 W | 96.95% |

**Answer:** (i) **97.89%**, (ii) **96.95%**.

Note: maximum efficiency occurs at $x = \sqrt{960/1200} = 0.894$, i.e. at about 89.4% of full load, where copper loss equals 960 W.

## Check yourself

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

1. For fraction x of load, efficiency is:
   A. x S cos(phi) / (x S cos(phi) + Pi + Pcu,FL)
   B. x S cos(phi) / (x S cos(phi) + x Pi + x Pcu,FL)
   C. x S cos(phi) / (x S cos(phi) + Pi + x^2 Pcu,FL)
   D. S cos(phi) / (S cos(phi) + x^2 Pi + Pcu,FL)
   Answer: C. Iron loss is constant, copper loss scales with x^2. (Key point: Efficiency formula)

2. At full load upf, the efficiency is:
   A. 96.95%
   B. 97.89%
   C. 98.9%
   D. 95.4%
   Answer: B. 100000/(100000 + 960 + 1200) = 97.89%. (Key point: Full load result)

3. At half load, the copper loss is:
   A. 600 W
   B. 1200 W
   C. 300 W
   D. 150 W
   Answer: C. 0.5^2 x 1200 = 300 W. (Key point: Half-load copper loss)

4. At half load and 0.8 pf the output used is:
   A. 50 kW
   B. 80 kW
   C. 100 kW
   D. 40 kW
   Answer: D. Output = 0.5 x 100 x 0.8 = 40 kW, giving efficiency 40000/41260 = 96.95%. (Key point: Half-load 0.8 pf)

## Related in Electrical Machines I

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## Related in other subjects

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- [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
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- [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