---
title: "100 kVA transformer efficiency at full and half load"
subject: "Electrical Machines"
chapter: 2
key: "semester_4/electrical_machines:tf-100kva-efficiency-full-half-load"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-100kva-efficiency-full-half-load
site: BCT Engineering
---

# 100 kVA transformer efficiency at full and half load

Electrical Machines · Chapter 2: Transformer

- Asked in: 2073 Bhadra (old course), 8 marks
- Video: [Efficiency Of Transformer | Transformer | Problem 1 | BEE](https://www.youtube.com/watch?v=r2hY-GtWGwA) — Aryabhat Institute
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-100kva-efficiency-full-half-load
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/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

(i) Full load, unity pf efficiency = 97.89%. (ii) Half load, 0.8 pf efficiency = 96.95%.

## Answer

**Given:** 100 kVA, $P_{cu,fl} = 1200$ W, $P_i = 960$ W.

The efficiency at a load fraction $x$ and power factor $\cos\phi$ is

$$
\eta = \frac{xS\cos\phi}{xS\cos\phi + P_i + x^2P_{cu,fl}}
$$

#### (i) Full load, unity power factor ($x = 1$)

$$
\begin{aligned}
P_{out} &= 100\times1 = 100\ \text{kW} \\
\text{Losses} &= 0.96 + 1.2 = 2.16\ \text{kW} \\
\eta &= \frac{100}{100 + 2.16}\times100 = 97.89\%
\end{aligned}
$$

#### (ii) Half load, 0.8 power factor ($x = 0.5$)

$$
\begin{aligned}
P_{out} &= 0.5\times100\times0.8 = 40\ \text{kW} \\
P_{cu} &= 0.5^2\times1.2 = 0.3\ \text{kW} \\
\text{Losses} &= 0.96 + 0.3 = 1.26\ \text{kW} \\
\eta &= \frac{40}{40 + 1.26}\times100 = 96.95\%
\end{aligned}
$$

**Answer: (i) 97.89%; (ii) 96.95%**

## Check yourself

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

1. Efficiency at load fraction x and power factor cos(phi) is
   A. xS cos(phi) / (xS cos(phi) + Pi + x Pcu,fl)
   B. xS / (xS + Pi + x^2 Pcu,fl)
   C. xS cos(phi) / (xS cos(phi) + Pi + x^2 Pcu,fl)
   D. xS cos(phi) / (xS cos(phi) + x^2 Pi + Pcu,fl)
   Answer: C. Output is xS cos(phi); iron loss is constant while copper loss varies as x^2. (Key point: Efficiency formula)

2. At full load, unity pf (100 kW output, 0.96 kW iron, 1.2 kW copper), the efficiency is
   A. 97.89%
   B. 98.92%
   C. 96.95%
   D. 97.84%
   Answer: A. Losses = 2.16 kW, so eta = 100/102.16 = 97.89%. (Key point: Full-load unity pf efficiency)

3. What is the copper loss at half load?
   A. 600 W
   B. 1200 W
   C. 300 W
   D. 960 W
   Answer: C. Copper loss varies with the square of the load: 0.5^2 x 1200 = 300 W. 960 W is the constant iron loss. (Key point: Copper loss at half load)

4. At half load, 0.8 pf, the output and efficiency are
   A. 50 kW and 97.5%
   B. 40 kW and 96.95%
   C. 80 kW and 97.9%
   D. 40 kW and 97.89%
   Answer: B. Output = 0.5 x 100 x 0.8 = 40 kW; losses = 0.96 + 0.3 = 1.26 kW; eta = 40/41.26 = 96.95%. (Key point: Half-load 0.8 pf efficiency)

## Related in Electrical Machines

- [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
- [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 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
- [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×
- [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×

## Related in other subjects

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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
- [8 kVA 400/120 V tests, half-load efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Atransformer-8kva-400-120-hv-side) — Electrical Machine, chapter 2
- [Energy loss of 100 kVA transformer in third year](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Atransformer-energy-loss-third-year) — Transmission and Distribution System Design, chapter 7