---
title: "Efficiency of 5 kVA transformer at 2 kW load"
subject: "Electrical Machines I"
chapter: 2
key: "semester_4/electrical_machines_i:5kva-1100-110-max-efficiency-40a"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/5kva-1100-110-max-efficiency-40a
site: BCT Engineering
---

# Efficiency of 5 kVA transformer at 2 kW load

Electrical Machines I · Chapter 2: Transformer

- Asked in: 2068 Bhadra, 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_i/5kva-1100-110-max-efficiency-40a
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2

## Question

A 5 kVA, 50 Hz, 1100V/110V single phase transformer has equivalent resistance of 0.04 Ohm and equivalent reactance of 0.24 Ohms referred to secondary side. When it delivers a current of 40 A to the load at 110 V, its efficiency is maximum. Calculate the efficiency of the transformer when it is delivering 2 kW to the load at 0.85 power factor lagging.

## Short answer

Iron loss Pi = 40^2 *0.04 = 64 W. At 2 kW, 0.85 pf, load current I2 = 2000/(110*0.85) = 21.39 A, copper loss = 21.39^2 *0.04 = 18.30 W. Efficiency = 2000/(2000+64+18.30)*100 = 96.05%.

## Answer

Given (secondary side): $R_{02} = 0.04\ \Omega$, $X_{02} = 0.24\ \Omega$. Maximum efficiency occurs at $I_2 = 40$ A.

#### Iron loss
At maximum efficiency, copper loss = iron loss:
$$
P_i = I_2^2R_{02} = 40^2 \times 0.04 = 64\ \text{W}
$$

#### Load current for 2 kW at 0.85 pf (at 110 V)
$$
I_2 = \frac{P}{V_2\cos\phi} = \frac{2000}{110 \times 0.85} = 21.39\ \text{A}
$$
(Rated current is $5000/110 = 45.45$ A, so this is 47% load.)

#### Copper loss at this load
$$
P_{cu} = I_2^2R_{02} = 21.39^2 \times 0.04 = 18.30\ \text{W}
$$

#### Efficiency
$$
\begin{aligned}
\eta &= \frac{2000}{2000 + 64 + 18.30}\times 100 \\
&= \frac{2000}{2082.30}\times 100 = 96.05\%
\end{aligned}
$$

The reactance $X_{02}$ is not needed because it causes no power loss.

| Quantity | Value |
|---|---|
| Iron loss | 64 W |
| Load current | 21.39 A |
| Copper loss | 18.30 W |
| Efficiency | 96.05% |

**Answer:** Efficiency at 2 kW, 0.85 pf lagging = **96.05%**.

## Check yourself

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

1. At maximum efficiency with I2 = 40 A and R02 = 0.04 ohm, the iron loss is:
   A. 16 W
   B. 64 W
   C. 160 W
   D. 1.6 W
   Answer: B. At maximum efficiency Pi = Pcu = 40^2 x 0.04 = 64 W. (Key point: Iron loss at max efficiency)

2. Current for 2 kW at 0.85 pf and 110 V is:
   A. 18.18 A
   B. 21.39 A
   C. 40 A
   D. 45.45 A
   Answer: B. I2 = P/(V2 cos phi) = 2000/(110 x 0.85) = 21.39 A. (Key point: Load current)

3. Why is X02 not used in the efficiency calculation?
   A. Reactance causes no power loss
   B. It is equal to R02
   C. It is referred to primary
   D. It is much larger than Z
   Answer: A. Only resistance dissipates power; reactance stores and returns energy. (Key point: Role of reactance)

4. The efficiency at 2 kW, 0.85 pf is about:
   A. 96.05%
   B. 98.2%
   C. 92.4%
   D. 94.5%
   Answer: A. Pcu = 21.39^2 x 0.04 = 18.3 W; eta = 2000/(2000 + 64 + 18.3) = 96.05%. (Key point: Efficiency)

## Related in Electrical Machines I

- [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
- [Efficiency and loading of 50 kVA transformer at 40 kW](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A50kva-200-2200-40kw-overload) — Electrical Machines I, chapter 2
- [Equivalent circuit, efficiency and regulation of 5 kVA transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aoc-sc-5kva-220-400-lv) — Electrical Machines I, chapter 2
- [Equivalent impedance and copper loss of 50 kVA transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A50kva-4400-220-equivalent-copper-loss) — 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
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## Related in other subjects

- [50 kVA transformer: equivalent parameters, copper loss](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-50kva-4400-220-equivalent-resistance) — Electrical Machines, 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
- [3-phase 500 kVA transformer primary current](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-500kva-11kv-400v-primary-current) — Electrical Machines, chapter 2
- [Design: 100 kVA 2200/480 V single phase transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-100kva-single-phase-ducts) — Electric Machine Design, chapter 3, asked 3×