---
title: "150 kVA transformer: max efficiency load"
subject: "Electrical Machines"
chapter: 2
key: "semester_4/electrical_machines:tf-11000-230-max-efficiency-kva"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-11000-230-max-efficiency-kva
site: BCT Engineering
---

# 150 kVA transformer: max efficiency load

Electrical Machines · Chapter 2: Transformer

- Asked in: 2078 Baisakh, 8 marks
- Video: [43. Transformer Efficiency, Load KVA, Maximum Efficiency Condition, Maximum Efficiency  Calculation.](https://www.youtube.com/watch?v=EsDMulvvKHo) — Electrical Tutorial
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-11000-230-max-efficiency-kva
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2

## Question

An 11000/230 V, 150 kVA, 50 Hz, single-phase transformer has a core loss of 1.4 kW and full load copper loss of 1.6 kW. Determine (i) the kVA load for maximum efficiency and the maximum efficiency (ii) the efficiency at half load and full load at 0.8 p.f. lagging.

## Answer

**Given:** 150 kVA, 11000/230 V, core loss $P_i = 1.4$ kW, full-load copper loss $P_{cu,fl} = 1.6$ kW.

#### (i) Load for maximum efficiency

At maximum efficiency, copper loss equals core loss:

$$
x^2P_{cu,fl} = P_i \Rightarrow x = \sqrt{\frac{1.4}{1.6}} = 0.9354
$$

$$
\text{kVA at } \eta_{max} = 0.9354\times150 = 140.3\ \text{kVA}
$$

The power factor is not given for this part; taking the same 0.8 lagging as in part (ii):

$$
\begin{aligned}
P_{out} &= 140.3\times0.8 = 112.25\ \text{kW} \\
\text{Losses} &= 2P_i = 2.8\ \text{kW} \\
\eta_{max} &= \frac{112.25}{112.25 + 2.8}\times100 = 97.57\%
\end{aligned}
$$

(At unity power factor the same load would give $\eta_{max} = \dfrac{140.31}{140.31 + 2.8}\times100 = 98.04\%$.)

#### (ii) Efficiency at 0.8 pf lagging

**Full load** ($x = 1$):

$$
\eta = \frac{150\times0.8}{150\times0.8 + 1.4 + 1.6}\times100 = \frac{120}{123}\times100 = 97.56\%
$$

**Half load** ($x = 0.5$): output $= 0.5\times150\times0.8 = 60$ kW, copper loss $= 0.25\times1.6 = 0.4$ kW.

$$
\eta = \frac{60}{60 + 1.4 + 0.4}\times100 = \frac{60}{61.8}\times100 = 97.09\%
$$

**Answer: (i) 140.3 kVA, $\eta_{max} = 97.57\%$ (at 0.8 pf); (ii) full load 97.56%, half load 97.09%**

## Check yourself

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

1. What is the kVA load for maximum efficiency?
   A. 140.3 kVA
   B. 150 kVA
   C. 160 kVA
   D. 120 kVA
   Answer: A. x = sqrt(1.4/1.6) = 0.9354, so 0.9354 x 150 = 140.3 kVA. (Key point: Max efficiency load)

2. What is the total loss at maximum efficiency?
   A. Pi = 1.4 kW
   B. 2 x Pi = 2.8 kW
   C. 3.0 kW
   D. Pi + Pcu,fl = 3.0 kW
   Answer: B. At that point copper loss equals core loss, so the total is 2.8 kW. (Key point: Losses at max efficiency)

3. What is the efficiency at half load, 0.8 pf?
   A. 97.56%
   B. 96.2%
   C. 97.09%
   D. 98.0%
   Answer: C. Output 60 kW; losses 1.4 + 0.4 kW; eta = 60/61.8. (Key point: Half load efficiency)

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