---
title: "Iron loss and maximum efficiency of 20 kVA 250/2500 V transformer"
subject: "Electrical Machines I"
chapter: 2
key: "semester_4/electrical_machines_i:20kva-250-2500-iron-loss-max"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/20kva-250-2500-iron-loss-max
site: BCT Engineering
---

# Iron loss and maximum efficiency of 20 kVA 250/2500 V transformer

Electrical Machines I · Chapter 2: Transformer

- Asked in: 2071 Magh, 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_i/20kva-250-2500-iron-loss-max
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2

## Question

A 20 kVA, 250 V / 2500 V, 50 Hz single phase transformer has the following parameters: Ro = 600 ohm, Xo = 180 ohm, R01 = 0.05 ohm and X01 = 0.15 ohm. i) Calculate the iron loss of the transformer. ii) Calculate the primary current at which the efficiency of the transformer will be maximum and also calculate the maximum efficiency.

## Short answer

Iron loss = V1^2/R0 = 250^2/600 = 104.17 W. Maximum efficiency occurs when I1^2 R01 equals iron loss, giving I1 = sqrt(104.17/0.05) = 45.64 A. At unity pf output is 11410.9 W, losses 208.33 W, so maximum efficiency is 98.21%.

## Answer

Given (referred to the 250 V primary): $R_0 = 600\ \Omega$, $X_0 = 180\ \Omega$, $R_{01} = 0.05\ \Omega$, $X_{01} = 0.15\ \Omega$; $V_1 = 250$ V.

#### (i) Iron loss
Only the core-loss resistance $R_0$ absorbs real power in the shunt branch, so at rated voltage:
$$
P_i = \frac{V_1^2}{R_0} = \frac{250^2}{600} = 104.17\ \text{W}
$$

#### (ii) Primary current for maximum efficiency
Maximum efficiency occurs when copper loss = iron loss:
$$
\begin{aligned}
I_1^2R_{01} &= P_i \\
I_1 &= \sqrt{\frac{104.17}{0.05}} = \sqrt{2083.3} = 45.64\ \text{A}
\end{aligned}
$$
Full-load primary current $= 20000/250 = 80$ A, so this is $45.64/80 = 0.571$, i.e. about **57% of full load**.

#### Maximum efficiency
Taking unity power factor (not stated; maximum efficiency is normally quoted at upf):
$$
\begin{aligned}
\text{Output} &= 250 \times 45.64 \times 1 = 11410.9\ \text{W} \\
\text{Losses} &= P_i + P_{cu} = 104.17 + 104.17 = 208.33\ \text{W} \\
\eta_{max} &= \frac{11410.9}{11410.9 + 208.33}\times 100 = 98.21\%
\end{aligned}
$$

**Answer:** Iron loss = **104.17 W**; maximum efficiency at a primary current of **45.64 A**; $\eta_{max} \approx$ **98.21%** (upf).

## Check yourself

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

1. Iron loss is found from the shunt branch as:
   A. V1^2 / X0
   B. V1^2 / R0
   C. V1^2 / (R0 + X0)
   D. I0^2 R0 at rated current
   Answer: B. Only R0 absorbs real power: 250^2/600 = 104.17 W. (Key point: Iron loss)

2. Maximum efficiency occurs when:
   A. Copper loss equals iron loss
   B. Copper loss is zero
   C. Iron loss is zero
   D. Copper loss is twice iron loss
   Answer: A. Max efficiency requires I1^2 R01 = Pi. (Key point: Max efficiency condition)

3. The primary current at maximum efficiency is:
   A. 80 A
   B. 57 A
   C. 104 A
   D. 45.64 A
   Answer: D. I1 = sqrt(104.17/0.05) = 45.64 A, about 57% of the full-load 80 A. (Key point: Current for max efficiency)

4. At maximum efficiency (upf) the total loss is:
   A. Pi only
   B. 2 Pi
   C. Pi/2
   D. 3 Pi
   Answer: B. Copper loss equals iron loss, so total loss = 208.33 W and efficiency is 98.21%. (Key point: Max efficiency)

## Close variants set in other exams

- [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×

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