---
title: "Condition for maximum transformer efficiency"
subject: "Basic Electrical and Electronics Engineering"
chapter: 2
key: "semester_1/basic_electrical_and_electronics_engineering:max-efficiency-condition-transformer"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/max-efficiency-condition-transformer
site: BCT Engineering
---

# Condition for maximum transformer efficiency

Basic Electrical and Electronics Engineering · Chapter 2: Electrical Machines

- Asked in: 2081 Baishakh (new course), 4 marks
- Video: [Efficiency of a Transformer - Single Phase Transformer - Basic Electrical Engineering](https://www.youtube.com/watch?v=GgckE4H5AJE) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/max-efficiency-condition-transformer
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_and_electronics_engineering/chapter-2

## Question

Derive the condition for which the efficiency of the transformer is maximum.

## Answer

**Efficiency of a transformer** at load fraction $x$ (of full-load current), power factor $\cos\phi$:

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

where $S$ is the rated VA, $P_i$ the iron loss (constant) and $P_{cu}$ the full-load copper loss.

For simplicity, write it with load current $I_2$ and secondary voltage $V_2$ (assumed constant):

$$
\eta=\frac{V_2I_2\cos\phi}{V_2I_2\cos\phi+P_i+I_2^2R_{2e}}
$$

where $R_{2e}$ is the equivalent resistance referred to the secondary.

**Condition for maximum.** Divide numerator and denominator by $I_2$:

$$
\eta=\frac{V_2\cos\phi}{V_2\cos\phi+\dfrac{P_i}{I_2}+I_2R_{2e}}
$$

$\eta$ is maximum when the denominator is minimum. Since $V_2\cos\phi$ is constant, differentiate $\left(\frac{P_i}{I_2}+I_2R_{2e}\right)$ with respect to $I_2$ and equate to zero:

$$
-\frac{P_i}{I_2^2}+R_{2e}=0 \ \Rightarrow\ P_i=I_2^2R_{2e}
$$

**Result:** the efficiency is maximum when the **iron loss equals the copper loss** (variable loss).

**Load at maximum efficiency:**

$$
x=\sqrt{\frac{P_i}{P_{cu}}}, \qquad \text{kVA at } \eta_{max}=\text{kVA}_{rated}\times\sqrt{\frac{P_i}{P_{cu}}}
$$

and

$$
\eta_{max}=\frac{xS\cos\phi}{xS\cos\phi+2P_i}
$$

## Check yourself

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

1. Which losses are constant and which vary as I2^2 in the efficiency expression?
   A. Copper loss constant; iron loss varies as I2^2
   B. Both constant
   C. Both vary as I2^2
   D. Iron loss constant (P_i); copper loss varies as I2^2 R2e
   Answer: D. Iron loss depends on flux (supply voltage), constant for fixed V; copper loss depends on load current squared. (Key point: Loss components)

2. Why is the numerator and denominator divided by I2?
   A. Efficiency becomes maximum when the denominator term Pi/I2 + I2 R2e is minimum
   B. To remove the iron loss
   C. To make the power factor zero
   D. To make the voltage constant
   Answer: A. After dividing, V2 cos(phi) is constant, so maximising efficiency means minimising Pi/I2 + I2 R2e. (Key point: Why divide by I2)

3. What does differentiating Pi/I2 + I2 R2e and equating to zero give?
   A. Pi = 2 I2^2 R2e
   B. Iron loss equals copper loss, Pi = I2^2 R2e
   C. Pi = 0
   D. Copper loss is twice iron loss
   Answer: B. -Pi/I2^2 + R2e = 0 gives Pi = I2^2 R2e. (Key point: Condition)

4. At what fraction of full load does maximum efficiency occur?
   A. x = Pcu/Pi
   B. x = Pi/Pcu
   C. x = Pi x Pcu
   D. x = sqrt(Pi/Pcu)
   Answer: D. I2^2 R2e = x^2 Pcu = Pi so x = sqrt(Pi/Pcu). (Key point: Load at max efficiency)

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