---
title: "Efficiency and condition for maximum efficiency"
subject: "Electrical Machines"
chapter: 2
key: "semester_4/electrical_machines:transformer-efficiency-condition-current"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/transformer-efficiency-condition-current
site: BCT Engineering
---

# Efficiency and condition for maximum efficiency

Electrical Machines · Chapter 2: Transformer

- Asked in: 2076 Baisakh, 8 marks; 2065 Chaitra (old course), 4+4 marks (asked 2 times)
- Video: [Transformers 12 Maximum Efficiency [Electric Machinery]](https://www.youtube.com/watch?v=TVmbgRRy85E) — Ognjen Grozdanovic
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/transformer-efficiency-condition-current
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2

## Question

Define the efficiency of transformer and find the condition for maximum efficiency of transformer and the current at maximum efficiency.

## Short answer

Efficiency η = output / (output + Pi + I2^2 R02). With constant V2 and cosφ2, condition for maximum is derivative zero: d/dI2 (Pi/I2 + I2 R02) = 0 → Pi = I2^2 R02. Thus I2 at max efficiency = sqrt(Pi/R02) = I2,fl * sqrt(Pi/Pcu,fl).

## Answer

**Efficiency** is the ratio of output power to input power for a transformer:

$$
\eta = \frac{\text{output}}{\text{input}} = \frac{\text{output}}{\text{output} + P_i + P_{cu}}
$$

where $P_i$ is the core loss (constant) and $P_{cu} = I_2^2R_{02}$ is the copper loss (varies as the square of the load).

For load current $I_2$, secondary voltage $V_2$ and power factor $\cos\phi_2$:

$$
\eta = \frac{V_2I_2\cos\phi_2}{V_2I_2\cos\phi_2 + P_i + I_2^2R_{02}}
$$

#### Condition for maximum efficiency

Dividing by $I_2$ gives

$$
\eta = \frac{V_2\cos\phi_2}{V_2\cos\phi_2 + \dfrac{P_i}{I_2} + I_2R_{02}}
$$

With $V_2$ and $\cos\phi_2$ constant, $\eta$ is maximum when the denominator is minimum:

$$
\frac{d}{dI_2}\left(\frac{P_i}{I_2} + I_2R_{02}\right) = -\frac{P_i}{I_2^2} + R_{02} = 0
$$

$$
P_i = I_2^2R_{02}\quad\Rightarrow\quad \text{iron loss} = \text{copper loss}
$$

#### Current at maximum efficiency

$$
I_{2,\eta max} = \sqrt{\frac{P_i}{R_{02}}}
$$

In terms of the full-load current $I_{2,fl}$ and full-load copper loss $P_{cu,fl} = I_{2,fl}^2R_{02}$:

$$
I_{2,\eta max} = I_{2,fl}\sqrt{\frac{P_i}{P_{cu,fl}}}
$$

The corresponding kVA load is $\text{kVA}_{\eta max} = \text{kVA}_{rated}\sqrt{P_i/P_{cu,fl}}$ and the maximum efficiency is

$$
\eta_{max} = \frac{V_2I_2\cos\phi_2}{V_2I_2\cos\phi_2 + 2P_i}
$$

Distribution transformers (on load all day) are designed with a small $P_i$ so that maximum efficiency occurs near the usual load, which is below full load.

## Check yourself

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

1. Efficiency of a transformer is:
   A. Input power divided by output power
   B. Output power divided by input power (output + Pi + Pcu)
   C. Copper loss divided by output
   D. Output divided by iron loss
   Answer: B. eta = output/input = output/(output + losses). (Key point: Definition)

2. To find the maximum, eta is rewritten with denominator V2cosphi + Pi/I2 + I2 R02. What is minimised?
   A. Pi * I2 + R02/I2
   B. V2 cosphi
   C. Pi/I2 + I2 R02
   D. Pi + R02
   Answer: C. With V2 and pf constant, eta is maximum when the denominator is minimum, i.e. when Pi/I2 + I2R02 is minimum. (Key point: Why divide by I2)

3. Differentiating gives:
   A. Pi = 2 I2^2 R02
   B. Pi = I2 R02
   C. Pi = 0
   D. Pi = I2^2 R02, iron loss equals copper loss
   Answer: D. -Pi/I2^2 + R02 = 0 gives Pi = I2^2 R02. (Key point: Condition)

4. Current at maximum efficiency in terms of full-load current is:
   A. I2,fl * sqrt(Pi/Pcu,fl)
   B. I2,fl * Pi/Pcu,fl
   C. I2,fl * sqrt(Pcu,fl/Pi)
   D. I2,fl * 2 Pi
   Answer: A. I = sqrt(Pi/R02) = I2,fl sqrt(Pi/(I2,fl^2 R02)) = I2,fl sqrt(Pi/Pcu,fl). (Key point: Current at max efficiency)

5. At maximum efficiency the total loss equals:
   A. Pi
   B. 2 Pi
   C. Pi/2
   D. 4 Pi
   Answer: B. Copper loss equals iron loss, so losses = 2Pi. (Key point: Max efficiency value)

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