---
title: "Transformer losses, efficiency and maximum efficiency condition"
subject: "Electrical Machines I"
chapter: 2
key: "semester_4/electrical_machines_i:transformer-losses-efficiency-max-condition"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/transformer-losses-efficiency-max-condition
site: BCT Engineering
---

# Transformer losses, efficiency and maximum efficiency condition

Electrical Machines I · Chapter 2: Transformer

- Asked in: 2070 Bhadra, 8 marks
- Video: [13 Losses and Efficiency of Transformer](https://www.youtube.com/watch?v=jShXjmkMyB8) — SP Engg. ACADEMY
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/transformer-losses-efficiency-max-condition
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2

## Question

Explain the different power losses in the transformer and how the efficiency is calculated? Derive the condition at which the efficiency of transformer will be maximum.

## Short answer

Core (iron) loss is constant; copper loss varies with load current squared. Efficiency = output/(output + iron loss + copper loss). Maximum efficiency occurs when copper loss equals iron loss, i.e., at load fraction x = √(Pi/Pcu,FL).

## Answer

A transformer has no moving parts, so it has no friction or windage loss. Its losses are of two kinds: **core (iron) loss** and **copper loss**.

#### 1. Core or iron loss ($P_i$)
- **Hysteresis loss:** energy lost in repeatedly magnetising the core in opposite directions. $P_h = \eta_h B_m^{1.6} f V$ (Steinmetz). Reduced by using silicon steel (CRGO) with a narrow hysteresis loop.
- **Eddy current loss:** circulating currents induced in the core. $P_e = k_e B_m^2 f^2 t^2 V$, where $t$ is lamination thickness. Reduced by thin, insulated laminations (0.35-0.5 mm).
- Since $V$ and $f$ are constant, $B_m$ is constant, so **iron loss is constant** at all loads. It is measured by the **open-circuit test**.

#### 2. Copper loss ($P_{cu}$)
- $I^2R$ loss in primary and secondary windings: $P_{cu} = I_1^2R_1 + I_2^2R_2 = I_1^2R_{01} = I_2^2R_{02}$.
- **Varies with the square of load**: at fraction $x$ of full load, $P_{cu} = x^2P_{cu,FL}$.
- Measured by the **short-circuit test**.

Minor losses: stray loss (leakage flux in tank and clamps) and dielectric loss in insulation, usually neglected.

#### Calculating efficiency
$$
\eta = \frac{\text{Output}}{\text{Output} + \text{Losses}} = \frac{x\,S\cos\phi}{x\,S\cos\phi + P_i + x^2P_{cu,FL}}\times 100
$$
where $S$ = rated VA. Because efficiency is very high (95-99%), direct measurement of input and output is inaccurate; it is therefore found indirectly from the OC and SC test losses.

**All-day efficiency** (for distribution transformers) = output energy in kWh / input energy in kWh over 24 hours.

#### Condition for maximum efficiency
With $V_2$ and $\cos\phi$ constant:
$$
\eta = \frac{V_2I_2\cos\phi}{V_2I_2\cos\phi + P_i + I_2^2R_{02}}
$$
Dividing by $I_2$:
$$
\eta = \frac{V_2\cos\phi}{V_2\cos\phi + \dfrac{P_i}{I_2} + I_2R_{02}}
$$
$\eta$ is maximum when the denominator is minimum, so
$$
\begin{aligned}
\frac{d}{dI_2}\left(\frac{P_i}{I_2} + I_2R_{02}\right) &= 0 \\
-\frac{P_i}{I_2^2} + R_{02} &= 0 \\
P_i &= I_2^2R_{02} = P_{cu}
\end{aligned}
$$
So **efficiency is maximum when copper loss equals iron loss**.

The load and current at maximum efficiency:
$$
x = \sqrt{\frac{P_i}{P_{cu,FL}}}, \qquad I_2 = I_{2,FL}\sqrt{\frac{P_i}{P_{cu,FL}}}
$$
$$
\eta_{max} = \frac{x\,S\cos\phi}{x\,S\cos\phi + 2P_i}
$$
For a given load current, efficiency is highest at **unity power factor**.

**Example:** $S = 100$ kVA, $P_i = 500$ W, $P_{cu,FL} = 2000$ W: $x = \sqrt{0.25} = 0.5$, so $\eta_{max}$ at upf $= \frac{50000}{50000 + 1000} = 98.04\%$.

## Check yourself

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

1. Why is the iron loss of a transformer practically constant from no load to full load?
   A. Because V and f are constant, so Bm is constant
   B. Because the load current is constant
   C. Because copper loss cancels it
   D. Because the core has moving parts
   Answer: A. Iron loss depends on Bm and f; with constant supply voltage and frequency, Bm stays constant, so iron loss does not vary with load. (Key point: Constant iron loss)

2. At a fraction x of full load, the copper loss of a transformer is:
   A. x * Pcu,FL
   B. x^2 * Pcu,FL
   C. Pcu,FL / x
   D. Pcu,FL (constant)
   Answer: B. Copper loss is I^2R and current is proportional to load, so it varies as the square of the load fraction. (Key point: Copper loss variation)

3. Efficiency of a transformer is maximum when:
   A. Copper loss is twice the iron loss
   B. Copper loss equals iron loss
   C. Iron loss is zero
   D. Load power factor is zero
   Answer: B. Setting d/dI2 (Pi/I2 + I2 R02) = 0 gives Pi = I2^2 R02, i.e. iron loss = copper loss. (Key point: Condition for maximum efficiency)

4. With Pi = 500 W and Pcu,FL = 2000 W, the load fraction at maximum efficiency is:
   A. 0.25
   B. 0.5
   C. 2
   D. 4
   Answer: B. x = sqrt(Pi/Pcu,FL) = sqrt(0.25) = 0.5. (Key point: Load at maximum efficiency)

## Related in Electrical Machines I

- [Transformer losses and condition for maximum efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Atransformer-losses-max-efficiency) — Electrical Machines I, chapter 2, asked 3×
- [Efficiency variation with load and maximum efficiency condition](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aefficiency-variation-max-condition) — Electrical Machines I, chapter 2
- [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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