---
title: "Why Cu loss/iron loss neglected in tests"
subject: "Electrical Machine"
chapter: 2
key: "semester_4/electrical_machine:cu-loss-noload-iron-loss-sc"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/cu-loss-noload-iron-loss-sc
site: BCT Engineering
---

# Why Cu loss/iron loss neglected in tests

Electrical Machine · Chapter 2: Transformer

- Asked in: 2073 Bhadra, 8 marks
- Video: [Open Circuit Test & Short Circuit Test: Calculating Iron and Copper Losses of a Transformer](https://www.youtube.com/watch?v=rtn25-k5Vs4) — ELECTRICAL VEDA
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/cu-loss-noload-iron-loss-sc
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-2

## Question

Copper loss is assumed to be negligible in no load test and iron loss is assumed negligible in short circuit test. Explain why it is so.

## Short answer

In OC test, secondary open, primary carries no-load current (2–5% rated), so copper loss I₀² R is <0.1% of full-load value, negligible compared to iron loss. In SC test, applied voltage is 5–10% rated, flux and iron loss are ~1% rated, negligible; full-load current makes copper loss dominant.

## Answer

Both losses are neglected because of how the test conditions make one of them very small compared with the other.

#### Copper loss in the no-load (OC) test
- In the OC test, rated voltage is applied to the primary with the secondary open. The secondary carries no current, so there is no secondary copper loss.
- The primary takes only the no-load current I0, which is about 2–5 % of the full-load current.
- Copper loss varies as the square of current: Pcu(no-load) = I0² R1 ≈ (0.03)² × Pcu(full load) ≈ 0.001 × full-load copper loss, which is less than 0.1 %.
- The iron loss at rated voltage is much larger than this small copper loss. So the wattmeter reading W0 is taken as entirely the iron loss.

#### Iron loss in the short-circuit (SC) test
- Core losses depend on the flux, which is proportional to the applied voltage: Pi ∝ Φm² ∝ V².
- In the SC test, the secondary is short-circuited, and the voltage applied to the primary only needs to circulate the rated current through the series impedance of the windings. This is only 5–10 % of the rated voltage.
- At about 5–10 % voltage, flux is about 5–10 % of normal, and the iron loss falls to about (0.1)² = 1 % or less of its rated value, which is very small.
- Also, almost the whole supply is used to drive the full-load current through the windings, so the magnetising branch carries negligible current.
- The wattmeter reading Wsc therefore equals the full-load copper loss.

| Test | Voltage | Current | Dominant loss | Neglected |
|---|---|---|---|---|
| OC | Rated | 2–5 % of rated | Iron | Copper |
| SC | 5–10 % of rated | Rated | Copper | Iron |

## Check yourself

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

1. Why is copper loss negligible in the no-load test?
   A. The core has no resistance
   B. The wattmeter does not measure copper loss
   C. Only a small no-load current (2-5 % of full load) flows, so I0² R is tiny
   D. The secondary is short-circuited
   Answer: C. Copper loss varies as current squared; at about 3 % current it is about 0.1 % of full-load copper loss. (Key point: Copper loss in OC test)

2. Why is iron loss negligible in the short-circuit test?
   A. Because the applied voltage is only 5-10 % of rated, so flux and iron loss (∝ V²) are very small
   B. Because the current is rated
   C. Because the core is removed
   D. Because the frequency is zero
   Answer: A. Pi ∝ Φm² ∝ V², so at 5-10 % voltage iron loss is about 1 % or less. (Key point: Iron loss in SC test)

3. How does iron loss depend on applied voltage?
   A. Proportional to V
   B. Independent of V
   C. Proportional to 1/V
   D. Proportional to V²
   Answer: D. Flux is proportional to voltage and core loss to flux squared. (Key point: Iron loss dependence)

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