---
title: "Loss assumptions in OC and SC tests"
subject: "Electrical Machines I"
chapter: 2
key: "semester_4/electrical_machines_i:oc-sc-test-loss-assumptions"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/oc-sc-test-loss-assumptions
site: BCT Engineering
---

# Loss assumptions in OC and SC tests

Electrical Machines I · Chapter 2: Transformer

- Asked in: 2071 Bhadra, 8 marks
- Video: [Open Circuit and Short Circuit Test](https://www.youtube.com/watch?v=3Iz05L3KIJI) — Electric Doctor of Machines.
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/oc-sc-test-loss-assumptions
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2

## Question

While performing transformer test, copper loss is assumed to be negligible during no-load test and iron loss is assumed to be negligible during short circuit test. Justify that these assumptions are correct with detail explanation.

## Short answer

In the OC test current is only 2-5% of rated value, so I0^2R is negligible and the wattmeter reads full iron loss. In the SC test applied voltage is only 5-10% of rated value, so flux and iron loss are negligible while rated current gives full copper loss. Hence the assumptions are valid.

## Answer

In the OC test the measured power is taken as iron loss, and in the SC test as copper loss. Both assumptions are justified because in each test the neglected loss is a tiny fraction of the measured one.

#### No-load (OC) test: copper loss is negligible
```text
 rated V  -->[W]--(A)--+ LV winding || HV winding  (open)
```
- The test is done on the LV side at **rated voltage**, with the HV side open.
- Rated voltage gives **rated flux** ($\phi_m = \frac{V}{4.44fN}$), so the **full normal iron loss** occurs (hysteresis loss $\propto B_m^{1.6}$, eddy-current loss $\propto B_m^2$, so $P_i$ varies roughly as $B_m^2$).
- The only current is the no-load current $I_0$, which is only **2-5% of rated current**. The copper loss in the energised winding is $I_0^2R_1$.
- Since copper loss varies as the square of current, it is only about $(0.02$ to $0.05)^2 = 0.04\%$ to $0.25\%$ of the full-load copper loss, which is negligible. The HV winding carries no current at all.
- Hence $P_0 = P_i + I_0^2R_1 \approx P_i$.

**Example:** in a 10 kVA, 250 V transformer with $I_0 = 0.8$ A and $R_1 = 0.04\ \Omega$, the copper loss is $0.8^2 \times 0.04 = 0.026$ W against a wattmeter reading of 80 W, i.e. 0.03%.

#### Short-circuit (SC) test: iron loss is negligible
```text
 low V  -->[W]--(A)--+ HV winding || LV winding (shorted)
```
- The LV side is shorted and a **low voltage** is applied to the HV side, raised until **rated current** flows. This voltage is only **5-10% of rated voltage**, because only the small leakage impedance $Z_{01}$ limits the current.
- Rated current in both windings gives the **full-load copper loss** $I_1^2R_{01}$.
- The flux is proportional to the applied voltage, so the core works at only 5-10% of its normal flux density. Iron loss varies roughly as $B_m^2$, so it is only $(0.05$ to $0.1)^2 = 0.25\%$ to $1\%$ of the normal iron loss, which is negligible compared with the full-load copper loss.
- Hence $P_{sc} = P_{cu,FL} + P_{i,sc} \approx P_{cu,FL}$.

#### Summary
| Item | OC test | SC test |
|---|---|---|
| Applied voltage | Rated (100%) | 5-10% of rated |
| Current | 2-5% of rated | Rated (100%) |
| Flux level | Normal | 5-10% of normal |
| Iron loss | Full value | ~0.25-1% of normal |
| Copper loss | ~0.1% of FL value | Full-load value |
| Wattmeter reads | Iron loss | Copper loss |

Because each neglected loss is far below 1% of the measured one, the error is within instrument accuracy. This lets the two losses be found **separately** while using little energy, so the efficiency and regulation at any load can be predicted.

## 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. Because voltage is low
   B. Because the winding has no resistance
   C. Because no-load current is only 2-5% of rated, and loss goes as I^2
   D. Because the HV winding carries full current
   Answer: C. Copper loss is I0^2 R1, a tiny fraction (0.04-0.25%) of full-load copper loss. (Key point: OC copper loss)

2. In the SC test, the applied voltage is only 5-10% of rated because:
   A. Only the small leakage impedance limits the current
   B. The wattmeter is limited
   C. The core is saturated
   D. Iron loss must be maximum
   Answer: A. With LV shorted, rated current flows at a small voltage across the leakage impedance. (Key point: SC applied voltage)

3. Iron loss in the SC test is small because it varies roughly as:
   A. Flux density to the power zero
   B. Current squared
   C. Flux density squared, and flux is only 5-10% of normal
   D. Frequency squared only
   Answer: C. Iron loss goes as Bm^2, giving (0.05 to 0.1)^2 = 0.25-1% of normal iron loss. (Key point: SC iron loss)

4. Why must the OC test be done at rated voltage?
   A. To obtain rated current
   B. To get rated flux and so full normal iron loss
   C. To get full-load copper loss
   D. To short the HV side
   Answer: B. Flux is proportional to voltage, so rated voltage gives rated flux and the full iron loss. (Key point: Rated voltage in OC)

## Related in Electrical Machines I

- [OC and SC tests to build transformer equivalent circuit](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aoc-sc-test-10kva-6600-220) — Electrical Machines I, chapter 2
- [OC/SC test of 2200/110 V, 40 kVA transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Atransformer-oc-sc-2200-110-40kva) — Electrical Machines I, chapter 2
- [No-load current components, iron loss and core flux](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ano-load-current-10a-0p3pf-500turns) — Electrical Machines I, chapter 2
- [OC and SC tests for efficiency and regulation](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aoc-sc-test-procedure-efficiency-regulation) — Electrical Machines I, chapter 2
- [Induction motor equivalent circuit from no-load and blocked rotor tests](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aim-noload-blocked-400v-equivalent) — Electrical Machines I, chapter 5
- [No-load and blocked rotor tests of induction motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aim-noload-blocked-rotor-test-explain) — Electrical Machines I, chapter 5

## Related in other subjects

- [Why Cu loss/iron loss neglected in tests](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Acu-loss-noload-iron-loss-sc) — Electrical Machine, chapter 2
- [Why iron loss neglected in SC test; parameters](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Asc-test-iron-loss-400-200) — Electrical Machines, chapter 2
- [OC and SC tests, efficiency and regulation](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Aoc-sc-test-efficiency-regulation) — Electrical Machine, chapter 2
- [No load current of single phase transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Atransformer-no-load-current) — Electric Machine Design, chapter 3