---
title: "Shunt motor losses and efficiency"
subject: "Electrical Machine"
chapter: 4
key: "semester_4/electrical_machine:shunt-motor-230v-5a-noload"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/shunt-motor-230v-5a-noload
site: BCT Engineering
---

# Shunt motor losses and efficiency

Electrical Machine · Chapter 4: DC Motor

- Asked in: 2075 Baisakh, 8 marks
- Video: [Module 1:problems on DC shunt motors](https://www.youtube.com/watch?v=5CRQsnBzGYw) — Rekha Murthy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/shunt-motor-230v-5a-noload
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-4

## Question

A 230 V shunt motor takes 5 A at no load. The resistance of armature and field circuit are 0.25 Ω and 115 Ω respectively. If the motor is loaded so as to carry 40 A, determine (i) Iron and friction losses (ii) efficiency.

## Short answer

No-load test: I_sh=2A, armature current 3A, input 1150W. Iron and friction losses = 1150 – 460 – 2.25 = 687.75W. At 40A load: input 9200W, armature current 38A, total losses 1508.75W, output 7691.25W, efficiency = 83.6%.

## Answer

At no load the motor input only supplies the iron, friction and windage losses (together called rotational loss) plus the small copper loss. This fixed (constant) loss is the same at 40 A load.

#### Equivalent circuit
```text
 I_L --+--> Ia --[Ra=0.25]--(Eb)--+
       |                          |
       +--> Ish --[Rf=115]--------+--- V = 230 V
```

#### i) Iron and friction losses (from no-load test)
- Field current: I_sh = 230 / 115 = 2 A (constant)
- No-load armature current: I_a0 = 5 − 2 = 3 A
- No-load input = 230 × 5 = 1150 W
- Field copper loss = 230 × 2 = 460 W
- Armature copper loss (no load) = 3² × 0.25 = 2.25 W
- **Iron + friction loss = 1150 − 460 − 2.25 = 687.75 W ≈ 688 W**

#### ii) Efficiency at 40 A
- Input = 230 × 40 = 9200 W
- Armature current: I_a = 40 − 2 = 38 A
- Armature copper loss = 38² × 0.25 = 361 W
- Field copper loss = 460 W
- Constant (iron + friction) loss = 687.75 W
- Total losses = 361 + 460 + 687.75 = 1508.75 W
- Output = 9200 − 1508.75 = 7691.25 W (about 10.3 hp)

η = 7691.25 / 9200 = 0.836

| Item | Value |
|---|---|
| Input | 9200 W |
| Arm. Cu loss | 361 W |
| Field Cu loss | 460 W |
| Iron + friction | 687.75 W |
| Output | 7691.25 W |

**Answer: (i) iron and friction losses = 687.75 W; (ii) efficiency = 83.6 %.**

## Check yourself

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

1. What is the shunt field current?
   A. 2 A
   B. 5 A
   C. 3 A
   D. 0.5 A
   Answer: A. I_sh = 230/115 = 2 A. (Key point: Field current)

2. How is the iron + friction loss found from the no-load data?
   A. Input minus output at 40 A
   B. No-load input minus field copper loss minus no-load armature copper loss
   C. Equal to the full-load armature copper loss
   D. Equal to the field copper loss only
   Answer: B. At no load, input = 1150 W; subtract 460 W and 2.25 W to get 687.75 W. (Key point: Iron and friction loss method)

3. How is the iron + friction loss treated at 40 A?
   A. It is zero
   B. It increases with the square of current
   C. It is assumed the same as at no load
   D. It doubles
   Answer: C. Iron and friction losses are constant, so 687.75 W is used at load. (Key point: Constant loss)

4. What is the efficiency at 40 A?
   A. 68.8 %
   B. 75 %
   C. 90 %
   D. 83.6 %
   Answer: D. Output = 9200 - 1508.75 = 7691.25 W; η = 7691.25/9200 = 0.836. (Key point: Efficiency)

## Related in Electrical Machine

- [Shunt motor speed with 0.1 Ω armature resistance, less load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-240v-5a-armature-0p1) — Electrical Machines I, chapter 4, asked 3×
- [Shunt motor speed from no load to load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-250v-light-5a-25a) — Electrical Machines I, chapter 4
- [Shunt motor loaded speed at 50 A](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Adcm-shunt-1-25kw-loaded-speed) — Electrical Machines, chapter 4
- [Shunt motor speed with 0.05 Ω armature resistance added](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-220v-25a-0p05ohm-torque) — Electrical Machines I, chapter 4, asked 3×
- [Shunt motor speed with 0.1 Ω armature resistance added](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-220v-20a-0p1ohm-armature) — Electrical Machines I, chapter 4, asked 4×
- [Shunt motor speed with 50 Ω field resistance and lower torque](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-220v-25a-50ohm-torque) — Electrical Machines I, chapter 4, asked 4×

## Related in other subjects

- [Shunt motor: field resistance to raise speed to 1100 rpm](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Adcm-shunt-field-resistance-1100rpm) — Electrical Machines, chapter 4
- [Shunt motor speed with 30 Ω field resistance added](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-220v-30ohm-field-torque) — Electrical Machines I, chapter 4, asked 5×
- [Shunt motor speed with 75 Ω field resistance added](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-250v-75ohm-torque-20pc) — Electrical Machines I, chapter 4, asked 2×
- [Shunt motor speed with 250 ohm in field circuit](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Adcm-shunt-field-resistance-250) — Electrical Machines, chapter 4