---
title: "Choice of number of poles in DC machine"
subject: "Electric Machine Design"
chapter: 5
key: "semester_5/electric_machine_design:dc-number-of-poles"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-number-of-poles
site: BCT Engineering
---

# Choice of number of poles in DC machine

Electric Machine Design · Chapter 5: DC Machine Design

- Asked in: 2081 Bhadra, 8 marks; 2079 Bhadra, 8 marks; 2076 Chaitra, 8 marks; 2076 Asoj, 6 marks; 2075 Asoj, 4 marks; 2073 Shrawan, 4 marks; 2072 Chaitra, 8 marks (asked 7 times)
- Video: [Selection of poles in dc machine](https://www.youtube.com/watch?v=B-J4eKirRXI) — Anand Melinamane
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-number-of-poles
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-5

## Question

Explain the factors to be considered while selecting the number of poles in a dc machine.

## Short answer

Choice balances frequency (25–50 Hz), iron and copper weight, armature reaction per pole, current per brush arm (≤200 A), flashover voltage between segments (<20 V), and labour/cost. More poles lower material and armature reaction but raise frequency and manufacturing effort.

## Answer

The number of poles of a dc machine is chosen after the output equation gives $D^2L$. More poles reduce some material but increase other costs and losses, so a compromise is made.

#### 1. Frequency of flux reversal
$$
f = \frac{Pn}{2}
$$
The armature iron sees alternating flux at this frequency. Iron loss rises with $f$, so $f$ is normally kept between 25 and 50 Hz. This limits the number of poles at high speed.

#### 2. Weight of iron
For the same total flux, the flux per pole $\Phi \propto 1/P$. So the yoke and armature core depth (which carry $\Phi/2$) become thinner, and the iron weight falls as $P$ increases.

#### 3. Weight of copper
- **Armature copper**: the end connections are shorter for a smaller pole pitch, so armature copper decreases with more poles.
- **Field copper**: each pole needs its own coil. The total field copper stays roughly constant or falls a little.

#### 4. Armature reaction (ampere-turns per pole)
$$
AT_a = \frac{ac\,\pi D}{2P}
$$
More poles lower the armature mmf per pole, so the distortion and the field mmf needed to overcome armature reaction are less. A typical limit is armature AT per pole below about 5000–10000, depending on output.

#### 5. Current per brush arm / per path
For a lap winding, the current per path $= I_a/P$. It should not exceed about 200 A (and about 400 A per brush arm), otherwise the commutator becomes long and commutation becomes difficult. Large-current machines therefore need more poles.

#### 6. Commutator length and brushes
More poles mean more brush arms with a smaller current in each, so the commutator is shorter, but there is more brush gear.

#### 7. Overall size, labour and cost
More poles give a smaller overall diameter and less material, but they need more coils, more parts and more labour. Frequency and losses also rise.

#### 8. Voltage between commutator segments (flashover)
For a lap winding, the terminal voltage $V$ is spread over $N_{seg}/P$ segments in series, so the average voltage between adjacent segments is $e_s = VP/N_{seg}$. This must be kept below about 20 V to avoid flashover. For the same terminal voltage, more poles therefore need more commutator segments, which makes the commutator larger and more costly.

#### Typical choice
| Output | Usual poles |
|---|---|
| Up to about 5 kW | 2 |
| 5–100 kW | 4 |
| 100–500 kW | 6 to 8 |
| Above 500 kW | 8 or more |

Example: a 37 kW, 1400 rpm motor with 4 poles has $f = 4 \times 1400/120 = 46.7$ Hz and a current per path of about 45 A, so 4 poles is suitable. Six poles would give 70 Hz, which is too high.

## Check yourself

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

1. What is the frequency of flux reversal in a dc armature and its usual limit?
   A. f = Pn, kept below 10 Hz
   B. f = Pn/2, kept between about 25 and 50 Hz
   C. f = Pn/2, kept above 100 Hz
   D. f = n/P, kept between 25 and 50 Hz
   Answer: B. Iron loss rises with frequency, so f = Pn/2 is limited to about 25-50 Hz, limiting poles at high speed. (Key point: Frequency of flux reversal)

2. What happens to iron weight as the number of poles increases?
   A. It rises, because there are more poles
   B. It falls, because flux per pole is smaller so yoke and core are thinner
   C. It stays the same
   D. It rises, because flux per pole increases
   Answer: B. Flux per pole varies as 1/P, so yoke and armature core depth reduce and iron weight falls. (Key point: Iron weight)

3. How does the armature mmf per pole change with the number of poles?
   A. It rises with P, so armature reaction is more
   B. It is independent of P
   C. ac pi D / 2P falls, so armature reaction is less
   D. It falls only if D is raised
   Answer: C. AT per pole = ac pi D/(2P), inversely proportional to P. (Key point: Armature reaction)

4. For a lap winding, why do large-current machines need more poles?
   A. Current per path Ia/P must stay below about 200 A
   B. Current per path Ia x P must stay low
   C. Frequency must stay low
   D. Brush arms are fewer with more poles
   Answer: A. Lap paths equal P, so more poles reduce current per path and ease commutation. (Key point: Current per path)

## Close variants set in other exams

- [Field form factor and choice of poles](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Afield-form-factor-poles) — Electric Machine Design, chapter 5, asked 8×

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- [Choice of ampere conductors per meter in DC machine](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-ampere-conductors-selection) — Electric Machine Design, chapter 5, asked 4×
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