---
title: "EMF equation of DC generator"
subject: "Electrical Machines I"
chapter: 3
key: "semester_4/electrical_machines_i:dc-generator-emf-equation"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/dc-generator-emf-equation
site: BCT Engineering
---

# EMF equation of DC generator

Electrical Machines I · Chapter 3: DC Generator

- Asked in: 2069 Bhadra, 4 marks
- Video: [EMF Equation of DC Generator - DC Machines - Basic Electrical Engineering](https://www.youtube.com/watch?v=S8VIewhwqZs) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/dc-generator-emf-equation
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-3

## Question

Derive emf equation of a dc generator.

## Short answer

Average emf per conductor is PφN/60, and with Z/A conductors in series per path, generated emf Eg = (P φ Z N)/(60 A). For lap winding A=P, for wave A=2.

## Answer

The emf generated in a DC generator is the emf of one parallel path of the armature winding, found from Faraday's law.

Let
- $P$ = number of poles
- $\phi$ = flux per pole (Wb)
- $Z$ = total number of armature conductors
- $A$ = number of parallel paths ($A = P$ for lap, $A = 2$ for wave)
- $N$ = armature speed (rpm)

#### Derivation
Flux cut by one conductor in one revolution $= P\phi$ Wb.

Time taken for one revolution $= \dfrac{60}{N}$ s.

Average emf in one conductor:

$$
e = \frac{P\phi}{60/N} = \frac{P \phi N}{60}\ \text{V}
$$

Number of conductors in series in one path $= Z/A$. Emf between brushes = emf of one path:

$$
E_g = \frac{P \phi N}{60} \times \frac{Z}{A} = \frac{P \phi Z N}{60 A}\ \text{V}
$$

#### Special cases
- Lap winding ($A = P$): $E_g = \dfrac{\phi Z N}{60}$
- Wave winding ($A = 2$): $E_g = \dfrac{P \phi Z N}{120}$

For a given machine, $E_g = k \phi N$, so emf is directly proportional to flux per pole and speed.

**Example:** 4-pole, wave wound, $Z = 378$, $\phi = 0.02$ Wb, $N = 1000$ rpm: $E_g = \dfrac{4 \times 0.02 \times 378 \times 1000}{120} = 252$ V.

## Check yourself

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

1. What flux does one conductor cut in one revolution?
   A. phi
   B. P phi
   C. P phi/60
   D. phi/P
   Answer: B. It passes under P poles each of flux phi. (Key point: Flux cut per revolution)

2. What is the time for one revolution at N rpm?
   A. N/60 s
   B. 60/N s
   C. 60 N s
   D. 1/(60N) s
   Answer: B. N rpm is N/60 rev/s, so one revolution takes 60/N s. (Key point: Time for one revolution)

3. What is A for lap and wave winding?
   A. Lap A = 2, wave A = P
   B. Lap A = P, wave A = 2
   C. Both A = P
   D. Both A = 2
   Answer: B. Lap has as many paths as poles; wave always has two. (Key point: Parallel paths)

4. What is the emf equation?
   A. P phi Z N/(60 A)
   B. P phi N/(60 A Z)
   C. phi Z N/(60 P)
   D. P Z N/(60 A phi)
   Answer: A. E_g = (P phi N/60)(Z/A). (Key point: Final result)

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