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

# EMF equation of a DC generator

Electrical Machines · Chapter 3: DC Generator

- Asked in: 2070 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/dc-generator-emf-equation
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-3

## Question

Derive an emf equation for a dc generator.

## Short answer

Average emf per conductor is PΦN/60. With Z/A conductors in series per parallel path, generated emf is Eg = (Φ Z N P)/(60 A) V. For lap winding A=P, Eg = ΦZN/60; for wave winding A=2, Eg = ΦZNP/120.

## Answer

The emf equation gives the average emf generated in the armature of a DC machine.

Let
- $\Phi$ = flux per pole (Wb), $P$ = number of poles
- $Z$ = total armature conductors, $A$ = number of parallel paths
- $N$ = speed in rpm

In one revolution a conductor cuts the flux of all poles, $P\Phi$ weber, in time $60/N$ second.

$$
\text{average emf per conductor} = \frac{d\phi}{dt} = \frac{P\Phi}{60/N} = \frac{P\Phi N}{60}\ \text{V}
$$

The $Z$ conductors form $A$ parallel paths, so each path has $Z/A$ conductors in series. The generated emf equals the emf of one path:

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

- Lap winding: $A = P$, so $E_g = \dfrac{\Phi Z N}{60}$
- Wave winding: $A = 2$, so $E_g = \dfrac{\Phi Z N P}{120}$

In terms of angular speed, $E_g = \dfrac{ZP}{2\pi A}\Phi\omega = K_a\Phi\omega$.

## Check yourself

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

1. The average emf generated in one conductor is
   A. Phi N/60
   B. P Phi N/60
   C. P Phi N/A
   D. Phi Z N/60
   Answer: B. Flux cut per revolution is P Phi in a time of 60/N s, so e = P Phi N/60 volts. (Key point: Emf per conductor)

2. Why is the emf per conductor multiplied by Z/A?
   A. A conductors are in series in each path
   B. Each of A parallel paths has Z/A conductors in series, and the generated emf equals that of one path
   C. Z conductors are in parallel
   D. Because A is the number of poles
   Answer: B. Parallel paths do not add voltage; the terminal emf equals the emf of one path, Z/A series conductors. (Key point: Role of parallel paths)

3. For a lap winding, the number of parallel paths A is
   A. 2
   B. P (the number of poles)
   C. P/2
   D. Z
   Answer: B. Lap winding has A = P, giving Eg = Phi Z N/60. (Key point: Lap winding)

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