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

# EMF equation of a DC generator

Electrical Machine · Chapter 3: DC Generator

- Asked in: 2070 Bhadra, 4 marks
- Video: [EMF equation of dc generator with animation | Derive EMF equation of dc generator | Mruduraj](https://www.youtube.com/watch?v=m7lxUqVyAyg) — Learning Vibes
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/dc-generator-emf-equation
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-3

## Question

Derive an emf equation for a dc generator.

## Short answer

E = PΦZN/(60A) volts. In one revolution each conductor cuts flux ΦP; average conductor emf = ΦPN/60. With Z/A series conductors per parallel path, the terminal emf becomes E = (ΦPN/60)×(Z/A). For lap winding A=P, E = ΦZN/60; for wave winding A=2, E = PΦZN/120.

## Answer

**Statement:** the average emf generated in a DC generator is E_g = PΦZN / (60A) volts.

#### Derivation
Let
- P = number of poles
- Φ = flux per pole (Wb)
- Z = total number of armature conductors
- N = speed of the armature (rpm)
- A = number of parallel paths (lap: A = P, wave: A = 2)

1. In one revolution, a conductor cuts the flux of all poles: total flux cut = Φ × P (Wb).
2. Time taken for one revolution = 60/N seconds.
3. By Faraday's law, average emf induced in one conductor = flux cut / time:

   e = ΦP / (60/N) = ΦPN/60 volts

4. The Z conductors are divided into A parallel paths, so each path has Z/A conductors in series.
5. The emf of the generator equals the emf across one path:

```text
E_g = e x (Z/A)
    = (ΦPN/60) x (Z/A)
```
**E_g = PΦZN / (60A) volts**

#### Special cases
- Lap winding (A = P): E_g = ΦZN/60
- Wave winding (A = 2): E_g = PΦZN/120

For a given machine the quantity PZ/(60A) is constant, so E_g = KΦN. Hence the emf is directly proportional to the flux per pole and the speed. Terminal voltage V = E_g − I_a R_a − brush drop.

## Check yourself

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

1. How much flux does one conductor cut in one revolution?
   A. Φ x P
   B. Φ / P
   C. Φ x Z
   D. Φ x A
   Answer: A. It passes all P poles, each with flux Φ. (Key point: Flux cut)

2. Why is the per-conductor emf multiplied by Z/A?
   A. All Z conductors are in series
   B. The generator emf equals the emf of one path, which has Z/A conductors in series
   C. Conductors in parallel add their emfs
   D. A is the number of poles only
   Answer: B. Emf of parallel paths is not added; only series conductors in one path add. (Key point: Parallel paths)

3. Which is the final emf equation?
   A. E_g = PΦZN/(A)
   B. E_g = ΦZN/(60P)
   C. E_g = PΦZN/(60A)
   D. E_g = PΦN/(60Z)
   Answer: C. Combine ΦPN/60 with Z/A. (Key point: Result)

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