---
title: "EMF equation of an alternator"
subject: "Electrical Machines"
chapter: 6
key: "semester_4/electrical_machines:alternator-emf-equation"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/alternator-emf-equation
site: BCT Engineering
---

# EMF equation of an alternator

Electrical Machines · Chapter 6: Three Phase Synchronous Machines

- Asked in: 2078 Baisakh, 6 marks
- Video: [EMF Equation of Alternator](https://www.youtube.com/watch?v=4831aKM2qfg) — Electrical Gurukul
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/alternator-emf-equation
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-6

## Question

Derive emf equation of an alternator.

## Short answer

Derivation yields induced emf per phase E_ph = 4.44 f φ T_ph K_p K_d volts, where f = PN_s/120, φ is flux per pole, T_ph turns per phase, and K_p, K_d are pitch and distribution factors.

## Answer

#### Derivation of the emf equation
Let
- $P$ = number of poles, $\phi$ = flux per pole (Wb)
- $N_s$ = speed in rpm, $f = \frac{PN_s}{120}$ = frequency
- $Z$ = total conductors in series per phase, $T_{ph} = Z/2$ turns per phase

**Average emf per conductor.** In one revolution a conductor cuts $P\phi$ weber in time $\frac{60}{N_s}$ s. So the average emf per conductor is
$$
e_{avg} = \frac{P\phi}{60/N_s} = \frac{P\phi N_s}{60} = 2f\phi\ \text{V}
$$
since $\frac{PN_s}{60} = 2f$.

**Per turn (two conductors):** $e_{turn} = 4f\phi$.

**Per phase** with $T_{ph}$ turns in series:
$$
E_{avg} = 4 f \phi T_{ph}
$$
Form factor for a sinusoidal emf $= \frac{\text{rms}}{\text{avg}} = 1.11$:
$$
E_{ph} = 1.11\times 4f\phi T_{ph} = 4.44\, f\,\phi\, T_{ph}\ \text{V}
$$

**Winding factors.** In a practical winding, the coils are short-pitched and the phase belt is distributed over several slots. This reduces the emf by two factors:
- Pitch factor $K_p = \cos(\alpha/2)$, where $\alpha$ is the angle by which the coil is short-pitched (electrical degrees).
- Distribution factor $K_d = \dfrac{\sin(m\gamma/2)}{m\sin(\gamma/2)}$, where $m$ is slots per pole per phase and $\gamma$ the slot angle.

$$
\boxed{E_{ph} = 4.44\,K_pK_d\, f\,\phi\, T_{ph}\ \text{V}}
$$
For a star connection, line emf $E_L = \sqrt3E_{ph}$.

## Check yourself

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

1. What is the average emf per conductor?
   A. 2 f phi
   B. 4.44 f phi
   C. 4 f phi
   D. f phi/2
   Answer: A. P phi N/60 = 2 f phi, since PN/60 = 2f. (Key point: Average emf per conductor)

2. What is the average emf per phase with Tph turns?
   A. 2 f phi Tph
   B. 4 f phi Tph
   C. 4.44 f phi Tph
   D. 8 f phi Tph
   Answer: B. Each turn has two conductors, giving 4 f phi per turn. (Key point: Per phase average)

3. What does the factor 1.11 represent?
   A. Pitch factor
   B. Form factor of sinusoid (rms/average)
   C. Distribution factor
   D. Efficiency
   Answer: B. E_rms = 1.11 x E_avg gives 4.44 f phi Tph. (Key point: Form factor)

4. What is the effect of the winding factors Kp and Kd?
   A. Increase emf
   B. Reduce emf because short-pitching and distribution reduce the phasor sum
   C. No effect
   D. Change frequency
   Answer: B. E = 4.44 Kp Kd f phi Tph is less than the full-pitch concentrated value. (Key point: Winding factors)

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