# EMF Equation of DC Generator: Derivation with Examples

- Link: https://studyelectrical.com/2022/09/emf-equation-of-dc-generator-examples.html
- Published: 2022-09-26T07:07:02+00:00
- Author: Tesla

A DC generator is a machine which converts mechanical energy (or power) into electrical
energy (or power).

The energy conversion is based on the principle of the production of dynamically
induced EMF. According to Faraday’s Laws of Electromagnetic Induction, when a moving
conductor cuts magnetic flux, dynamically induced EMF is produced in the conductor.
If the conductor circuit is closed, this EMF causes current to flow.

The EMF generated is different in lap wound and wave wound DC generators. So finding
the amount of generated EMF is essential in the design of different [types of DC generator](https://studyelectrical.com/2014/08/types-of-dc-generators.html).
In this article we will derive the expression of generated EMF in DC generator.

## Derivation of EMF Equation of DC Generator

Let ,
_Φ = Flux/pole in weberZ = Total number of armature conductors = No. of slots
× No. of conductors/slotP = No. of generator polesA = No. of parallel paths in armatureN
= Armature rotation in revolutions per minute (RPM)E = EMF induced in any parallel
path in armature

Generated e.m.f. $E_g=$ e.m.f. generated in any one of the parallel paths i.e. $
E$.
Average e.m.f. generated/conductor $=\frac{d \Phi}{d t}$ volt $\quad(\because
n=1)$

Now, flux cut/conductor in one revolution $d \Phi=\Phi P \mathrm{~Wb}$

No. of revolutions/second $=N / 60$
$\therefore$ Time for one revolution, $d t=60/
N$ second

Hence, according to Faraday’s Laws of Electromagnetic Induction,

**EMF generated/conductor $=\frac{d \Phi}{d t}=\frac{\Phi P N}{60}$ volt**

Number of conductors in series per parallel path = Z/A

Total EMF of generator, E_(g) = EMF Per Parallel Path

= EMF generated/conductor × No. of conductors in series per parallel path

E_(g) = $\frac{\mathrm{P} \phi \mathrm{N}}{60} \times \frac{\mathrm{Z}}{\mathrm{
A}}$

**$$\therefore \ Total \ Generated \ EMF, \quad \mathrm{E}_{\mathrm{g}}=\frac{\mathrm{
P} \phi \mathrm{ZN}}{60 \mathrm{~A}}$$**

### EMF Equation a simplex wave-wound generator

For a simplex wave wound dc generator,

No. of parallel paths = 2
No. of conductors (in series) in one path = Z/2

**$\therefore \quad$ E.M.F. generated $/$ path $=\frac{\Phi P N}{60} \times \frac{
Z}{2}=\frac{\Phi Z P N}{120}$ volt**

### EMF Equation a simplex Lap-wound generator

For a simplex lap-wound generator,

No. of parallel paths = P
No. of conductors (in series) in one path = Z/P

$\therefore \quad$ E.M.F. generated $/$ path $=\frac{\Phi P N}{60} \times \frac{
Z}{P}=\frac{\Phi Z N}{60}$ volt

## Conclusion

In general generated e.m.f. $E_g=\frac{\Phi Z N}{60} \times\left(\frac{P}{A}\right)
$ volt

A = 2 for simplex wave-winding
A = P for simplex lap-winding

Also,

$E_g=\frac{1}{2 \pi} \cdot\left(\frac{2 \pi N}{60}\right) \Phi Z\left(\frac{P}{A}\
right)=\frac{\omega \Phi Z}{2 \pi}\left(\frac{P}{A}\right)$ volt $-\omega$ in rad
$/ \mathrm{s}$

For a given DC machine, Z, P and A are constant. Hence, putting K_(a) = ZP/A, we
get

$E_g=K_a \Phi N$ volts-where $N$ is in r.p.s.

## Example:

**A four-pole generator, having wave-wound armature winding has 51 slots, each slot
containing 20 conductors. What will be the voltage generated in the machine when
driven at 1500 rpm assuming the flux per pole to be 7.0 mWb ?**

Solution:

$E_g=\frac{\Phi Z N}{60}\left(\frac{P}{A}\right)$ volts

Here, Φ = 7 × 10-3 Wb, 
Z = 51 × 20 = 1020, A = P = 4, N = 1500 r.p.m.

$\therefore \quad E_g=\frac{7 \times 10^{-3} \times 1020 \times 1500}{60}\left(\
frac{4}{2}\right)=178.5 \mathrm{~V}$

[Click here to practice problems with solutions on Emf equation of DC Generator](https://studyelectrical.com/2022/09/dc-generator-problems-and-solution-emf-equation-of-dc-generator.html)

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