EMF Equation of DC Generator

The expression which gives the magnitude of EMF generated in a DC generator is called EMF equation of DC generator. We shall now drive the expression for the EMF induced in a DC generator.

Let,

  • $\phi $ = flux per pole

  • P = number of poles in the generator

  • Z = no.of armature coductors

  • A = no.of parallel paths

  • N = speed of armature in RPM

  • E = EMF generated

Thus, the magnetic flux (in weber) cut by a conductor in one revolution of the armature is given by,

$$\mathrm{\mathit{d\phi \:=\:P\times \phi }}$$

If N is the number of revolution per minute, then the time (in seconds) taken complete one revolution is,

$$\mathrm{\mathit{dt \:=\frac{60}{N}}}$$

According to Faraday’s law of electromagnetic induction, the EMF induced per conductor is given by,

$$\mathrm{\mathrm{EMF/conductor}\:=\:\mathit{\frac{d\phi }{dt}}\:=\:\frac{\mathit{P\phi }}{\mathrm{\left ( {60/\mathit{N}} \right )}}\:=\:\frac{\mathit{P\phi N}}{\mathrm{60}}}$$

The total EMF generated in the generator is equal to the EMF per parallel path, which is the product of EMF per conductor and the number of conductors in series per parallel path, i.e.,

$$\mathrm{\mathit{E}\:=\:\left ( EMF/Conductor \right )\times \left ( No.\:of\:conductors/parallel\:path \right )}$$

$$\mathrm{\Rightarrow \mathit{E}\:=\:\frac{\mathit{P\phi N}}{60}\times \frac{\mathit{Z}}{\mathit{A}}}$$

$$\mathrm{\therefore \mathit{E}\:=\:\frac{\mathit{NP\phi N}}{60\mathit{A}}\:\cdot \cdot \cdot \left ( 1 \right )}$$

Equation (1) is called the EMF equation of DC generator.

For wave winding,

$$\mathrm{\mathrm{Number\:of\:parllel\:paths,}\mathit{A}\:=\:2}$$

$$\mathrm{\therefore \mathit{E}\:=\:\frac{\mathit{NP\phi Z}}{\mathrm{120}}}$$

For lap winding,

$$\mathrm{\mathrm{Number\:of\:parllel\:paths,}\mathit{A}\:=\:\mathit{P}}$$

$$\mathrm{\therefore \mathit{E}\:=\:\frac{\mathit{N\phi Z}}{\mathrm{60}}}$$

For a given DC generator, Z, P and A are constant so that the generated EMF (E) is directly proportional to flux per pole ($\phi$) and speed of armature rotation (N).

Numerical Example

A 6-pole dc generator has 600 armature conductors and a useful flux of 0.06 Wb. What will be the EMF generated, if it is wave connected and lap connected and runs at 1000 RPM?

Solution:

Given data,

  • No.of poles,P = 6

  • No.of armature conductors,Z = 600

  • Flux per pole,$\phi$ = 0.06 Wb

  • Speed of armature,N = 1000 RPM

For wave-connected generator,

$$\mathrm{\mathit{E}\:=\:\frac{\mathit{NP\phi Z}}{\mathrm{120}}}$$

$$\mathrm{\Rightarrow \mathit{E}\:=\:\frac{1000\times6\times 0.06\times 600}{120}}$$

$$\mathrm{\therefore \mathit{E}\:=\:1800\:V}$$

For lap-connected generator,

$$\mathrm{\mathit{E}\:=\:\frac{\mathit{N\phi Z}}{\mathrm{60}}}$$

$$\mathrm{\Rightarrow \mathit{E}\:=\:\frac{1000\times 0.06\times 600}{60}}$$

$$\mathrm{\therefore \mathit{E}\:=\:600\:V}$$

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