# EMF Equation of DC generator – Derivation and Examples

Digital ElectronicsElectronElectronics & Electrical

When the armature of a DC generator rotates in magnetic field, an emf is induced in the armature winding, this induced emf is known as generated emf. It is denoted by Eg.

## Derivation of EMF Equation of DC Generator

Let

$$\mathrm{\varphi = Magnetic\: flux \:per\: pole\: in\: Wb}$$

$$\mathrm{𝑍 = Total \:number \:of \:armature\: conductors}$$

$$\mathrm{𝑃 = Number \:of \:poles\: in\: the \:machine}$$

$$\mathrm{𝐴 = Number\: of \:parallel \:paths}$$

$$\mathrm{Where, \:𝐴 = 𝑃\: … for \:LAP \:Winding \:= 2\: … for \:Wave \:Winding}$$

$$\mathrm{𝑁 = Speed \:of \:armature\: in \:RPM}$$

$$\mathrm{E_{g} = Generated \:EMF = EMF \:per\: parallel \:path}$$

Therefore, the magnetic flux cut by one conductor in one revolution of the armature being,

$$\mathrm{\varphi = Magnetic\: flux \:per\: pole\: in\: Wb𝑑\varphi = 𝑃 × \varphi Wb}$$

Time taken in completing one revolution is given by,

$$\mathrm{𝑑𝑡 =\frac{60}{𝑁}seconds}$$

Hence, according to law of electromagnetic induction, the emf generated per conductor is,

$$\mathrm{E_{g}/Per conductor =\frac{𝑑\varphi}{𝑑𝑡} =\frac{𝑃\varphi}{60⁄𝑁}=\frac{𝑃\varphi𝑁}{60}}$$

Since, the number of conductors in series per parallel path is,

$$\mathrm{No.\:of \:Conducters⁄Parallel \:Path =\frac{𝑍}{𝐴}}$$

Therefore,

$$\mathrm{Total\: Generated\: EMF,\:E_{g} = EMF\: Per\: Parallel\: Path}$$

$$\mathrm{⇒ E_{g} = (E_{g}⁄Per \:conductor) × (No. of \:Conducters⁄Parallel \:Path)}$$

$$\mathrm{⇒ E_{g} =\frac{𝑃\varphi𝑁}{60}×\frac{𝑍}{𝐴}}$$

Hence, the EMF equation of a DC generator is,

$$\mathrm{E_{g} =\frac{𝑃\varphi𝑁𝑍}{60𝐴}\:… (1)}$$

It is clear from eqn. (1), that for any dc generator Z, P and A are constant so that Eg ∝ Nϕ. Therefore, for a given DC generator, the induced EMF in the armature is directly proportional to the flux per pole and speed of rotation.

Case 1 – For Lap winding, number of parallel paths A = P. Thus,

$$\mathrm{E_{g} =\frac{\varphi𝑁𝑍}{60}… (2)}$$

Case 2 – For Wave winding, number of parallel paths A = 2. Thus,

$$\mathrm{E_{g} =\frac{P\varphi𝑁𝑍}{120}… (3)}$$

## Numerical Example

A 6-pole, DC generator has 800 conductors on its armature. The flux per pole is 0.035 Wb. The speed of rotation of the armature is 1500 RPM. Calculate the generated EMF when the armature is, (a) Lap wound, (b) Wave wound.

## Solution

(a). For Lap wound armature −

$$\mathrm{E_{g} =\frac{P\varphi𝑁𝑍}{60A}=\frac{6 × 0.035 × 1500 × 800}{60 × 6}= 700 V}$$

(b). For wave wound armature −

$$\mathrm{E_{g} =\frac{P\varphi𝑁𝑍}{60A}=\frac{6 × 0.035 × 1500 × 800}{60 × 2}= 2100 V}$$

Published on 20-Aug-2021 06:54:54