Question

If the speed of rotation of a separately excited DC machine is halved and the field strength is also reduced to half of the original. What is the induced emf in the armature?​

Answer 1

Given :

Speed of rotation is reduced to half.

Field strength is reduced to half.

To find :

Induced emf in the armature.

Solution :

According to faraday ' s law:

Change in flux with respect to time in a loop is equal to emf induced.

Or, If in a loop the quantity of flux is changed with respect to time then an emf is induced in a direction that can stop the change in flux.

So according to faraday 's law :

$\epsilon \: = - \frac{ d \phi}{dt}$

ϵ = EMF induced.

Negative sign represents that emf is induced in the direction opposite to the change in flux.

In an AC generator,

EMF induced :

$\epsilon \: = NBA \omega \:$

where

N = Number of turns in a loop,

B = Magnetic field strength,

A = Area of cross section of loop,

ω = Angular velocity.

In first case when nothing was changed then,

EMF induced in the armature was :

$\epsilon _ 1 \: = NBA \omega \: = \epsilon$

In second case when field strength B is halved and speed of rotation ω is halved.

So

EMF induced in the armature is :

$\epsilon _ 2 \: = N A (\frac{B}{2} ) (\frac{\omega}{2} ) \: = \frac{\epsilon}{4}$