SSE JE Electrical Engineering 24 March 2021 Shift 2 Paper

Question : 175 of 200

Marks: +1, -0

A 200-V Dc generator supplies 4 kW at a terminal voltage of 200 V, the armature resistance being 0.5 Ω. If the machine is operated as a motor at the same terminal voltage with the same armature current, find the ratio of the generator speed

\N_\g

, to the motor speed

\N_\m

${ \N _{ \g }}/{ \N _{ \m }}=1.25$
${ \N _{ \g }}/{ \N _{ \m }}=1.105$
${ \N _{ \g }}/{ \N _{ \m }}=0.905$
${ \N _{ \g }}/{ \N _{ \m }}=0.833$

Validate

Solution:

\V =200, \R _{ \a }=0.5 \Ω, \I _{ \a }=

constant,

\ϕ_{2}=\ϕ_{1}

Let

\ϕ_{1}=

Generator flux,

\ϕ_{2}=

Motor flux,

\N _{ \g }=

Generator speed,

\N _{ \m }=

Motor speed

∴\E _{ \g }=200+20 ×0.5=210 \V

\E _{ \b }=200-20 × 0.5=190 \V

{ \N _{ \m }}/{ \N _{ \g }}={ \E _{ \b }}/{ \E _{ \g }} × {\ϕ_{1}}/{\ϕ_{2}}={190}/{210}=0.904

{ \N _{ \g }}/{ \N _{ \m }}=1.105

Go to Question:

Prev Question

Next Question