wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A 230 V, dc series has 500 lap connected armature conductor. It is supplying a shaft load of 1.5 kW. Armature and field winding resistances are 0.02 Ω/phase and 0.04Ω/phase respectively. Friction and windage losses are 100 W. At this load flux per pole is 0.035 Wb. Then speed of motor at this load is

A
787.13 rpm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
813.22 rpm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
790.1 rpm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
900.32 rpm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 787.13 rpm

Armature power developed,

= shaft power + rotational losses

= 1.5 kW + 0.1 kW

EbIa=1.6kW

Back emf is given by,

Eb=NPϕZA×60=N×0.035×50060

(For lap A = P)

and armature current, Ia=230Eb0.06

Eb(230Eb0.06)=1600

230EbE2b96=0

Eb=229.58V

Speed, N = 60EbϕZ=60×229.580.035×500

N=787.13rpm

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Cell and Cell Combinations
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon