Question

# A 230 V, 250 rpm, 100 A separately-excited dc motor has an armature resistance of 0.5 Ω. The motor is connected to 230 V dc supply and rated dc voltage applied to the field winding. It is driving a load whose torque-speed characteristic is given by TL=500−10 ω. The steady state speed at which the motor will drive the load is

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

## The correct option is A 300.75 rpmAt rated load, Motor counter emf, Ea=Vt−Iara or Km.ωr=230−100×0.5=180V Where, ωr= rated motor speed in rad.sec ∴ Motor constant, Km=180×602π×20,V−s/rad Armature current at any speed ω is given by, Ia=Vt−Eara=230−Kmω0.5 ∴ Motor torque, Te=KmIa=Km0.5[230−Kmω] Under steady state, Motor torque, Ts= load torque, TL (or) Km0.5[230−kmω]=500−10ω (or) 2300.5×180×602π×250−10.5[180×602π×250]2.ω=500−10ω (or) 3162.73−94.545ω=500−10ω (or) ω=3162.73−50084.545=31.495 rad/sec Motor speed Nm=31.495×602π=300.75 rpm

Suggest Corrections
2
Join BYJU'S Learning Program
Related Videos
Work Energy and Power
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program