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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=50010 ω. The steady state speed at which the motor will drive the load is

A
300.75 rpm
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B
428 rpm
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C
157 rpm
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D
600 rpm
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Solution

The correct option is A 300.75 rpm
At rated load,
Motor counter emf, Ea=VtIara
or
Km.ωr=230100×0.5=180V

Where,
ωr= rated motor speed in rad.sec

Motor constant,
Km=180×602π×20,Vs/rad

Armature current at any speed ω is given by,
Ia=VtEara=230Kmω0.5

Motor torque,
Te=KmIa=Km0.5[230Kmω]

Under steady state,
Motor torque, Ts= load torque, TL
(or)
Km0.5[230kmω]=50010ω
(or)
2300.5×180×602π×25010.5[180×602π×250]2.ω=50010ω
(or)
3162.7394.545ω=50010ω
(or)
ω=3162.7350084.545=31.495 rad/sec

Motor speed
Nm=31.495×602π=300.75 rpm

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