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A 50 Hp, 250 V, 1200 rpm dc shunt motor without compensating windings has an armature resistance (including brushes and interpoles) of 0.06 Ω. Its field circuit has a resistance of 50 Ω, which produces a no load speed of 1200 rpm. There are 1200 turns per pole on the shunt field winding, and the armature reaction produces a demagnetizing magneto motive force of 840 AT at a load current of 200 A. The magnetizing curve of this machine is shown below,

The speed of this motor when its input current is 200 A is

A
1160 rpm
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B
1412 rpm
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C
1227 rpm
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D
11200 rpm
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Solution

The correct option is C 1227 rpm
Ia=ILIf=20025050=195 A

Ea=VtIaRa=250(195×0.06)=238.3 V

At IL=200 A, the demagnetizing mmf is 840 AT.
So, effective shunt field current of the motor is

If=IfmmfNf=58401200=4.3 A

From the given graph, if If=4.3 A
then Eno load=233 V

E1,nlE2=N1,nlN2

233238.3=1200N2

Speed of the motor, N = 1227.3 rpm

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