Question

# 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 rpmIa=IL−If=200−25050=195 A Ea=Vt−IaRa=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 I∗f=If−mmfNf=5−8401200=4.3 A From the given graph, if I∗f=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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