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A 100KW, 250 V, 400 A, a long shunt compound generator has an armature resistance of 0.025 ohms. There are 1000 shunt fields turns per pole and 3 series field turns per pole. The series field is connected in a such a fashion that positive armature current produces direct-axis MMF which adds to that of the shunt field. Compute the gross MMF at the rated terminal current when shunt field current is 4.7A and speed is 1150 rpm.

A 100KW, 250 V, 400 A, a long shunt compound generator has an armature resistance of 0.025 ohms. There are 1000 shunt fields turns per pole and 3 series field turns per pole. The series field is connected in a such a fashion that positive armature current produces direct-axis MMF which adds to that of the shunt field. Compute the gross MMF at the rated terminal current when shunt field current is 4.7A and speed is 1150 rpm. Correct Answer 5.9AT

Series field current = Is = I(l)+I(f) = 400+4.7 = 405 A(approx) Main field mmf = I(f)+(Ns/Nf)*I(s) = 4.7+(3/1000)*405 = 5.9 AT.

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