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The generated e.m.f from 25-pole armature having 200 conductors driven at 10 rev/sec having flux per pole as 20 mWb, with two parallel paths is ___________

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The generated e.m.f from 25-pole armature having 200 conductors driven at 10 rev/sec having flux per pole as 20 mWb, with two parallel paths is ___________

(a) 400 V

(b) 500 V

(c) 200 V

(d) 300 V

I have been asked this question in homework.

This question is from Dynamics of Motor-Load Combination in portion Dynamics of Electrical Drives of Electric Drives

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Correct answer is (b) 500 V

Best explanation: The generated can be calculated using the formula Eb = Φ×Z×N×P÷60×A, Φ represent flux per pole, Z represents the total number of conductors, P represents the number of poles, A represents the number of parallel paths, N represents speed in rpm. Eb = .02×25×200×600÷60×2= 500 V.

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