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 VI 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

The generated e.m.f from 25-pole armature having 200 conductors driven

1.	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 VI 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
Answer» 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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