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What is the total ampere turns per pole for 720 lap wounded conductors with carrying armature current equal to 625A in a 6-pole machine?(a) 6252 AT/pole(b) 625.2 AT/pole(c) 62.52 AT/pole(d) 8252 AT/poleThis question was addressed to me in an online quiz.This intriguing question originated from Armature Reaction topic in chapter Armature Reaction of DC Machines

Answer»

Correct OPTION is (a) 6252 AT/pole

To EXPLAIN: For a given machine number of parallel paths is equal to 6. So, conductor CURRENT will be equal to ARMATURE current divide by no. of parallel paths i.e. 625/6. Conductor current = 104.2 A. Total armature ampere-turns, ATa = ½(720*104.2/6)= 6252 AT/pole.



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