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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/pole

This question was addressed to me in an online quiz.

This intriguing question originated from Armature Reaction topic in chapter Armature Reaction of DC Machines

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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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