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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?

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? Correct Answer 6252 AT/pole

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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