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Q128HardDC Machines

A 400-V, 1000-A, lap wound DC machine has 10 poles, 860 armature conductors. The number of conductors in the pole face to give full compensation if pole face covers 70% of pole span is

Answer & step-by-step solution
Correct answer: A. 3010
Quick explanation

For full compensation, the compensating ampere-turns under a pole face must cancel the armature ampere-turns in that region. Using the lap-wound armature data and 70% pole-arc coverage gives the keyed requirement of 3010 ampere-conductors/ampere-turn equivalent under the pole face.

Armature AT/pole = Z Ia /(2 A P) = 860×1000/(2×10×10) = 4300 AT/pole = 0.7×4300 = 3010

Concept & Reasoning

In a lap-wound DC machine the number of parallel armature paths A equals the number of poles P. Armature ampere-turns per pole are proportional to ZIa/(2AP). Only the portion of armature reaction under the pole face must be neutralized by the compensating winding, so the pole-arc/pole-pitch ratio multiplies the result. With Z=860, Ia=1000 A, A=P=10 and pole arc ratio 0.7:
AT/pole = 860×1000/(2×10×10) = 4300 AT/pole.
Compensating requirement under the pole face = 0.7×4300 = 3010 AT/pole. The wording says 'number of conductors', but the numerical option corresponds to the compensating ampere-turn requirement.

How to Solve It in the Exam

Compensating winding cancels armature reaction only under the pole arc.

Important Exam Point

Use A=P for lap winding before substituting.

Quick Trick

Compensating winding cancels armature reaction only under the pole arc.

Common Mistake

The option value is an ampere-turn figure; do not silently drop the armature current from the calculation.

Solution diagram – A 400-V, 1000-A, lap wound DC machine has 10 poles, 860 armature conductors.…
Exam tipUse A=P for lap winding before substituting.
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