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A 500 kW, 500 V, 10-pole, dc generator has a lap wound armature with 800 conductors. If the pole face covers 75% of pole pitch, the number of pole-face conductors in each pole of a compensating winding will be
1. 12
2. 10
3. 8
4. 6

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Correct Answer - Option 4 : 6

No. of conductors per pole in compensating winding, ZC \(= \frac{Z}{{PA}}\;\left( {\frac{{Pole\;arc}}{{Pole\;pitch}}} \right)\)

Where P = no. of poles

A = no. of parallel paths

For lap winding, the numbers of the parallel path are equal to the total of number poles i.e. A = P

No. of compensating winding = \(\frac{{800}}{{10\; \times 10}}\; \times 0.75 = 8\; \times 0.75 = 6\)

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