ARMA-TURE REACTIONS 27 



12. A 2-pole armature has 64 segments, two turns per segment . 

 The total current is 120; angle of lead 25 degrees; brushes opposite 

 pole tips. The gap density is 10,000 and the leakage coefiV 

 1.35. Required: (a) The cross ampere-turns. (6) Tin -hark am- 

 pere-turns, (c) The series turns necessary to balance the back 

 ampere-turns if 0.8 of the total current be used in the series wind- 

 ing, (d) The least thickness of air gap that could be used without 

 reversing the field under the pole tip (cm.), (e) The same quan- 

 tities if the polar angle be increased to 145 degrees. (/) Also if, 

 with 290 degrees covered by the pole pieces, the machine be made 

 a 4-pole machine with a 2-path winding, find values for each 

 magnetic circuit, (g) Also the gap thickness in the first case if 

 the gap density be reduced to 80QO. (12 min.) 



13. In a smooth-core machine, if the density due to the field 

 winding be 7000 and the length of one air gap 0.8 cm.; if there be 

 260 conductors and the polar angle be 125 degrees; plot the curves 

 of density across the pole face with 0, 50 and 200 amperes given 

 by the armature. (12 min.) 



14. If in the machine of problem 3 the slots be 2.4 cm. deep and 

 the tooth density 20,000, also if the clearance be 2 mm. and the 

 density above the teeth be on account of the spreading of the lines 

 80% of the tooth density, also neglecting all other conditions than 

 the effect of reluctance on the circuit of the cross-magnetization 

 (through the teeth under the tips), what current would reduce the 

 density under the leading pole tip to 75% of its no-load value? 

 (5 min.) 



Note. The effect of the unequal reluctance under the two tips 

 would effect the distribution of the main field, thus requiring a still 

 larger current to produce the proposed change in density. 



