136 THE MAGNETIC CIRCUIT [ART. 45 



in the rotor, or with a squirrel-cage winding. In these two cases 

 the individual coils or bars in the rotor are all in parallel, while the 

 stator coils of a phase are usually all in series, or in two parallel 

 groups. In the case of a squirrel-cage winding the resistance r 2 

 includes that of a bar, of two contacts with the end-rings, and of 

 the equivalent resistance of a section of the two end-rings. 1 



Prob. 12. In a 300 horse-power, Y-connected, 14-pole induction 

 motor the full-load current is estimated to be 310 amp. The primary 

 winding consists of 336 turns placed in 168 slots; the winding pitch is 

 0.75. What is the minimum number of bars in the squirrel-cage second- 

 ary winding, if the current per bar must not exceed 800 amp.? The 

 secondary counter-m.m.f. is equal to about 90 per cent of the primary 

 m.m.f. Ans. 208. 



Prob. 13. What must be the resistance of each secondary bar in the 

 preceding problem (including the equivalent resistance of the adjoining 

 segments of the end-rings and also of the contacts) if the slip at full load 

 is to be about 4 per cent.? Hint: The per cent slip is equal to the i 2 r 

 loss in the rotor, expressed in per cent of the power input into the second- 

 ary. If x is the i 2 r loss in the rotor, expressed in horse-power, we have 

 that x = 0.04 (300 + x) . Ans. 70 microhms. 



Prob. 14. The motor with the individually short circuited second- 

 ary coils, that is used as an illustration in the text above, is to be 

 investigated with respect to its performance. By what factor must the 

 actual resistance and inductance of each secondary coil be multiplied 

 in order to obtain the equivalent resistance and inductance per primary 

 phase? Also by what factor must the equivalent current be multiplied 

 in order to obtain the actual current in each secondary coil? 



Ans. 297; 0.402. 



Prob. 15. Prove formula (64a) directly, by considering the m.m.fs. 

 of the individual bars. Solution : At any instant the currents in the bars 

 under a pole are distributed in space according to the sine law, because 

 the gliding flux which induces these currents is sinusoidal. The average 

 current per bar is f\X2X(2/*) = 0.9i. The number of turns per pole is 

 C 2 /2p, and all these turns are active at the crest of the m.m.f. wave. 

 Therefore, M = 0.9i(C 8 /2p). 



45. The Higher Harmonics of the M.M.FS. In the preceding 

 study, the effect of the higher harmonics in the m.m.f. wave was dis- 

 regarded. In fact, these harmonics usually exert a negligible 

 influence upon the operation of a good polyphase induction motor, 

 under normal conditions. These m.m.f. harmonics move at lower 

 speeds than the fundamental field ; therefore, the fluxes which they 



1 See the author's Electric Circuit; also E. Arnold, Die Wechselstromtechnik, 

 Vol. 5, Part I (1909), p. 57. 



