132 THE MAGNETIC CIRCUIT [ART. 44 



primary winding, the same as in a transformer. Therefore, the 

 net number of exciting ampere-turns, M, is approximately the 

 same as at no load. This means that the geometric sum of 

 the m.m.fs. produced by the primary and the secondary currents at 

 any load is nearly equal to the m.m.f . due to the primary winding 

 alone at no load. In this respect the induction motor is similar to 

 a transformer. 



(a) Calculation of the Secondary Current. Knowing the primary 

 full-load current, the secondary full-load current can be calcu- 

 lated from the required counter-m.m.f.; the procedure can be best 

 illustrated by an example. In the motor given in prob. 4 above, 

 the full-load current is estimated at 57 amp.; taking the direction 

 of the vector of the applied voltage as the axis of reference, the 

 full-load current can be represented as 51.3 /24.8 amp. The 

 magnetizing current, 0.25X57= 14.25, is practically in quadrature 

 with the applied voltage, because it is in quadrature with the 

 induced counter e.m.f., the same as in a transformer. The full- 

 load current of 57 amp. contains a component which supplies the 

 iron loss in the stator; we estimate it to be equal to about 1.1 amp. 

 (2 per cent of the input). Thus, the component of the primary 

 current, the action of which must be compensated by the second- 

 ary currents, is (51.3 -/24.8) - (1.1 -/14) = 50.2 -/10.8 amp., or its 

 absolute value is 5.14 amp. This is called the current transmitted 

 into the secondary, or the secondary current reduced to the primary 

 circuit. This current produces a maximum m.m.f. of 0.9 X3 X0.958 

 X 42X51. 4 = 5580 amp.-turns. 



Let the rotor be provided with a three-phase winding, with 5 

 slots per pole per phase, and let the winding pitch be 13/15. The 

 number of slots is selected so as to be different from that in the sta- 

 tor, in order to insure a more uniform torque, and to reduce the 

 fluctuations in the reluctance of the active layer. We have, 

 according to eq. (64), that 5580 = 0.9X3X0.935X (ra), from which 

 ni=2210 amp.-turns. Certain practical considerations, for 

 instance, the value of the induced secondary voltage, usually 

 limit the choice of one of these factors; then the other factor 

 also becomes definite. If, for instance, the rotor is to have 10 

 conductors per slot, the secondary current will be about 89 amp. 

 The secondary i 2 r loss is determined by the desired per cent slip; 

 knowing the secondary current and the number of turns, the 

 necessary size of the conductor can easily be calculated. 



