THE SINGLE-PHASE MOTOR. 



THE CURRENTS IN THE ARMATURE. 



100. The current in the armature of the single-phase current motor is 



equal to the vector sum of the secondary currents - . B B' and 



v \ 



. A' B' of the two polyphase motors, hence equal to . A' B. 



*l Vi 



A glance at Fig. 28 shows us that only one-half of the energy dissi- 

 pated in the armature can be utilized for the production of the torque. 

 We see, namely, that at all loads the secondary currents represented by 

 A' B' and B' B remain very nearly equal, and as only motor 7 is do- 



FIG. 28. 



ing useful work, the armature currents in motor II represent a loss 

 almost exactly equal to the loss in the armature of the working motor 

 /. The slip in a single-phase motor indicates, therefore, only one- 

 half of the energy dissipated in the armature, hence the loss in the 

 armature of a single-phase motor is twice as large as that in a poly- 

 phase motor, provided the slip be equal in the two motors. 



Torque. 101. The torque can be calculated as follows : Suppose the* 

 armature is wound in three phases, each having the resistance r y 

 The output of the motor is then : 



P = *, ;, . cos <t> i, . r, 3 /, r tt 



57 



