8-AJ OPERATION AND LOAD TEST. 275 



the necessary increased torque ; it has a greater slip, greater secondary 

 current and greater secondary loss. The secondary input is EJ 2 cos 6 2 . 

 A certain part of this, RJ 2) supplies the secondary loss; the remainder 

 is available for useful torque or mechanical output. 



It can be shown * that slip depends directly upon the secondary 

 copper loss; per cent, slip is equal to secondary copper loss RJ 2 , 

 divided by secondary input E 2 I 2 cos0 2 . Expressed in another way: 

 the actual speed of an induction motor is the same percentage less than 

 the synchronous speed as the useful mechanical output is less than the 

 secondary input. The difference between the synchronous and the 

 actual speeds is the slip; the difference between the secondary input 

 and the mechanical output is the secondary loss. 



Hence, since in general torque is equal to mechanical output divided 

 by actual speed (3b, Exp. 2-A), it follows that the torque of an 

 induction motor is equal to secondary input divided by synchronous 

 speed. Since synchronous speed is constant, torque is directly propor- 

 tional to secondary input. 



55. A comparison between a shunt motor and an induction motor 

 is of interest. In the rotor or armature of each there is an impressed 

 electromotive force E 2 , whether at standstill or rotating. At standstill 

 this is all available for causing current to flow through the impedance 

 of the windings and supplying Rl z losses. When running, a counter- 

 electromotive force is generated in proportion to the useful work or 

 mechanical output; the electromotive force sE 2 which is available for 

 overcoming the impedance and supplying losses is the difference be- 

 tween the impressed and counter-electromotive forces. For any given 

 load, the rotor input is proportional to E 2 and the rotor losses are pro- 

 portional to sE 2 , or 



losses -=- input = sE 2 -^E 2 = s = slip ; 



that is, slip is equal to rotor losses divided by input, as already shown. 

 This is equally true for a shunt motor or for an induction motor. 



*(54a). This is shown as follows. For a slip s, the secondary current 

 is 



VXS + s*X* 

 Hence, R 2 l2* = sEJ 2 cos 0; or, 



s = RJ?-t-EJ 2 cos 



