THE INDUCTION* MOTOR. 413 



output of the stationary motor is zero under the con- 

 ditions represented by the position of B l (i.e., when the 

 motor is stationary). 



In order to find the "torque" circle A H F, the scale 

 to which B l A represents volts must be determined. 

 It is known that O A is proportional to the stator flux 

 N a and that this is proportional to the impressed voltage 

 E. 



volts represented by B l A E 

 Hence, - Rl A - = Q-^ Thus the 



line A represents the primary voltage to the same scale 

 as B l A represents volts. Hence a scale of volts can be 

 constructed by dividing the line A into E parts, each 

 corresponding to one volt. 



From B l mark off the length B l H l equal to the drop 

 in the stator windings, due to the static current. This 

 length is equal to C a x E s (i.e., product of stator resistance 

 and static current measured on the scale of volts just 

 determined). The point H l divides the line B l A into 

 two parts, which are proportional respectively to the 

 stator and rotor resistances. Draw the " torque " 

 circle A H F through the points A v H\ and F with its 

 centre f l on the line p g. 



In order to obtain the slip line join A f l , and draw 

 B l T perpendicular to A /', cutting A O at T. The scale 

 of slip is obtained from the fact that the slip corresponding 

 to the position B l , i.e., the length .B 1 T, is 100 per cent. slip. 



The following example of an actual test of a three- 

 phase induction motor is taken from Mr. Eborall's paper, 

 read before the Society of Arts in 1901. 



The motor was rated to give 80 b.h.p. at 600 revs. 

 per minute at 40 cycles and 350 volts. 



Fig. 197 gives the results of the static and no-load 

 tests from which the diagram Fig. 198 was constructed. 



The upper curve gives the relation between voltage 

 and static current with the rotor rigidly clamped, and 

 short-circuited; the second curve shows the relation of 

 no-load current to voltage, the motor running light ; 

 the third curve gives the corresponding watts input, 

 with the motor running light. 



From the curves we see that at a pressure of 202 

 volts per phase (i.e., a terminal pressure of 350 volts with 



