404 



THE INDUCTION MOTOR. 



Then, for a given applied voltage, the primary back voltage 

 will be constant and equal to the applied voltage. Con- 

 sequently, the primary flux, producing the back voltage, 

 will be constant and at right angles to it in phase. 

 Let Z s = stator flux . 

 Z r = rotor flux. 



z = leakage flux. 

 In Fig. 194 draw O E s = primary voltage to scale of volts. 



= back E.M.F. 



A = stator flux = Z a to scale of flux. 



O (7 S = stator current to scale of amps., 



making an angle 4> with O E s equal to the angle of lag of 



current in primary winding. The line A is, of course, 



perpendicular to the voltage line O E a , since the voltage 



AEs 



FKJ. 194. SIMPLE DIAGRAM OF INDUCTION MOTOK. 



is proportional to " rate of change of flux," and is conse- 

 quently perpendicular in phase to the flux. 



As in the case of a transformer, the primary flux 

 partly enters the secondary winding, producing the 

 secondary voltage, and partly forms the stray field. For 

 constant primary volts the total primary flux is constant, 

 and we may write 



Z s = Z r + z = constant. 



magnetising current 



Also z is proportional to ; ^-f r 



reluctance of leakage path 



hence z will always be in phase with the magnetising 

 current. Mark off a length B along O C a , to represent 



