246 



ALTERNATING CURRENTS 



by an increase in slip, and therefore, by a decrease in torque. 

 As the motor now develops a decreased torque with an increased 

 load it must come to a standstill, unless the load is removed. At 

 standstill (s= 1.0), the torque is comparatively small. 



The underlying cause of this small starting torque is the 

 reactance of the stator and of the rotor. The rotor reactance is 

 proportional to the rotor frequency (x'z = ZirfzLz). The rotor 

 frequency /2 is proportional to the slip. As the rotor slip in- 

 creases, the rotor reactance increases proportionately, whereas 

 the resistance does not change materially. The effect of this 

 increased reactance is to produce a greater phase difference be- 

 tween the rotor currents and their induced voltages (tan a = 

 As these currents at the same time differ in space- 



Slip 1. 



FIG. 234. Slip-torque curves for squirrel-cage motor. 



phase with the flux, less torque per ampere is developed (see 

 Par. 103). In fact the current and the flux may get so far out 

 of space-phase with each other that, even with four or five times 

 the rated current, only a small fraction of the full-load torque is 

 developed. It can be shown that the break-down torque of an 

 induction motor is decreased by an increase in the rotor react- 

 ance (z 2 = 27r/L 2 ) where x% is the rotor reactance at standstill. 

 Therefore, it is desirable that the rotor reactance, x 2) and hence 

 rotor inductance, be as low as possible (see page 247, equation 67). 

 It can also be shown that the torque of an induction motor for a 

 given slip is proportional to the square of the line voltage. If the 

 line voltage is halved the flux is halved, neglecting the stator 

 impedance drop, and the rotor current for a given value of slip 

 is halved. Therefore, the torque is quartered, the torque being 



