236 ALTERNATING CURRENTS 



For example, the slip in the above motor is 



The rotor speed is 



N 2 = N(l - s) (from equation 64) (65) 



The full-load slip in commercial motors varies from 1 to 10 

 per cent., depending upon the size and the type of motor. 



102. Rotor Frequency and Induced Emf. If the rotor of a 

 two-pole, 60-cycle motor is at standstill and voltage is applied to 

 the stator, each rotor conductor will be cut by a north pole 60 

 times per second and by a south pole 60 times per second, as this 

 is the speed of the rotating field. If the stator be wound for 

 four poles, the speed of the rotating field is halved, but each 

 conductor is then cut by two north and two south poles per 

 revolution of the field and therefore by 60 north and 60 south 

 poles per second, the same as in the two-pole motor. Conse- 

 quently, the frequency of the rotor currents at standstill (s = 

 1.0) will be the same as the stator frequency. This holds true 

 for any number of poles. At standstill the motor is a simple 

 static transformer, the stator being the primary and the rotor 

 being the secondary. 



If the rotor of the above 60-cycle motor revolves at half speed 

 in the direction of the rotating field (s = 0.5), the rotor con- 

 ductors are cut by just one-half as many north and south poles 

 per second as when standing still and the frequency of the rotor 

 currents is therefore 30 cycles per second. 



By taking other rotor speeds, it can be shown that the rotor 

 frequency 



/ 2 = sf (66) 



where / 2 is the rotor frequency, s the slip, and / the stator fre- 

 quency. The rotor frequency is equal to the stator frequency 

 multiplied by the slip. 



Example. What is the frequency of the currents in the rotor of a 60-cycle, 

 six-pole induction motor, if the rotor speed is 1,164 r.p.m. 

 The synchronous speed 



60 X 120 

 N = - - = 1,200 r.p.m. (Equation 63, page 235) 



