262 ALTERNATING CURRENTS 



156. Effect of Varying Number of Rotor 

 Conductors. Speed Control 



First, let C r < C*. Assume the rotor to be originally at rest and 

 unloaded, and consider the resultant e.m.f. acting in each of the three 

 closed circuits formed by the short-circuited rotor windings (Fig. 160). 

 This resultant e.m.f. is the arithmetical difference of the e.m.f. induced 

 in the rotor windings and the p.d. of the mains. Now, if V = p.d. 

 across mains, and E g , E,. be the e.m.f.'s induced in the stator and 

 rotor windings respectively, we may write, neglecting the stator 

 resistance drop 



V = E = kC 8 , and E r = k C, 

 k being a constant. Hence 



resultant rotor e.m.f. = E,. V = k(C r C s ) . . . (1) 



But by supposition C r < C, so that the resultant rotor e.m.f. is 

 negative, i.e. the rotor current flows in the direction of the p.d., and 

 against the induced e.m.f. of the rotor. Now we know that if the 

 rotor current were allowed to flow in the direction of the induced 

 e.m.f. (as happens in an ordinary induction motor), such a current 

 would develop a torque in the direction of the rotating field. A 

 reversal of the rotor current (or a change of TT as regards phase) 

 will reverse the torque. We thus see that our commutator rotor 

 will experience a torque opposed to the direction of rotation of the 

 field. The rotor will accordingly begin to run in a direction opposed 

 to that corresponding to an ordinary induction motor, and as its speed 

 increases, E r will increase and the rotor current decrease, until it is 

 just sufficient to provide the small torque required to make up for 

 the losses due to friction, etc. 



If a load be now applied to the rotor, its speed will slightly 

 decrease, and the torque and current will increase. To balance the 

 increased rotor ampere-turns due to the load, an equivalent number 

 of ampere-turns having the opposite direction must be provided in the 

 stator.* Hence, the stator current will flow against the p.d. of the 

 mains, i.e. the stator will act as a generator, supplying power to 

 the mains. The stator of the motor now acts, in fact, as if it were 

 the secondary of a transformer of which the rotor is the primary. 



Let us next suppose that C,. = C g , the rotor being as before 

 originally at rest and unloaded. A reference to equation (1) shows 



* The supply p.d. being constant, and magnetic leakage assumed to be negligible, 

 it follows that the magnetic flux, which is common to stator and rotor, must also 

 remain constant thus involving a constancy of the resultant ampere-turns. 



