PROBLEMS. 407 



(b) Torque 1 3.72 kg.-meters, 92 per cent, of the power is avail- 

 able at the armature belt and 8 per cent, is lost in heating the 

 armature. 



Note. This problem may be solved by the use of the ordinary equations of the 

 direct current dynamo, namely : 



T= 1. 4 i^Z'J? pound inches, 

 where Z f in the present case is 260 -4- io 8 , and n is the relative speed. 



129$. The synchronous speed n of an induction motor is 100. 

 The motor develops its maximum torque when the rotor speed 

 is 80. How many times must the rotor resistance be increased 

 in order that the motor may develop its maximum torque at 

 stand-still ? Explain. 



130. An ideal three-phase induction motor takes 5 amperes 

 of current into each phase of its delta-connected primary member 

 at 200 volts and 60 cycles per second. The rotor, which has a 

 three-phase winding, delta-connected to collecting rings, supplies 

 25 amperes of current to each of three similar circuits, each hav- 

 ing a power factor equal to 0.75. The rotor runs at f syn- 

 chronous speed, (a) What is the ratio of the stator to the rotor 

 turns ? (b) What is the rotor terminal voltage ? (c) What is 

 the total intake of power ? (d) What is the electrical output of 

 power? (e) What is the mechanical output of power? Ans. 

 (a) Ratio of turns is 5 : i. (b) Rotor terminal voltage is 

 J X \ X 200 volts, (r) Total intake of power is 2,250 watts. 

 (d) Total electrical output is 750 watts, (e) Total mechanical 

 output is 1,500 watts. 



131. A 3 -phase 4-pole induction motor has its stator windings 

 delta-connected to 1,100 volt 25 cycle 3 -phase mains. The 

 rotor has a 3 -phase winding which is Y-connected to collector 

 rings, the number of conductors in the rotor winding being J as 

 great as the number in the stator winding. The rotor is driven 

 backwards at a speed of 1,050 revolutions per minute. Neglect- 



