134 ELEMENTS OF ELECTRICAL ENGINEERING. 



R t = 5 5 ohms (hot) 

 R c = 0.02 ohm (hot) 

 R a = 0.14 ohm (hot) 



Stray power loss at given speed and voltage = 700 watts. 

 Solution : 



(a) Power output = 1 10 volts x 50 amperes = 5,500 watts 



(\2 

 *) =220 watts 



(c) Series field loss = R c ($o + 7) 2 = 54. i watts 



(d) Armature loss = ^ a (5O + /) 2 =379 watts 



(e) Efficiency = - 5>5 =0.803 



5,500+ 220+ 54. i +379'+ 700 



Example j. The compound generator (short-shunt). Given a 

 short-shunt compound generator which delivers 50 amperes of 

 current at 1 10 volts between its terminals. It is required to cal- 

 culate the efficiency of the generator under these conditions, hav- 

 ing given the following data : 



R s = 55 ohms (hot) 

 R c = 0.02 ohm (hot) 

 R a = 0.14 ohm (hot) 



Stray power loss at given speed and voltage = 700 watts. 

 Solution : 

 (d) Power output = no volts x 50 amperes = 5,500 watts 



(b) Shunt field loss = RJf =224 watts 



[E t = E x + RJ c = 111 volts so that 7 = 2.018 

 amperes] 



(c) Series field loss = RJ* =50 watts 



(d) Armature loss = R a l* = 379 watts 



(e) Efficiency = - 5>5 O - - =0.803 



5,500 -f 224 + 50 + 379 4- 700 



Example 4.. The series generator. Given a series generator 

 which delivers 50 amperes of current at 1 10 volts between its 



