* 

 The total coil power is now 



P = 3E co il I coil COS C oil 



The system power is 



P = V3E line / cos e eoa (30) 



and the system kw. is equal to En n Jn nf cos 6 CO ii 



1 , 



Therefore, in a balanced three-phase system, the system power- 

 f actor is the cosine of the angle between the coil current and the coil 

 voltage. 



The angles between the line currents and the line voltages 

 arc not power-factor angles, for they involve the factors (6 30) 

 (Fig. 81) and also (0 + 30), 6 being the coil power-factor angle. 



Obviously the system power-factor, which is the coil power- 

 >r, is 



where P is the total system power. 



If the system is unbalanced, that is, if the currents or vol 

 are not equal or are not 120 apart, the question arises as to 

 just what the system power-factor is under these conditions. 

 Where such unbalancing is not very great, equation (31) is used, 

 the line currents and voltages being averaged. The system 

 power-factor has little significance when the unbalancing is con- 

 siderable. 



nple. A three-phase alternator has three <-iis each rand \\\ 

 volts and l.~>() amp. \Vh:it is the voltaic, kv-a., and current ratiim of this 



tor if the throe coils are connected in Y1 



, = V8 X 1330 = 2,300 volts. Ans. 

 Rating = v/3 X 2300 X 150 - 600 kv-a. Ans. 

 Current rating = 150 amp K 



46. Delta-connection. The three coils of Fiir. 7r eaa !>< Mu- 



ted as shown in Fig. 82 (a), the diagram leiim -implified in 



The end of each coil, which, in Fig. 78, was con- 



nected to the neutral i< now connected to the miter end of t he 



oil. afl -hown in Fig. 82 (a). As points o and a are now 

 connected directly together. /:,',. = <#, etc. The o's are Ii< 

 supertluotis and are drop: 



