96 TRANSMISSION LINE FORMULAS 



also IA'+IB'+IC =, (4) 



since they are currents flowing in a three-phase line. 

 From equations (i) and (3) 



o T ' T ' T ' J2irfE(- 1-7327) / e \ 



2l A -lB-lc=- -^ (S) 



Adding (4) and (5), we have 



T , i 27T/EX LOO 



IA = -- = X " - - -- 



Fromthis, V = - - X 



2 log ~ 

 P 



, I vx 2 yfE (~ 0.50 ~ 0.866^) 



and l c = -- ~^ X - 



\A i s 



V ^ 2log e - 



P 



The vectors for //, /a' and /</ may now be plotted as 

 in Fig. 27, and it is seen that the vectors are the same 

 , length and are at 120 to each other. 

 Thus the charging current is a bal- 

 anced three-phase current. 



The power factor of the charging 



current is zero, since the current in 

 any wire I A ' (Fig. 27) is at right angles 

 I' B to the direction OP (Fig. 26) of the 

 Fig. 27. corresponding star voltage or in-phase 



current. 

 The total amperes of charging current are 



I A = ^ per centimeter 



2log 6 - 

 P 



