THREE-PHASE POWER SYSTEMS 299 



and 



Iac = Isa = Isb = Ibc = (13) 



Striking out the columns of (1) containing the currents in (13) and 

 the corresponding rows {Ac, Ba, Bb and Be) the following two equations 

 remain : 



AiilAa + AillAh = SE 



AnlAa + A22lAb^ 3a'E ^ ^ 



from which on substituting the numerical values for the yl's from 

 Table I the two currents Iau and 7^16 can be found. The total fault 

 current to ground at -4 is : 



IaF = Iao + lAb (15) 



In the special case where Rao, RAb and Raf are zero the expression 

 for Iaf can be reduced to the following expression after a direct sub- 

 stitution for the A's in (14) is made: 



T ^ — ZaZ^E . . 



■'■■A.F 7 7 .±.7 7 _1_77 \^^) 



where : 



Z\ — Za\ + Zci 



Z2 = Za2 -\- Zc2 (17) 



Z(j = Zao + Zco 



3.3 Simultaneous Double Line-to-Ground Fault at A and Double 

 Line-to-Ground Fault at B 

 Consider a fault-to-ground on phases "a" and "6" at yl and phases 

 "a" and "c" at 5. Then: 



Rac = RBb= ^ (18) 



Hence : 



I Ac = lBb = (19) 



Striking out the two columns containing Iac and Isb and the two 

 corresponding rows {Ac and Bb), the four following equations remain: 



A n^Aa + A ulAb + A u^Ba + A \&I Be = 3£ 

 AiilAa + ^22/46 + AiilBa + Aid Be = Za'^E 

 AiilAa + AiilAb + AiilBa + A id Be = 3£ 



AqiIao -\- A&ilAb 4" AeilBa + Addsc = 3aE 

 A symbolic solution in terms of the sequence impedances for these 



