(47) 



304 BELL SYSTEM TECHNICAL JOURNAL 



Vsa = Vbo + Vbi + Vb2 



= {Rsa -\- RsFjlBa + RsplBb + RbfI Be 



Vbi, =, Vbq + a^VBi + aVBa 



= RbfIbu + {RBb + RsFjlBh + RbfI Be 



Vbc = Vbo + aVBi + a'^VB2 



= RBF^Ba + RbfIbI) + (-Rfic + Rbf)Ibc 



Consider the positive sequence diagram in Fig. 2. Evidently: 



Vai — B, — {Za\ + Zc\)Ia\ — ZciIbi ,.. 



Vbi = E — ZciIai — {Zbi + Zc\)Ibi 



where Va\ and Vbi are the positive sequence voltages to ground at the 

 two fault locations. Similar expressions can be obtained for Va2, 

 Vb2, Vao and Vbo, except for the fact the E is zero in these cases and 

 the impedances and currents are the negative and zero sequence 

 quantities. They are : 



Va2 = — {Za2 + Zc2)Ia'i ~ Zc^Ibt. 



Vb2 = ~ Zc2^A2 — \,Zb2 + Zc2)Ib2 

 Vao = ~ {Zao ~\~ ZcojIaO — Z col BO 



Vbo = ~~ ZcoIao ~ \Zbo + Zco)Ibo 

 Solving (43) and (44) for the sequence currents the result is : 



(49) 

 (50) 



IaO = ai^Aa + ^Ab + I Ac) 



Ia\ = l{lAa + alAh + o^Iac) (51) 



Ia2 = \{lAa + a^lAb + alAc) 



I BO = 'si^Ba + ^Bb + -^Bc) 



IbI = K^Ba + alsb + o^Ibc) (52) 



Ib2 = 1(1 Ba + a^lBb + alBc) 



Substituting the expressions for the sequence currents in (48), (49) 

 and (50) the result is: 



Va1 = E - \{ZaX + Zci){lAa + alAb + O^Iac) 



— lZci{lBa -\- aiBb + a^Isc) 



Va2 = — KZa2 + Zc2)(lAa + O^/ylb + alAc) (c-l\ 



— \Zc2{lBa + a^Bb + alBc) ^ ^ 



Vao = — \{ZaO 4" Zco){.lAa + lAb + I Ac) 



— \Zco{lBa + iBb + I Be) 



