THREE-PHASE POWER SYSTEMS 301 



3.5 Phase-to-Phase Fault at A 

 Consider a fault between phases "a" and "Z>" at A. Then let: 



I Ac = Isa = iBb ^ Ibc = (29) 



Striking out the columns in (3) containing the currents in (29) and 

 the corresponding rows (all rows containing Ac, Ba, Bb and Be) only- 

 one equation is left : 



CiJau + CidAh = 3(1 - a'~)E (30) 



It is further known from {Za) that: 



Iai = — Iao (31) 



Substituting (31) and the constants Cn and Cn from Table I in 

 (30) the result is : 



T - - r - (1 - ^')-S HON 



-"" ~ ^' ~ Z^ + Z.-i- i?.4. + Ra, ^"^^^ 



Z\ = Za\ + Zc\ 



Zi = Za2 -\- Zc2 



which is a well-known expression for a phase-to-phase fault. 



{^3) 



3.6 Three-Phase Fault at A 

 For this case : 



iBa = iBb = Ibc = (34) 



Striking out the columns in (3) containing the currents in (34) and 

 the corresponding rows, three equations remain, any two of which 

 together with the equation from (Sa) relating to the currents at A give : 



CiJac + CulAb + CuIac = 3(1 - a^)E 



C2i/.4a + C22/^6 + C^sIac = 3(1 — a)E (35) 



lAa + lAb + Iac — 



from which the currents can be found. 



In the special case where the fault resistances are all zero, the three 

 currents are equal in magnitude and related as follows : 



lAa = alAb = a^lAc (36) 



The rank of the system determinant in (35) is therefore 1. Using 

 any of the first two equations in (35) in connection with (36) and the 

 constants in Table I, the result is: 



lAa = alAb = a^lAc = ^ (37) 



Z\ = Zai -\- Zci (38) 



