THREE-PHASE POWER SYSTEMS 



291 



can be reduced is given by the expression ^n{n — 1). For w = 3, 

 the case treated in this paper, three impedances are required which 

 necessitate six equations for the general solution of the six fault 

 currents. For w = 4, which would be the case for three points of 

 fault (or two points of fault and two generating voltages), six im- 

 pedances would appear in the reduced network and this would necessi- 

 tate twelve equations for the general solution of the fault currents. 

 For larger values of n, the necessary number of equations increases 

 rapidly, thus making the solution impractical. Such problems 

 usually, as a practical matter, are more readily solved by the use of a-c. 

 calculating boards. 



While no departure from the general methods of symmetrical com- 

 ponents is made in the present development, a systematic method of 

 handling the equations is presented and means of determining the 

 coefficients given so that numerical calculations can be directly carried 

 out when the constants of the network are known. 



2. General Solution 

 The equations developed in this paper are based on the sequence 

 impedances looking into a three-phase network from two points of fault. 

 Consider the network shown in Fig. 1. This system can be reduced 



CONTAINS ENTIRE NETWORK 



FAULT A FAULT B 



Fig. 1 — General network diagram. 



to an equivalent star for each of the positive, negative and zero se- 

 quence networks, with legs to the points of generation and to the faults 

 at A and B. Figure 2 shows the reduced positive sequence network. 

 Similar diagrams can be made for the negative and zero sequence 

 systems except for the fact that in these cases there are no generated 



