388 ALTERNATING CURRENTS 



If this system be split along the line CD, two systems result, 

 one of which is shown in Fig. 352. Each of these two systems 

 transmits one-half the total power and the sending-end and 

 receiving-end voltage of each system is half the voltage between 

 conductors. The voltage at each end is now the voltage to neu- 

 tral. The ground is assumed to be the return conductor. The 

 return conductor need be merely hypothetical, however, for under 

 balanced conditions, Fig. 351, no current flows back through the 





R x 



-VWVWSA ntfft 



=- Ground Ground 



FIG. 352. Single-phase line and voltages to neutral. 



ground, as each half of the system acts as a return for the other 

 half. Therefore, the voltage drop through the ground is zero. 

 That is, Fig. 352, for purposes of calculation, the ground may be 

 considered as having zero resistance and zero reactance. 



Let it be required, in Fig. 352, to determine the generator 

 voltage EG when the load voltage E R , the current /, and power- 

 factor cos are given. The vector diagram is shown in Fig. 

 353 (a). The component of voltage to supply the IR drop is 

 laid off in phase with the current I; the component to supply 



-I 

 (a) 



FIG. 353. Vector diagrams for single-phase transmission line. 



the IX drop is laid off 90 ahead of the current 7. The resultant 

 of these two components is the component to supply the IZ 

 drop, or to supply the actual voltage drop per conductor. The 

 voltage at the generator E G is the vector sum of E R and IZ. In 

 Fig. 353(6) the IR and IX components are added to E R vec- 

 torially. It will be seen that this figure is similar to Fig. 145, 

 Chap. VI, page 139, and its geometrical solution is identical. 



E = V(E R cos $ + IR) 2 + (E R sin 6 + IX) 2 (90) 



