350 



ELEMENTS OF ELECTRICAL ENGINEERING. 



prescribed (see Fig. 298). Then 



P= 



from which the full-load line current / may be calculated. 



The numerical difference E E l is the electromotive force 

 drop in one wire. Therefore, looking upon the problem as one 

 in direct currents, we have Q E l = r f I, where r' is the ap- 

 proximate resistance of one wire. From this the approximate 

 size of the wire may be found from the table. 



Consider one of the wires, say wire number 2. The other two 

 wires together constitute the return circuit for this wire, and, the 

 three wires being arranged as indi- 

 cated in Fig. 299, the distance from Jrf 



wire number 2 to each of the other 

 wires is equal to d, which is given. 

 Find the reactance x of a pair of 

 wires at the prescribed distance apart 

 center to center, from the above table. 



The component of E l parallel to 

 / is E^ cos 0, and the component of 

 E l perpendicular to / is E l sin 0. 



The resistance drop in one main is 

 rl and the reactance drop in one main 

 is \xl t the former being parallel to / 

 and the latter being perpendicular to /. Then the components 

 of E Q are E^ cos 6-\-rI, and E l sin 6-\- \xl, respectively, so that 





whence 



E* = (E l cos 6 + r/) 2 + (E l sin + 



(E I sin e -f \xij E, cos e 



which gives the true resistance r of one wire from which the 

 correct size of wire is easily found. 



Example. The electromotive force between mains at the 

 receiving stations is to be 20,000 volts. Therefore, the electro- 

 motive force between terminals of F-connected receiving circuits 



