TRANSMISSION OF POWER BY ALTERNATING CURRENT 3&1 



It can l>e shown that t he inductance of such a loop is 



L = 2/(0.080 + 0.741 lo glo y) mil-hcnrys, (84) 



where D is the distance between conductor centers, and r is the 

 radius of each conductor. both expressed in the same units. I is 

 tin- lenirth of the line in miles. The reactance of the loop is 



A = 27T/L (85) 



where /is the frequency in cycles per second. 



It is usually more convenient to consider the inductance of a 

 single conductor only. The inductance per single conductor is 

 obviously one-half the value given in equation (84), which applies 

 to the two conductors of the circuit. 



The reactance per mile then becomes 



X = 27r/(80 + 741 logic-) 10- 6 ohms per mile. (86) 



Table I in the Appendix gives values of the reactance at 60 

 cycles per second for solid and stranded conductors, at various 

 spacing. The reactance for stranded conduct <>?> is >lightly less 

 than the corresponding values given for solid conductors. The 

 nee at other frequencies may be found by direct propor- 

 tion. (For more complete tables see Sec. XI, Standard Hand- 

 book. fifth e<lit ion.) 



nple. A .sin^le-pha^e transmission line is 40 miles long ami n.n>iMs 

 f t\vu ooo'i <olid conductors spaced 4 ft. on centers. 



! md tin- inductance of the entire line and the reactance per conductor 



at 60 cycles per second; "'-amp., <> |1 



hi- line (ind tin- total reactance drop. 

 The diameter of 0000 conductor is 460 mils: the radius, r 0.2 



48 



a 



,209 - -J 



Tin- induct'ince per mile 



r o.rii x 2.,'*2) - 3.60 mii-henrys (from equation 84). 



(a) The total ind 

 L 3.60 X 40 Ml rnil-li J mil-lien- nductor. Ana. 



<io-- 11 Ant, 



(6) The react 



X t - 2r60 X 72 X 10- - 27.1 ohms. Ana. 



(c) The- total drop 



V - 27.1 X 200 X 2 - 10,840 volts. Ant. 



