TRANSMISSION OF POWKlt BY ALTERNATING CURRENT 3&5 



thin conducting plate of infinite breadth without disturbing the 

 electrostatic field. Each conductor has the same capacitance 

 to this plate. This capacitance must be twice the capacit- 

 ance between the conductors themselves. That is, the capacit- 

 ance C between conductors, Fig. 348 (6), may be replaced by two 

 equal capacitances, Ci, Ci, connected in series, Fig. 348 (c), 

 where Ci = 2C. The joint capacitance of the two capacitances 

 ! in series is obviously just equal to that of the single 

 capacitance, C. The point is the neutral of the system, its 

 potential being the same as that of the plate xy. 



If the capacitance to neutral is used when calculating the 

 charging current, the voltage to neutral must also be used. With 

 half the voltage and twice the capacitance, the charging current 

 per conductor is the same as if the total voltage and the capaci- 

 tance between conductors had been used. 



The capacitance to neutral may be found by multiplying equa- 

 tion (87) by 2. 



0388 

 Ci = - inf. per mile to neutral. (88) 



logio- 



Obviously, the line charging current is 



I c = 2irfCiElQ-* amperes per mile of line. 



where / is the frequency in cycles per second, E is the voltage to 

 '//, and d is the capacitance to neutral in microfarads per 

 mile of line. 



Appendix J, page 466, gives amperes per mile of line, per 

 100,000 volts to neutral, at 60 cycles per second, for various 

 inductor and various spacings. 



nple. A 10-mile, 60-cycle, single-phase line consists of two 000 

 conductors spaced 5 ft. apart. What is tin- charKiriK current if the voltage 



n wires is iU.OOO volts? 

 Tin- di.-niK -NT of 000 wire is 410 mils. 

 The r:ul: 



r - 0.20o in 

 D/r - 60/0.205 - 293 

 I..KI.203 - 2.47 



?- 0.028 ml 



The charmiif 



1 1( n | 



/. - 260 X 0.628 IO- - 3.91 amp. Ant. 



25 



