CURRENTS AND POTENTIALS ALONG CONDUCTORS 



111 



For conductors with large leakance, these simplifying assumptions 

 are not justified. When the earth resistivity is uniform or varies 

 with depth only, the electric force may be formulated in terms of the 

 current along the entire length of the conductor, in which case the 

 usual differential equation for the current is replaced by an integro- 

 difterential equation. A general solution of this equation and for 

 conductor and earth potentials has been obtained. For homogeneous 

 earth, rigorous as well as approximate solutions of special cases of 

 interest in connection with the railway electrification problems men- 

 tioned above have also been derived. 



One of these cases is of general interest since it may be regarded 

 as fundamental to the solution of the general case of an arbitrary 

 impressed electric force along the conductor. In this case a voltage 

 2 F(0) is impressed across a break in the conductor at a certain point, 

 which may be taken as the origin. The conductor current and 

 potential are given by rather complicated integrals, which, in order 

 to obtain practical formulas, may be expanded in series as: 



I(x) = /i(x) + hix); hix) = Mx) + Inix), (1) 



V{x) = Fi(.r) + FsW; V^ix) = Vn(x) + V,2(x), (2) 



where To2(x) and Vo^ix) may again be written as the sum of two terms, 

 etc. The first terms in these expansions are: 



h(x) = hi- x) = F(0)^^g-' 



l{0)e- 



V,(x) = - Vi(- x) = V{0)e-~^- = 7(0) 



where x > and : 



GiT) 



(3) 

 (4) 



r = vz(r)G(r). 



„, s tcov 1.85- 



Z(T) = 2 + -— log, 



ZTT aa 



G(T) = 



, . p . 1.12- 

 g 1 + - log, —-—- 

 T or 



a = {iwvp-^ + T'^y-. 



CO = 2irf. 



f = frequency in cycles per 

 second. 



X — distance from origin, in 

 meters. 



a — conductor radius, in meters. 



2 = internal impedance in ohms 

 per meter. 



g = leakage conductance in ohms 

 per meter. 



p — earth resistivity in meter- 

 ohms. 



V = 47r-10~'^ henries per meter. 



