and Inductance of a Concentric Main. 543 



where 

 R = (pm/27raY a) {her ma bei' ma — hei ma ber' ma) 



pm bei' ma , A ber mh + B bei ^ + c kep mb + D kei m& j. 



27raY„ l 



-P mhev ' ma {Ahei mb-Bhev mb + G kei m&-D ker m&}, 

 27raY a ^ 89 ) 



and 



L = 2 u' loo- - + /f r - (ber ma ber' ma + bei ma bei' ma) 



^ & a Mifl 



2^Wma , A ber w6 + B bei mh + c ker w6 + D kei m ^ | 

 inaY a 

 + Vbei' ma ^ A bei m6 _ B ber w?> + c kei m6 _ D ker mh y . 



wYa .... (90) 



In (89) and (90), Y a stands for ber' 2 ma + bei' 2 ma, and 

 A, B, C, and D can be found as follows from the equations 

 (81)-(84), using the notation adopted in formulae (54) 

 and (55) : — 



AY C = -CS C + DT C (91) 



and BY C = -CT C -DS, (92) 



Hence substituting the values of A and B, given by (91) 

 and (94), in (83) and (84), we get 



C(Y c Sa - Y 6 S C ) + D (Y 6 T - Y.T,) 



= — j Y c (ber' ma ber' mb + bei' ma bei' mb) , (93) 



and __ C(YJc _ Yc t 6 ) + D(Y C S*- Y 6 S C ) 



= — Y c (ber' ma bei' m6 — bei' ma ber' m&). (94) 



From these equations C and D are easily found, and then 

 A and B can be found from (91) and (92). 



The above formulae give the complete solution of the 

 problem when the applied wave is sine-shaped and the 

 capacity and leakage effects are neglected. 



If the applied wave be not sine-shaped, but if it be a 

 periodic function of the time, it may be expanded in a series 

 of sines by Fourier's theorem ; and hence we could write 

 down the solution without difficulty, but it would be very 

 cumbrous. 



