678 Prof. J. Perry on 



Thus the common expression s + kqi becomes 



For preliminary calculations it is generally sufficiently 

 correct to assume R/L 2 ^ 2 to be zero. 



We may now consider that the telephone line has, per mile, 



a resistance r. an inductance l'=l— - — ^, a leakance 5 = 0, 



1 . k tf 



and a capacity k = k — . 2 - It will be found desirable in 



almost all cases to have V so large that we may take 



r f~l$ . 



Assuming these conditions to exist we have, very nearly, 



c =/y/|^-i"^^'sin (qt-qx</FF). 



For a line of, say, L miles in length the amplitude of the 

 arriving current is a maximum if a/ j r ? = JrL. In fact the 



arriving current is the sending current divided by e. It will 

 be found that in practice we never have so small a value of h 

 as this gives, and we generally try to make h as small as 

 possible. 



We can approximate to the above conditions by inserting 

 contrivances at equal distances in the line, I will take a 

 few examples. I prefer just now to consider a cable in which 

 I have recently taken an interest; r = 18 ohms per mile, 

 £=•055 X 10" 6 farad per mile, 5 = 0, 1 — 0. I take q to be 

 5000 (that is, a frequency of 800). 



1. This cable with no contrivances, 



h = g = V%/2 ='04975. 



So that if x is 1 mile, e~ hx = 0*9513. That is, there is an 

 attenuation of 5 per cent per mile. 



2. Let Z' = Z = 0. That is, there is neither inductance nor 

 capacity inserted in the line. Now let there be an inductance 

 leak to earth which causes k! to be 0*01375 x 10~ 6 . Then 



h = q = y/k'qr/% = '024875, 



