Telephone Circuits. 681 



It is easy to show that 



and 



r R 1 ?. 



*-*>*5- + '( l+ 2> 



and it follows that 



P = ( 1 + - 1 ) cosh mn -f ( r 2 -( 1 + £-) + £- - ) sinh mn. 

 V r 2 ' (. r\ 2r 3 / 2r 3 7iJ 



It may be well to work out one example. In the above- 

 mentioned cable r=lS, £ = 0055 x 10 -6 , s=0, 1 = 0; take 

 £=5000, 



n = s/r~k^i = -07036 (45°), - = '003909 (45°), 

 - = 255*8 (-45°). 



Let us place contrivances at m = 4*263 miles apart. I choose 

 this distance because mn = Q'X (45°), and this happens to be 

 given in Prof. Kennelly's table, which tells me that 



cosh mn = P0007 (2°. 35'), sinh mn = 0'29994 (45°. 51'). 



To imitate the continuous case given above we need an 

 inductance leak of 0*5925 henry per mile or L = '5925h-4-263 

 every 4*263 miles. Hence 



r z = Lqi = 694-7 i. 



i\ is a condenser of capacity K=, f -— ;a10~ 6 , so that as 



r, = r^-., ?•■ = — 971*13 i. Inserting these values in the 

 ] Ivy/' ' n 



above formula, I find 



P = -0*05384- -068078i 

 and 



C = 1*057 (183°. 54') or 0'946 (-183°. 54'). 



That is, there is an attenuation or increase of 5*5 per cent, 

 in 4*263 miles or 1*3 per cent, per mile. Observe how 

 different this is from the effect of the continuous distribution. 



As another case, a most important one, consider the 

 standard subterranean cable r = 88, k = *05 X 10~ 6 , s = 0, 

 1 = 0. 



I. Without contrivances It =g= y/rkg/'A. If g=5000 

 (a frequency of bOO). h = '1049 = £/. There is an attenuation 



