KENNELLY. — EQUIVALENT CIRCUITS OF COMPOSITE LINES. 53 



Single Lines Equivalent to a Dissymmetrical n or T. 



It is evident that formulas (21) to (24) apply only to a symmetrical 

 n or T. Moreover, it may be seen that no single smooth and uniform 

 line can correspond to a dissymmetrical n or T. This means that, in 

 general, no single smooth and uniform line can be the counterpart of a 

 composite line having sections of different surge-resistance. But if we 

 reduce a dissymmetrical fl to a symmetrical n and a terminal leak, we 

 may apply equations (23) and (24) to transform the symmetrical n into 

 an equivalent single line. It follows that any composite line may be 

 resolved into one and only one uniform smooth line of the same length 

 with a leak permanently applied to one end ; or to an infinitude of such 

 single uniform smooth lines having a leak at each end. 



Similarly, the T of a composite line may be reduced to a symmetrical 

 T plus a line-impedance at one end. By the use of equations (21) and 

 (22), we may substitute a single smooth uniform line for the symmetri- 

 cal T. Consequently, any composite line may be resolved into one and 

 only one uniform smooth line of the same length with a line-impedance 

 at one end ; or, to an infinitude of such single uniform smooth lines 

 having a line-impedance at each end. 



Composite Line with Three Sections of Different Surge- 

 Impedances. 



A three-section composite line is indicated in Figure 8. 



AB has a length L x of 100 km. 

 CD " " U of 300 km. 

 EF " " L 3 of 50 km. 



The respective linear constants are 



r\ — 20 ohms/km. ; r 2 = 10 ohms/km. ; r 3 = 25 ohms/km. 

 g x = 20 X 10~ 6 mho/km. ; g 2 = 2.5 X 10~ 6 mho/km. ; 



g 3 = 4 X 10~ 6 mho/km. 

 a x = 0.02 hyp /km. ; a 2 = 0.005 hyp/km. ; a 3 = 0.01 hyp/km. 

 X = 2 hyps ; 6 2 = 1.5 hyps ; S = 0.5 hyp. 

 zi = 1000 ohms ; z 2 = 2000 ohms ; z 3 = 2500 ohms. 



Equivalent (1. First Method. 



We proceed to compute the equivalent l"l of the composite line AF 

 in the same manner as in connection with Figure 7. Ground the end 

 Fi and develop the line-angles towards A v As before, 



( Ke 



and 8 B = tanh" 1 ( ^\ hyps. (76) 



