KENNELLY. — EQUIVALENT CIRCUITS OF COMPOSITE LINES. 45 



Then the architrave resistance AD of the composite l~I will be: 



P " = z x sinh 8 A ohms (40) 



= 1,000 sinh 3 = 10,017.87 ohms. 

 y " = i/p" = 0.998,212,5 X 10" 4 mho. 



The conductance g" A of the leak at A is, by (20), 



9"a = Vx coth S A - y" mho (41) 



= 9.051,49 X 10" 4 mho. 



If we ground the composite line at A instead of at D, the angle sub- 

 tended by the whole line at D will be 8 D = Q x + 2 = &a- The archi- 

 trave resistance DA will be the same as that given in (40). The 

 sending-end resistance R gD and conductance G gD will be identical with 

 R gA and G gA respectively, by (38) and (39) ; so that the leak-conduct- 

 tance g" D at D will be identical with g" A by (41). This completes 

 the hyperbolic n, ADGG of the composite line. 



Equivalent T. 



To find the hyperbolic equivalent T of the composite line AD (Figure 

 6), free the line at one end, say D. Then the angle subtended by the 

 line at A will be, as before, 8 A — 6 X + 6 2 hyps. 



The sending-end resistance of the line at A will be, by (4), 



RfA = 2 coth S A ohms (42) 



= 1,000 coth 3 = 1,004.97 ohms. 



The conductance of the leak OG is, by (17), 



g' = y sinh S A mhos (43) 



= 0.001 sinh 3 = 10.017,87 X 10 -3 mhos. 



and its resistance is 



R' = \lg' = 99.821,25 ohms. 



The resistance of the AO branch is, then, by (18), 



P = R fA - R' ohms (44) 



= 1,004.97 — 99.821 = 905.149 ohms. 



Similarly, if we free the composite line at A, instead of at D, the angle 

 subtended by the line at D will be 8 D . As before, S D = 6. 2 + B x = 8 A 



