KENNELLY. — EQUIVALENT CIRCUITS OF COMPOSITE LINES. 51 



P=R fA -R' ohms (59) 



= 1,013.897 — 79.762 = 934.135 ohms. 



In order to complete the equivalent T of the line AD, we must repeat 

 the above process from the opposite end, by freeing the end A, as shown 

 at A 4 D 4 (Figure 7). The line-angle at B is 8 B = 2.0. Across the junc- 

 tion BC this angle changes suddenly to 



S c = coth-(*^) hyps (60) 



= coth" 1 ( 1Q 2 8 q 3 15 ) = coth" 1 0.518,657,5. 



In order to avoid an impossible operation, apply formula (52) 

 8 C — j^ = tanbr 1 0.518,657,5 = 0.574,50 hyp 

 8 C = 0.574,5 +i^ hyps. 

 The line-angle at the end D 4 is thus 6 2 + 8 C = 2.074,5 +j- hyps. 

 The sending-end resistance at D 4 is finally, by (4) and (57), 

 R/d = *2 coth 8 D ohms (61) 



= 2,000 coth ( 2.074,5 +j^j = 2,000 tanh 2.074,5 = 1,937.873 



^ ' ohms. 



The conductance of the leak OG' is, therefore, by (43) and (58), 



. • i o coshS B . , . 



g =y 2 smh 8 D • -^j- mhos (62) 



= o.ooi sinh (2.074,5 +/§) cosh2 -° ^ 



V ' cosh (o.574,5 +>M 



cosh 2 

 = 0.001 cosh 2.074,5 • . , ', e = 12.537,3 X 10~ 8 mho. 

 sinh 0.574,5 



The resistance of the leak OG' is, therefore, R' = \/g' = 79.762 ohms. 

 The resistance of the DO branch is then, by (18) and (59), 



p ' = B fD - R' ohms (63) 



= 1,937.873 - 79.762 = 1,858.111 ohms. 



This completes the T of the composite line. 



