50 PROCEEDINGS OF THE AMERICAN ACADEMY. 



// • i o cosh S c i / - -n 



p =ClSinh ^'cosh^ ohms (55) 



= 1,000 cosh 2.621,818 



sinh 0.621,818 

 = 24,553.55 ohms 



and y" = l//>" = 0.407,273 X 10~ 4 mho. 



The conductance of the (1 leak at A is, as in (50), 



g" A = G b a — y" 



= 9.487,693 X 10~ 4 mho. 



Equivalent T. 



To compute the equivalent T of the composite line AD (Figure 7), 

 free the line at one end, say D 3 , and find the sending-end resistance at 

 C in this condition. It is, by (4), (36), and (42), 



R/c = ~2 coth 2 



= 2,000 coth 1.5 = 2,209.59 ohms. 



The line-angle changes abruptly at the junction BC from 8 C = 1.5 to 

 h B = 0.487,935 hyp, by the condition 



B B = coth" 1 f z * c °y> \ = coth' 1 f-A hyps (56) 



= coth" 1 2.209,59 = 0.487,935 hyp. 



The line-angle at the end A 8 is thus #1 + 8* = 2.487,935 hyps. 

 The sending-end resistance at A 3 is finally, by (4), 



R/A = zi coth 8^ ohms (57) 



= 1,000 coth 2.487,935 = 1,013.897 ohms. 

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



^ = * lSinh V^ mhos (58) 



= 0.001 X sinh 2.487,935 X — ^ ' <t> ' . = 12.537,3 X 10~ 8 mho. 



cosh 0.487,935 



The resistance of the leak OG' is, therefore, R' = l/(f = 79.762 ohms. 

 The resistance of the AO branch is then, by (18) and (44), 



