KENNELLY. — EQUIVALENT CIRCUITS OF COMPOSITE LINES. 71 



Consequently, so far as concerns the first method, we should make no 

 change in the formula for p" on account of (a), but introduce the ratio 

 R gF /R ga for (b) and substitute z for c 8 on account of (c). Conse- 

 quently, following (77) with these changes, 



• , s cosh<V, coshS fl R gF 



" - * smh s " ■ ciss; ' ™sj c ■ t<, ohms (132) 



= 1,936.87 si n h 1.535,312 . °°~ »•«»■»» . «**■» . H°Mi 



cosh 1.035,31 cosh 0.592,95 1565.14 

 = 51,615 ohms. 



The H-leak is then found in the usual way. 



Equivalent TT. Second Method. 



Similarly, by reference to the Table, for changes in the p" formula 

 under the second method, we should introduce the ratio R gC /R gB for 

 (a), make no change for (b), but introduce the ratio R gH /R gG for (c). 

 Consequently, following (83) with these changes, 



n • T. /i sma ^d sinh 8 F R aC R nH . 



p" = z x smh X • . ,/ • . , / • jp • -^ ohms (133) 

 sinh b c smh b E R gB R gG 



= 51,615 ohms. 



The A-leak is then computed in the regular manner. 



If now w T e ground the line at the H-end, we obtain similarly, by the 



first method, 



„ . , . coshSc cosh 8 F coshO R g0 , 



p —z x sinh 8 A ■ — p.- • — r— ■ — — - .— g — ohms (134) 

 cosh 6^ cosh b D cosh b F R gF 



= 51,615 ohms, 



and by the second method, 



„ sinhS* sinh 8 C sinh 8 A R gB G gF , , 



P —^3 • - s • ■ i g • • i g - • -w~ • -ft ohms. (135) 



smh b F smh b D smh b B R gC G g0 ' 



Equivalent T. First Method. 

 Freeing the line at H, as at A 3 H (Figure 14), we have 



, . , s cosh8 c cosh8 s cosh8 G fC 



g =yi smhS A • — j-s- • — r-g- • — r^ • 77- mhos (136) 



cosh b B cosh b D cosh b F G fB v ' 



aaai -u^oi cosh 2.233,54 cosh 1.049,31 

 = 0.001 • sinh 2.504,81 • — . „ „ ',. ■ — ^— — zrzhr. ■ 

 cosh 0.504,81 cosh 0.733,54 



1 2,146.46 



cosh 0.549,31 ' 2,046.46 

 = 28.851,7 X 10" 3 mho, 



