KENNELLY. — EQUIVALENT CIRCUITS OF COMPOSITE LINES. 55 



In order to complete the f~l, we ground the line at A 2 (Figure 8), and 

 develop the line-angles towards F 2 . The architrave resistance is then 



,, . , ., coshS/) cosh 8 B . , nnS 



P " = z a sinh 8 F x — r-^r X — r-^ ohms (80) 



cosh 8 B cosh 8 C 



= 44,247 ohms. 

 The sending-end resistance at F is 



B gF = c 8 tanh 8 r ohms (81) 



= 2,500 tanh 1.526,83 = 2,274.71 ohms. 

 Again, 



/', = l/B gP - 1/p" mhos. (82) 



Equivalent n . Second Method. 



An alternative method of arriving at the architrave resistance, 

 which we may call the second method, is by following (66) and (68). 

 Grounding at A 2 , we have 



sinh 8 D sinh 8 

 \ sinn v\ • 



and, grounding at F x , 



// • i i\ sum o D sum o„ . . . 



^^■"^•sa^-Shr; ohms ' (83) 



. , . sinhS c sinnS^ 

 P = c 3 sinh 3 • . , • . , - ohms (84) 



sinho B sinhS B 



= 44,247 ohms. 



Equivalent T. First Method. 



We proceed to compute the equivalent T of the composite line AF 

 in the same manner as the T in Figure 7. Free the end F 3 and develop 

 the line-angles towards A 3 . As before, 



8 D = coth" 1 (J^\ and 8 B = coth" 1 f^\ hyps. (85) 



The T leak conductance is then, following (58) and (75), 



, . , s cosh 8 C cosh 8, , 



" =2 " Smh8i -^h8,-SSs; mhos (86) 



= 0.001 sinh 2.519,86- COShl - 888 ' ft71 C ° sh °' 5 



cosh 0.519,860 cosh 0.388,071 

 = 19.2016 X 10" 3 mho 

 Bf = 52.079 ohms. 



