KENNELLY. — EQUIVALENT CIRCUITS OF COMPOSITE LINES. 59 



s , , _, / 100 + 964.026 \ „, QOQK i 

 8 C = tanh J I ^^ \= 0.592,95 hyp. 



Formulas (80), (81), and (82) then apply without change. 



Equivalent l~l . Second Method. 



The alternative method for computing the architrave resistance of 

 the line when grounded at A 2 , and developed in angles, is 



n ■ uo sinhS^ sinhS^ R g0 , . . 



P =zx sinh $i • . , • . , g • -^- ohms, (95) 



sum o c sinh b E li gB 



and when grounded at Fi it is 



n -1/3 sinhSp sinh 3^ R gB . 



p = 2 8 sinh 3 • • i g • ■ , « • -rr- ohms. (96) 



sinh dp smn b B H gG 



That is, the effect of the load is to increase the architrave impedance 

 in the ratio of the change of sending-end impedance across the load. 

 In (95) this ratio is 1,064.026/964.026, and in (96) it is 2,047.385/ 

 1,947.385. 



Equivalent T. First Method. 



To compute the equivalent T of the loaded line in Figure 9, free the 

 line at one end, as at F 8 , and develop the line-angles towards A 3 , as in 

 (85). The only change effected by the load is in the angles at and 

 beyond B. The sending-end impedance at C is 



B fC = 2 2 coth 8 C ohms (97) 



= 2,000 coth 1.888,071 = 2,093.82 ohms. 



The sending-end impedance at B is, therefore, 



B/b = o- + z 2 coth S a ohms (98) 



= 100 + 2,093.82 = 2,193.82 ohms. (98) 



The new line-angle at B is then 



Smooth- 1 3?) hyps (99) 



= coth"^^^) = 0.492,025 hyp. 



