KENNELLY. — EQUIVALENT CIRCUITS OF COMPOSITE LINES. 67 



the equivalent l~l of the same line unloaded, by preceding formulas, and 

 then to the proper leak of the l~l add the terminal load leak, numerically 

 in the D. C. case, vectorially in an A. C. case. 



Equivalent T. First Method. 



To compute the equivalent T, free one end of the line, say F, as at 

 F 3 (Figure 13), and develop the line-angles towards A. We commence 

 with 



A . ii^l , B c ?'~ hir .D 



«■ z. = 11100- ** ** 



^= Zirvo' 



E •SSoSF 



w *« 



Ah 



e 1= ? 



Z,s 1000 



BC 



JpFfa > 9 '- 2 



0, = is 



r, = 2CTOT)" 



<0; O « 



^.jro« 



A"' 



44-247' 



0-226004 xio"%. 



A - T 988 0S"O >04-0-y' l T / 



G G G' 



Figure 13. Composite line of three sections with terminal leak load. 



S P =coth" 1 [tU- 



>m 



hyps (122) 



= 1.098,6 +j- hyps, 



where y is the admittance of the load in an A. C. case or conductance 

 of the load in the D. C. case (mhos). 

 The T-leak admittance is then 



, . , . coshS c coshSjj coshO , ,,__,. 



g = vi smh 8 A • — p-^- • — 7-5- • — r-«- mhos (123) 

 y * cosh8 B cosh8i> coshoV v 



= 41.066 X 10 -8 mho, 

 and the line-branch impedance AO follows at once from (87). 



