40B TRANSMISSION LINE FORMULAS 



B = 2670+ 14,570- 43-o Q 

 = 17,240 - 43.0 Q. 



Now the voltage at the substation end of the ist section 

 of the line is 110,000; that is, 



A +- - = 110,000; 



2A 



squaring both sides, A 2 + B 2 = 121 X io 8 . 

 This gives a quadratic equation in Q, 



2-754 <2 2 + 3*986 - 

 from which Q= 27.53 total amps.; 



Power factor = 2jI = 95.7%, 

 94.99 



and, as Q is positive, this is a lagging power factor. 



The power factor obtained by the hyperbolic formulas 



is 95.9%. 



Using the above value of Q, we obtain 



C = +82.78, 

 D = + 90.59. 



In-phase current at substation end of ist section 



= +95-H. 

 Reactive current at substation end of ist section 



= ~ 77.53. 

 2nd section of line, / = 200 miles. 



