TRANSMISSION LINE PROBLEMS OI 



Reactive power 



= absolute value of 7,X absolute value of jE,Xsin (6<f>). 

 = VC 2 + D 2 X VA 2 + B 2 (sin 6 cos <t> - cos B sin 

 = VC 2 H- D 2 X 



( B X 



I VA*+B* 

 = BC - AD. 



When the expression BC AD has a positive value, the 

 current at the supply end is lagging behind the supply 

 voltage, and when the expression has a negative value, the 

 current leads the voltage in phase. 



We can now obtain the in-phase component of current, 

 which is equal to watts divided by voltage (equations 15 

 and 1 6), and in the same way the quadrature component 

 of current, which is equal to reactive power divided by 

 voltage (equations 17 and 18). The power factor at the 

 supply end is equal to watts divided by volt-amperes 

 (equations 13 and 14). Since the power supplied is known, 

 being AC + BD, and the power delivered at the receiver is 

 also known, being equal to EP, their difference represents 

 the loss of power in the line due to resistance of the con- 

 ductors, leakage over the insulators and corona loss. 



The equations in F, G, M and N are quite similar to the 

 above equations in their derivation, and they give the solu- 

 tions of similar problems when conditions are given at the 

 supply end of the line. 



