60 TRANSMISSION LINE FORMULAS 



and let its phase be denoted by the angle 6, Fig. n, where 



tan0 = =p 

 A 



T> 



and, therefore, sin 6 = 



and cos 6 



Similarly, let the current at the supply end be C + JD, 

 at a phase angle <, where 



tan <f> = 

 and, therefore sin < = 



VC 2 

 C 



and cos 4> = 



The watts at the supply end are equal to the current, 

 multiplied by the voltage, multiplied by the power factor; 

 that is, 



Watts = absolute value of I a X absolute value of E a 

 X cos (0 - 0) 



= VC 2 + D 2 X^A 2 + B 2 X (cos cos 0+sin sin 0) 



= VC 2 + D 2 X VA 2 +B 2 

 ( _ A _ C 



' 



BD, 



as in equation n, Table III. 



The quadrature volt-amperes, or reactive power, are 

 given by the following equation: 



