64 ^ELEMENTS OF ELECTRICAL ENGINEERING. 



so that 



Av. d = EI cos Av.(sinW) EI sin 6 Av. (sin cot cos cot) 



But the average value of sin 2 o>/ is J^, and the average value of 

 sin cot cos cot is zero. Therefore 



EI 



Av. ei = cos 6 

 2 



or, using equations (5) and (6), we have : 

 Av. ei 7 cos 6 



Power factor. By comparing equation (7) with equation (2), 

 it is evident that the power factor of a receiving circuit is equal to 

 the cosine of the angle of phase difference between the harmonic 

 electromotive force which acts on the circuit and the harmonic 

 current which is maintained in the circuit. 



Power-component and wattless-component of electromotive force. 



Let the clock diagram, Fig. 55, represent a given harmonic 

 electromotive force E and the harmonic current 7 which is 

 maintained by E, let the line 7 be chosen as the reference 

 axis, and let us consider the two components of E, namely, 

 ECOS& and EsrnO, as shown. The component, Ecos6, which 

 is parallel to 7 is called the power-component of E, and the 

 component, E sin 0, which is at right angles to 7 is called the 

 wattless-component of E. 



So-called power-component and wattless-component of current. 



Let the clock diagram, Fig. 56, represent a given harmonic elec- 

 tromotive force, E, and the har- 

 monic current, 7, which is main- 

 tained by E, let the line E be 

 chosen as the reference axis, and 

 let us consider the two compon- 



Fig. 56. 



ents of 7, namely, 7 cos 6 and 7 



sin 0, as shown. The component, 7 cos 6, which is parallel to 

 E is sometimes called the power component of the current, and 



