332 



ELECTRICAL MEASUREMENTS 



Reading of No. 2 = VI cos (30 - 6) = VI (sin 6 sin 30 + 

 cos 6 cos 30). 



Sum of No. 1 and No. 2 = VI (2 cos 30 cos 0) 



= VI A/3 cos 8 (29) 



which is the power applied to the circuit. 



As the lag angle, 6, increases, the reading of No. 1, which is 

 the smaller of the two, decreases and will become zero when 

 = 60 (corresponding to a power factor of 0.5), for then /i 

 and Vu are in quadrature. If 6 > 60, No. 1 will reverse and 



FIG. 196. Vector diagram for Fig. 195. 



its reading is taken as negative when adding the readings of 

 No. 1 and No. 2 to obtain the power. 



It will be noticed that the phase difference of V& and 7 3 is 

 the same as that of Vi 3 and /i and as the maximum values of the 

 current and voltage are the same in both cases, one wattmeter 

 will suffice for measurements on a balanced load. For example, 

 two readings may be taken with wattmeter No. 1, the first with 

 the potential terminals connected between mains 1 and 2, the 

 second with the potential terminals connected between mains 1 



