534 ELECTRICAL MEASUREMENTS 



rest, the average turning moment on coil A must equal that on 

 coil B, so 



rK IV~\ 1 C T 



- J sin D ^ I sin (cot - 0) sin (ut)dt = 



r V J~\7~i 1 /" 



j* M jcos D ^ I sin (J - 0) sin (orf - 90) <# 



r*^ri sin D co , , . r^zi cos D sin ,. 



L R J L Lo> J 



If, by the construction of the apparatu? 



KA KB ( . 



-TL'-LZ (a) 



then 



tan D = tan B 

 and 



D = 6. 



That is, the movable system turns through an angle equal to the 

 power-factor angle of the load. 



The assumptions made in order to obtain this result are that 

 the frequency is constant, that the coils A and B are small com- 

 pared with Fj that the planes of coils A and B are 90 apart in 

 space, that the coils are traversed by currents which are 90 

 apart in time phase and that the current in coil A is in phase 

 with the line voltage V. In practice, these current relations can- 

 not be attained ; the lag in the circuit B can never be exactly 90. 

 Nevertheless, by a proper adjustment of the angle between the 

 coils, the instrument can be made to read correctly. 



To investigate this matter, suppose the current in coil B lags 

 A behind V and that the mechanical angle between the coils is 

 ft instead of 90. 



Assuming that the crossed coils are alike and have the same 

 number of ampere turns, 



sin D cos B = sin (D + ft) cos (B A). 



The fiducial point on the scale corresponds to the reading when 

 the power factor of the load is unity; in that case the deflection, 

 Do, will be given by 



cot Do = - cot ft. 



sin ft cos A 



