METHODS OF TESTING ALTERNATORS 



275 



open circuit; ad then represents the reactive current Id- The value 

 of the reactive current delivered or received by the motor may then 

 be deduced from the V-curve for all values of excitation. If reference 

 is made to the characteristic in Fig. 24, on which OC represents the 

 normal e.m.f. at the terminals, the knowledge of Aa gives for each 

 value of the current Id the corresponding value of the total lost ampere- 

 turns OF=Aa. A curve of these ampere-turns may then be drawn 

 as a function of Id, and the total fall of potential CD thus deduced 

 from the chart for every value of the armature current. This method 

 does not separate the reaction into two parts according to theory, 

 but it gives exactly the required result which makes it possible to 

 calculate the fall of potential for each value of the current and of 

 the phase with respect to the mean e.m.f. CD which the armature 

 designed to supply. 



To complete this indication it is sufficient to know the transverse 

 reaction. This may be determined readly enough by the same 

 experiment if care be taken to measure the phase angle, </>, of the cur- 

 rent with respect to U, the mean e.m.f. at the terminals. Let us, then, 

 draw (Fig. 25) the synchronous motor diagram with OA to represent 

 the internal induced e.m.f., E, and OB the mean voltage, U, at 

 the terminals; then the geometrical difference, AB, will represent 

 the fall of potential due to impedance, which may be resolved into 

 two vectors, one of which (BH) represents the ohmic drop, and the 

 other (HA) the fall of potential wL'I calculated as a function of 

 the transverse reaction. Knowledge of the angle </> and the current 

 strength 7 enables the point H, and also the direction-line HA, to 

 be determined. Knowing, from the experiments set forth below, 

 the angle of lag, </>, of the internal e.m.f., E, (whose direction, OA, 

 can thus be drawn) it is only necessary to take the point of intersection 

 of OA with the 'direction-line HA to locate the point A, and, conse- 

 quently, to be able to complete the triangle OH A. The length of 



