146 THE MAGNETIC CIRCUIT [ART. 47 



When t and e are given, the vector E is easily found if the 

 resistance and the reactance of the armature winding are known. 

 The required net excitation, M n , is then taken from the no-load 

 saturation curve of the machine, and M a is figured out from eq. 

 (64). Then the required field ampere-turns, Mf, are found from 

 the diagram, either graphically or analytically. 



The diagram shown in Fig. 37 is known as the Potier diagram. 

 Strictly speaking, it is correct only for machines with non-salient 

 poles, but as an approximate semi-empirical method it is some- 

 times used for machines with projecting poles, in place of the more 

 correct diagram shown in Fig. 40. Fig. 37 represents the condi- 

 tions in the case of a generator with lagging currents. When the 

 current is leading the vector i is drawn to the left of the vector e, 

 with the corresponding changes in the other vectors. 



A similar diagram for a synchronous motor which draws a 

 leading current from the line is shown in Fig. 38. The vector e' 

 represents the line voltage, and e is the equal and opposite voltage 

 which is the terminal voltage of the machine considered as a gen- 

 erator. The rest of the diagram is the same as in Fig. 37. A lead- 

 ing current with respect to the line voltage e r is a lagging current 

 with respect to the generator terminal voltage e, so that the field 

 is weakened by the armature reaction in both cases (M n < M f in 

 both figures). The energy component i\ of the current is reversed 

 in the motor, therefore the field is shifted in the opposite direc- 

 tion; M n leads Mf in the motor diagram and lags behind it in the 

 generator diagram. The case of a synchronous motor with a 

 lagging current can be easily analyzed by analogy with the above- 

 described cases. 



In practice, it is usually preferred to represent the relations 

 shown in Figs. 37 and 38 analytically, rather than to actually con- 

 struct a diagram. The following relations hold for both the gen- 

 erator and the motor. Projecting all the sides of the polygon 

 OABD on the direction e and on the direction perpendicular to 

 and leading e by 90 degrees, we have 



E cos(f> z =e+ir cos<j)+ix sin (j>, . . . (71) 

 E sin <t> g = ix cos <f> ir sin <, .... (72) 



where <j) t is the angle between the vectors e and E, counted positive 

 when E leads e, as in Fig. 37. The subscript z suggests that the 



