THE SYNCHROXOIS MOTOR 



311 



drop. Therefore, the current / is lending with respect to the 

 terminal voltage of the motor. It follows then that a current 

 which is lagging when a machine is considered from the point of 

 view of a generator, is leading when the same machine is con- 

 sidered from the point of view of a motor. In a generator a 

 lagging current weakens the field. Consequently, in a motor a 



"t must weaken the field. 

 Thi< is further illustrated 

 as follows: Figure 287 

 shows a motor coil moving 

 from left to right. When 

 its axis is in the position Y, 

 shown dotted, the coil sides 



Current 



a Maximu 



t Terminal Voltage 

 a Maximum 



FIG. 287. Demagnetizing effect of leading 

 current on the field of a synchronous motor. 



are under the centers of 

 the poles and the induced 

 emf. is a maximum. As 

 the terminal voltage is 

 substantially 180 from the induced emf., it also will be a 

 maximum at this instant, its direction being indicated in the 

 dotted coil. If the current leads this terminal voltage by 90 

 it will reach its maximum value one fourth of a cycle ahead of 

 the voltage, or at a time when the axis of the coil is in position 

 A. It will be observed that for this position of the axis, the 

 ampere-turns of the coil act in direct opposition to those of the 

 N-pole. Therefore, the effect of the leading current in the 



synchronous motor is to weaken 

 the field. In other words, the arma- 

 ture reaction tends to annul the effect 

 of the increased field current on over- 

 excUal 



The second efTeet is illustrated 

 by the vector diagram in Fig. 288. 

 V is the terminal voltage and / 

 ' lie armature enrrent leading V 

 The retfetanoe drop in the armature i> hud oil' 



17. 



V-s-Termlnol Volti 



288. Induced amiaf un- 

 voltage greater rh.ui terminal volt- 



Miiin:il vi> 



by an aimle . 



in phase with the current / and the / A" drop in the armature 

 is laid off at right angles to the current / and leading. in the 

 usual manner. The impedance drop // j> the \, ctor sum of //.' 

 an 1 /A. Th. \ to balance ihr back emf.. is 



fo md by subtracting // vectmially from K, just as in the shunt 



