288 SYNCHRONOUS MOTORS. 



losses and in doing mechanical work owing to the inter- 

 actions between the two machines. 



The idle component of the resultant voltage is numeri- 

 cally equal to the product of current and the reactance of 

 the circuit. Several methods of determining the approxi- 

 mate value of the armature reactance have been given ; it is 

 therefore usually possible to determine the amount of 

 reactance voltage or idle voltage (= 2imLC). 



Describe a semi-circle on E r) and from E T describe an 

 arc with radius equal to 2 T n L C, the reactance voltage, to 

 cut the semi-circle in P. Then F is the energy com- 

 ponent of the resultant electromotive force in the circuit, 

 and is in phase with the current. 



The power given to the circuit by either machine = 

 current x voltage of machine x cos 0, where # is the angle 

 enclosed between the line OP and the voltage of the 

 machine. For the lagging machine, the value of cos will 

 be negative, hence the power given by this machine to the 

 circuit is negative. In other words, this machine absorbs 

 power and is driven by the other.* 



The difference between the positive and negative power 

 is seen at once by observing whether the projection of the 

 current vector on the voltage vector would have its direc- 

 tion in the same or the opposite direction as the voltage. 

 Thus in Fig. 136 the projection of the current on the line 

 EZ is Ce. 2 , which is directed in the opposite direction from 

 to the voltage E. z itself. 



The power given to the circuit by the first machine is the 

 product of O C^ amperes and E l volts. The power taken 

 by the second machine is the product of Cf 2 amperes and 

 E., volts. In each case this is the product of the machine 

 voltage and that component of the current which is in phase 

 with it. 



The fact that, when one machine lags behind the other, 

 it receives a driving current, is the reason why it is possible 

 to run alternators in parallel. Whichever machine tends to 

 fall behind is at once accelerated by power supplied by the 

 other. Thus machines will run in parallel and resist any 

 tendency to fall out of step. 



* If there is any doubt as to the side of OEr upon which the semi-circle 

 should be drawn, this can always be decided by remembering that the circuit 

 is inductive, and consequently the -current must lag behind the resultant 

 voltage ET in phase. If the semi-circle in Fig. 136 had been drawn below 

 E T instead of above it, the current would apgarently lead. 



