82 SYNCHRONOUS ALTERNATORS. . [Exp. 



18. Knowing the values of resistance drop RI, and reactance 

 drop XI, we may have either of two problems to solve : 



(a) Given the terminal voltage ET, to determine the open- 

 circuit voltage EQ; or, 



(b) Given the open-circuit voltage EQ, to determine the termi- 

 nal voltage ET. 



The following examples will make clear the solution of either 

 problem. 



(a) Given T = 575; #7 = 7.4; XI = 233.9. Required to 

 find EQ. 



Lay off to scale the values of ET, RI and XI, as in Fig. 3 ; by 

 construction EQ is found to be 627. Designating the total in- 

 phase voltage by Ep, and the quadrature voltage by EQ; we have, 

 by computation, 



EQ = Vp 2 + Q 2 = V(x + #/) 2 + (XI) 



2 = 627. 



The regulation is 9 per cent., EQ being 9 per cent, greater 

 than ET. 



(b) Given = 627; RI ==7.4; XI = 233.9. Required to 

 find ET. 



Lay off RI and XI to scale, as in Fig. 3. From A as a center 

 and radius EQ = 627, strike an arc cutting at O the line OB, 

 drawn as a continuation of BC. By this construction, ET is 

 found to be 575 ; by computation 



At unity power factor, it is seen that the terminal voltage is 

 always less than the generated or no-load voltage. 



19. Power Factor Less than Unity, Current Lagging. With 

 an inductive load, the power factor of the load is less than unity 

 and the current, accordingly, lags behind the terminal electro- 

 motive force. This is shown in Fig. 4 in which the current I 

 lags behind the terminal electromotive force ET by an angle = 

 30, the power factor of the load, in this case, being cos 30 = 

 0.866. 



