140 ALTERNATING CURRENTS 



Example. Determine E' for a load in which the power-factor is 0.7, 

 current lagging, using the constants of the example on page 138. 



The rating of an alternator, as has already been pointed out, depends 

 on the current or kilovolt-amperes rather than the kilowatts. Therefore, 

 the current rating of the generator will remain unchanged, although the 

 kilowatts in this problem are reduced to 0.7 of their former value. 



cos e = 0.70 IR = 4.37 volts as before. 



e = 45.6 



sin 6 = 0.714 IX = 19.1 volts as before. 



E' = V(220 X 0.70 + 4.4) 2 + (220 X 0.714 + 19.1) 2 = 237 volts. Ans. 



It is to be noted that the induced emf. is now higher than before, 

 although the value of the impedance drop itself is the same. 

 Therefore, for a fixed value of induced emf., the terminal volts 

 become less with increasing lag of the current, even though the 

 value of the current remains unchanged. This is due to the angle 

 at which the impedance drop subtracts from the induced emf. 

 It would be expected, therefore, that the regulation of an alter- 

 nator would be poorer for lagging current. 



At unity power-factor, the armature resistance drop is the 

 important factor in determinng the value of E'. With a lagging 

 current, the resistance drop plays but a small part and the 



armature reactance drop becomes the 

 important factor. 



Leading Current. Figure 146 shows 

 the alternator vector diagram when 

 the current leads the terminal voltage 

 by an angle 6. As the current changes 

 FIG. 146. Alternator vec- its phase relation with respect to the 



tor diagram for power-factor voltage y the i mpe dance triangle 

 cos 0, leading current. . 



swings with the current in a counter- 



clockwise direction about the end of V. E r is found in the same 

 manner as in Fig. 145. The voltage drop IR, parallel to the 

 current, is projected on the current vector. 



Oa = V cos 6 

 ab = IR 



aV = be = V sin 

 cd = IX 



E' = V06 2 + bd 2 = V(0a + a6) 2 + (be - cd)* = 



V(V cos 6 + IR) 2 + (V sin 6 - IX)* (40) 



