3-B] 



PREDETERMINATION. 



Fig. 4 is drawn by first constructing to scale the triangle ABC, 

 with two sides equal to RI and XI, respectively, and then laying 



Ccs0-1 



Coi 6 -.866 



Cos 6 = 0.5 



FIG. 4. Electromotive force diagram, at power factor 0.866; current lagging 

 3o a behind terminal voltage. 



off OB at an angle with BC, so that cos 6 equals the power 

 factor of the load. 



(a) Given ET = 575, we find by construction EQ = 726 ; or, 

 by computation 



Eo = 



cos 



( T sin 



= V(575 X .866 + 74) 2 + (575 X -5 + 233.9)* = 726. 



The regulation is 26.3 per cent. With inductive load, the 

 regulation is always poorer than with non-inductive load. The 

 dotted quadrant indicates the locus of the point O for different 

 power factors. 



(b) Given EQ and power factor; required the terminal voltage 

 ET. Lay off a line in the direction BO making the proper angle 6. 



