5-B] TEST BY LOSSES. 169 



factor (cos 6) less than unity and the current is lagging, the 

 regulation is practically* as follows: 



Per cent, regulation = r cos + x sin 0. 

 For example, let cos = 0.866; sin = 0.5; 



In-phase resistance drop = r cos # 2.57 X 0.866 =2.23 per cent. 

 In-phase reactance drop = -rsin 6= 1.76 X 0.500=088 per cent. 



Regulation = 3.11 per cent. 



38. For Leading Current. When the load has a power 

 factor (cos 6) less than unity and the current is leading, the 

 regulation is practicallyf 



Per cent, regulation = r cos 6 x sin 6. 

 For example, let cos = 0.866; sin = 0.5; 



In-phase resistance drop = r cos = 2.57X0.866=2.23 per cent. 

 In-phase reactance drop = x sin 0= 1.76 X 0.500=0.88 per cent. 



Regulation = 1.35 per cent. 



39. Proof. Fig. 9 shows a simple graphical method for obtain- 

 ing regulation at non-inductive load. (Compare also Fig. n, Exp. 

 5~C, in which the same lettering is used, and Fig. 3, Exp. 3-8.) 



Referring to Fig. 9, lay off AL equal to the secondary full-load 

 voltage 2 =ioo per cent. (A scale of volts could be used, if 



FIG. 9. Method for determining regulation ; r = per cent, resistance drop ; 

 4r = per cent, reactance drop; regulation = E , = 2.585. (This Fig. is not 

 drawn to scale.) 



*(37a). For greater accuracy, a term for effective quadrature drop 

 (g 2 H-2oo), should be added, 42. In the present example this term is 

 only .0003, making the regulation 3.1103. In any ordinary case, on lagging 

 current, this term can be neglected. 



t (3&0. For greater accuracy, a term q 2 -i- 200 should be added, 43; 

 in the present example, this term equals .039. 



