302 ELECTRICAL ENGINEERING 



primary current 



/i = V(6.36) 2 + (0.07) 2 = 6.36 amperes, 

 exciting current 



7 = CO.ll) 2 + (0.07) 2 = 0.13 amperes, 

 primary induced e.m.f. 



E' = V(16,375) 2 + (12,500) 2 = 20,600 volts. 



The primary impressed e.m.f. is inclined to the axis of coordi- 

 nates at an angle 0', where 



tan tf = = 0.7785 and therefore 0' = 3750'. 



16,699 



The primary current is inclined to the axis at an angle 0", where 

 tan 0" = - Z = _ o.Oll and therefore 0" = - 6 17'. 



The angle of phase difference between the primary current and 

 the primary impressed e.m.f. is 0i = 6' 6" = 44 7' and the 

 primary power factor is cos 0i = cos 44 7' = 0.722 or 72.2 per 

 cent. 



The regulation of the transformer under these conditions of 



,. . 21,160 - 20,000 v 

 loading is on noo " ^ er c = ^ er cen ^- 



Primary copper loss is 



JiVi = (6.36) 2 X 50 = 2020 watts. 

 Secondary copper loss is 



7 2 V 2 = (62.5) 2 X 0.6 = 2340 watts. 

 Iron loss is 



ESgo = (20,600) 2 X 2 X 10' 6 = 850 watts. 



The efficiency is therefore 



output 



TQ = - -100 per cent 



output + losses 



1 nn 



' X 10 Per Cent = 95 per Cent ' 



100,000+ 5210 



77. If the transformer in example I, with 2000 volts impressed, 

 is used as a step-up transformer to charge a cable system of 

 negligible resistance and supplies 5 amperes, determine the second- 

 ary terminal e.m.f. 



