5-B] TEST BY LOSSES. I 7 I 



voltage E r Fig. 9 would then be drawn as Fig. 4, Exp. 3-6, in 

 which the line EC makes an angle with the line OB. 



Resolve r into an in-phase component, r cos 0, and a quadrature 

 component, r sin ', resolve x into an in-phase component, x sin 0, and 

 a quadrature component, x cos 0. 



In-phase drop = p = r cos -f- # sin 0. 



Quadrature drop = q = x cos r sin 0. 



+ 



Regulation = 



/>-{- (cf -- 200) , approximately. 

 For practical purposes 



Regulation = p = r cos -f * sin 0. 

 43. Leading Current. Similarly, for a leading current, 



p = r cos x sin 0. 

 q = x cos + r sin 0. 



Regulation = 



-1-200), approximately. 



APPENDIX I. 

 MISCELLANEOUS NOTES. 



44. Selection of Instruments. In selecting instruments, the neces- 

 sary range can be approximately told by assuming some reasonable 

 value for efficiency or losses. Thus, in a 2 K.W. transformer, let 

 us assume that the efficiency is 95 per cent, and that the iron losses 

 and copper losses are equal. Assume the power factor in the open- 

 pircuit and short-circuit tests to be 0.66. 



In the open-circuit test the core loss will be 50 watts ; the ammeter 

 and current coil of the wattmeter must, accordingly, carry a current 

 of 1.5 amperes, if the test is made on a 50-volt coil; 0.75 amp. for a 

 loo-volt coil; 0.375 amp. f r a 2OO-volt coil, etc. 



In the short-circuit test, the copper loss will be 50 watts; if the 



