THE TRANSFORMER. 



223 



The current a in service main I is the vector difference 7 l 7 3 , and the current 

 b in service main 3 is the vector difference 7 2 7 3 , as shown in Fig. 188. The 

 arrows in Fig. 187 represent the directions which are chosen as positive. Of course 

 the currents in the secondary coils a and b are precisely the same as the currents in 

 mains I and 3 respectively. The voltage E^ which is generated in coil a develops 

 an amount of power equal to l times a times the cosine of the phase difference 

 between l and a, or E^a cos (0 -\- 30 ) ; and the voltage E^ which is gener- 



Fig. 188. 



ated in coil b develops an amount of power equal to E^b cos (6 30). Using, 

 therefore, E for the common numerical value of E l and E v and a for the common 

 numerical value of a and d, we have, as an expression for the total power deliv- 

 ered 



a\_cos (0 + 30) + cos (6 30)] 

 or 



za cos 6 cos 30 



Example. Consider three AA-connected loo-kilowatt transformers delivering 

 three-phase currents to balanced receiving circuits. These transformers can deliver 

 300 kilowatts if the receiving circuits are non-inductive or in general they can de- 

 liver 300 cos 6 kilowatts. If one transformer is put out of service, the two remaining 

 ing transformers will operate at full load, not on 200 cos 6 kilowatts, but on 

 o.86X 2 cos# kilowatts; that is, the two transformers have a full-load capacity 

 of only 0.577 of the full -load rating of the three transformers. 



111. Polyphase transformers. It is possible to combine the 

 magnetic circuits of transformers which are used for polyphase 

 transformation. Thus Fig. 189 shows what may be called a 

 two-phase transformer. It consists of two similar transformers 



