42 SYNCHRONOUS MOTORS 



[To make the diagram complete we should draw two similar diagrams, turned 

 in different directions, one corresponding to the volts, the other to the amperes, 

 each with its own special scale. However, this would be of no advantage, and 

 we will, therefore, retain a single diagram with the two scales, remembering that 

 the origin of phases is OA l for the E.M.F.'s, and /l^V for the currents. We shall 

 indicate, on that diagram, the expressions for the segments of lines, as if they 

 were all E.M.F.'s; and it will be sufficient to divide them all by Z to have their 

 expressions for currents. 



For example, the segments OA l and O.4 2 will represent, respectively, by the 

 scale of amperes, the two currents 



E \ 



which each one of the alternators would produce, respectively, in the circuit, 

 if the other were stopped.] 



The line A\N will therefore be the origin of phases for the current 

 with respect to the E.M.F. E\. 



If we project A\A%, on the line A\N and on the line AA 2 per- 

 pendicular thereto, we will obtain, by the scale of amperes, two 

 current-components : 



(a) The watted current i=AiA, or the current which is in phase 

 with the E.M.F. 



(b) The wattless current j=A 2 A, cr the current which is in quad- 

 rature with the E.M.F. 1 



The phases of the two currents are both referred to A^N as origin. 



The work of Dobrowolsky has shown that this decomposition of the 

 current is very interesting from a practical standpoint; it is more so, 

 in general, than the decomposition of the E.M.F. already indicated 

 (page 41). 



The power furnished by the active E.M.F. may be deduced from 

 the watted current, thus: 



The power actually utilized may be deduced from this, in the 

 following manner: 



in which r 2 =the armature-resistance of the motor. 



These terms, "watted" and "wattless," are adopted here to conform to the 

 international language of manufacturers, although they are objectionable. 



