292 ELEMENTS OF ELECTRICAL ENGINEERING. 



Choose the magnetizing cur- 

 rent OM as a reference axis. 

 Lay off the line OA representing 

 the value of the electromotive 

 force A[= M(r^ +jx l )~\, and lay 

 offAtf equal to M(R'+jX\ zero- 

 load value of M being used in 

 each case. Then OE' is the zero- 

 load position of the supply voltage 

 of one of the stator winding's. 

 Draw about O a circle of radius 

 E' , then a line drawn from to 

 any point P of this circle is a possible position of the supply- volt- 

 age vector with the motor running under load. 



Evidently the voltage A across ab, Fig. 250, decreases with 

 increase of motor load (increase of current through R f and X) 

 and of course any decrease of A produces a proportional de- 

 crease of the magnetizing current and of the voltage drop in R f 

 and X due to the magnetizing current, so that, when the voltage 

 across ab, Fig. 250, is reduced to OA f , Fig. 252, the voltage 

 drop in R f and X due to the reduced magnetizing current is 

 A'c which is parallel to AE' '. 



Now the total primary voltage (of one phase of course) OP 

 is equal to the vector sum of OA' y A'c, and the voltage 

 drop I'(R f + jX) in R' and X due to the load current ; but 

 P(R* +jX) is in the direction Ob which is ahead of /' 

 or ahead of A, since A and I 1 are in phase with each 

 other. Therefore, having drawn the line Ob, choose any point 

 P, draw PC parallel to Ob, draw Pb parallel to OE f t scale 

 ofT the value of Ob in volts, and the value of OA' in volts. 

 Dividing the value of Ob by V R' -f X 2 gives the value of I' 

 in amperes, whence, from equation (ii) s may be calculated, inas- 

 much as R" is supposed to be given, and A is known (= OA f ). 

 Knowing /' and s, the mechanical power output of the motor 

 may be calculated from equation ( iv ) for the chosen position of 



