156 ALTERNATING CURRENTS 



A certain field magnetomotive force FI is required to produce 

 this voltage FI. The value of this magnetomotive force in 

 terms of the field current is found on the saturation curve, Fig. 

 157. Corresponding to the value of FI the field current FI is 

 found. FI is laid off at right angles to V\ and leading it, as a 

 mmf . leads by 90 the emf . which its flux induces. In the short- 

 circuit test the field current is adjusted until the rated current 

 flows. The corresponding value of field current A (Fig. 157) is 

 then read. The magnetomotive force represented by this field 

 current is necessary to send rated current through the armature 

 reactance and at the same time overcome the armature reaction, 

 if the resistance be neglected. This magnetomotive force, A 9 

 replaces the combined effect of the armature reactance and the 

 armature reaction. It is laid off 180 from the current as shown 

 at A in Fig. 156. (The total mmf. which is assumed to pro- 

 duce the total voltage drop is +A. The component which must 

 balance this mmf. is A.) The resultant magnetomotive force 

 is F, which, at unity power-factor, is the square root of the sum 

 of the squares of FI and A. F is the mmf. which exists at no 

 load under the assumptions made. The no-load voltage E lags 

 F by 90, Fig. 156, and is found on the saturation curve corre- 

 sponding to field current F, Fig. 157. 



To summarize the method at unity power-factor, the IR drop 

 jsa,dc]dto the terminal voltage, and the field current correspond- 

 ing to this sum is found on the saturation curve. The machine 

 is then short-circuited and the field current necessary to send 

 rated current through the armature is determined. The square 

 root of the sum of the squares of these field currents is then found. 

 The value of emf. oi> the saturation curve corresponding to this 

 resultant field current is assumed to be the no-load voltage of 

 the machine. 



When the power-factor is less than unity, the diagram is similar 

 to that shown in Fig. 158. 



The voltage FI is the vector sum of F and IR. Its value is 

 readily found by projecting these voltages on the current vector. 



Thus, 



F! = V(F cos 6 + IR) 2 + (Fsin 0) 2 



In most cases a numerical addition of F and IR is sufficiently 

 accurate. 



