ALTERNATORS. 245 



assumed to be too small to seriously affect the result on account 

 of the low peripheral speed of the machine. 

 The idle voltage of the machine 



= \/(total voltage) 2 (energy voltage) 2 . 

 In this case idle volts = y' 52- 3 2 5-T9' 2 = 52-0 volts. 



The triangle of voltages for the figures just given is shown 

 in Fig. 116, the line a b showing the phase of the current in 

 the circuit. 



The total voltage shown by the line 6 c in Fig. 116 is the 

 voltage necessary to overcome the self-induction and resis- 

 tance of the armature and leads when the alternator is 

 giving a current of 25-8 amps. In a similar way the two 

 curves given in Figs. 92 and 115 will give the voltage taken 

 up. in the armature for any value of the current shown in 

 Fig. 115. It would not, however, be permissible to simply 

 subtract this voltage from the total no-load voltage of the 



10 20 30 40. 



SCALE OF VOLTS 



c 



FIG. 116. DIAGRAM OF ELECTROMOTIVE FORCE, EXPERIMENT XXXVIII. 



alternator at normal excitation in order to get the terminal 

 voltage available for sending current through the external 

 circuit, for the reason that the voltage spent in overcoming 

 armature impedance and the voltage applied to the external 

 circuit are not generally in phase with each other. It depends 

 upon the difference in phase between these two voltages, to 

 what extent the volts spent in the armature affect the external 

 voltage. 



