ALTERNATORS. 



205 



The value of the self-induction is directly obtained from 

 the value obtained for E.,, since E s = 2 TT n L G, 

 when n = periodicity of current, 

 and L = coefficient of self-induction of armature. 

 These results can be obtained directly from the readings 

 shown in Fig. 94, and the following example, taken from that 

 figure, is given to show the method of calculation. 



With an excitation of 1-64 amperes, the total voltage 

 generated at no-load is seen to be exactly 100 volts. With 



SCALE Or VOLTS 



10 20 30 40 50 60 70 80 90 100 

 FIG. 95. DIAGRAM OF VOLTAGES IN ARMATURE AND EXTERNAL CIRCUIT. 



a load of 20 amperes the terminal voltage is 93-7. We 

 have for the whole circuit, including external resistance and 

 armature, 



(Total volts generated) 2 = (idle volts) 2 + (energy volts) 2 , 



or 100 2 = (idle volts) 2 + (93-7 + 20 x -14) 2 , 

 since 93-7 is the energy voltage applied to the non-inductive 

 external circuit, and the armature resistance was -14 ohms. 

 Hence idle volts = 2 n L C = ylOO 2 (96-5) 2 



= x/10000 - 9312^= 26-2. 



The' periodicity was 40 and the value of the self-induction 

 becomes 



26-2 26-2 



L = 



= ' 0052 



~C 2 * x 40 x 20 



= 5-2 millihenry at the particular excitation and 

 current taken. 



26-2 

 The value of 2 * n L at this excitation is ^- = 1-31. 



43U 



Consequently the impedance of the armature 



t (2 T n L)-> = A/-14 2 + i-3i- = </rm 



= 1-317 apparent ohms. 



