THE CONTINUOUS CURRENT GENERATOR 101 



in this armature, and as each can safely carry 8.63 amperes, 

 the armature can carry about 17.3 amperes. If we had 

 supposed a poorly ventilated armature and allowed 900 

 circular mils per ampere, the safe current would have been 

 11.6 amperes. 



Calculation of Armature Resistance. The resistance of 

 the armature is obtained by calculating the resistance per 

 path and then dividing by the number of paths in parallel 

 in the armature. In the machine above there are in each 

 path 22 coils of 8 turns each and each turn is 2 feet long. 

 The length of wire per path is therefore 352 feet. The 

 resistance of 352 feet of No. 13 wire (at 50 C.) =0.786 ohms. 



786 

 As there are two paths, the armature resistance is ' or 



4 



0.393 ohms. 



Calculation of Terminal Voltage. The full load IR drop 

 m the windings is therefore 



0.393X17.3 = 6.8 volts (say 7 volts). 



The drop at the brush contacts (carbon brushes assumed) 



= 2 volts. Therefore 

 the total drop in the armature with full load current 



= 7+2-9 volts. 

 The full-load terminal voltage (voltage at brushes) 



= 491 volts (generated) -9 volts (IR 

 drop) =482 volts. 



Another Problem. Next consider a 12-pole, lap-wound 

 generator having 240 coils on the armature, 4 turns per coil 

 each turn consisting of two No. 8 wires in parallel. The 

 length per turn is 4 feet and the length of inductor per 

 turn is 20 inches. The armature is 4 feet in diameter and 

 makes 200 r.p.m. 50% of the inductors lie in a field of 9000 



