ELECTRIC CIRCUITS 81 



If the drop of voltage had been limited to 10 volts what size 

 of wire would have been required? 

 The drop in voltage is 



1000 



2 Ir = 2 X 90.9 X 10.55 - - = 10 volts; 



c.m. 



therefore the required section in circular mils is 



1000 

 A = 2 X 90.9 X 10.55^ = 192,000 c.m. 



58. Current-carrying Capacity of Wires. The energy con- 

 sumed in the resistance of a wire raises the temperature of the 

 wire until the point is reached where the heat radiated and con- 

 ducted from the wire is equal to the heat generated in it. When 

 the wire is bare the heat will escape easily into the air, but when 

 it is covered with insulating material the heat cannot escape so 

 easily and for a given current density the temperature rise will 

 be greater. This increase in temperature decreases the resist- 

 ance of the insulating material and so decreases its insulating 

 properties; in extreme cases the insulation may be charred and 

 rendered useless. The last two columns of the table in article 56 

 give the values of current which can be carried safely by differ- 

 ent sizes of wire. With rubber insulation the allowable current 

 is about 25 per cent less than with weatherproof insulation be- 

 cause the rubber is more easily affected by heat. When the in- 

 sulated wires are inclosed in conduits the current-carrying capacity 

 is less than that given in the table. 



59. Examples. (1) Determine the resistance of a copper wire 

 of 97 per cent conductivity, 100,000 c.m. in section and 50 ft. 

 in length at 50 C. 



The resistance is 



* - ao 605 ohms - 



(2) Determine the resistance of an aluminum bar 0.75 in. 

 X 0.375 in. X 100 ft. at 25 C., if the conductivity is 60 per cent. 

 The resistance is 



(3) If the resistance of the shunt-field winding of a generator 



