ELECTRIC DISTRIBUTION AND WIRING. 285 



117. Voltage regulation as a factor determining the size of 

 wires. The so-called constant-voltage system of current dis- 

 tribution is the system that is generally used in electric-light and 

 power installations, and the sizes of feeders, mains, and service 

 wires used in such a system are generally determined on the 

 basis of a prescribed voltage-drop. This is especially the case 

 when current is supplied to incandescent lamps, the satisfactory 

 operation of which depends upon approximate constancy of 

 voltage. The voltage-drop in the feeders which lead out from a 

 central station to a center of distribution is usually compensated 

 for by what is called feeder control at the station as explained on 

 page 195. On the other hand, the voltage-drop in the mains 

 which lead out from a center of distribution and the voltage-drop 

 in the service wires which lead from the street mains to the lamps 

 are not compensated for, and, as they affect the value of the voltage 

 at the lamps, these voltage-drops must be small. The total drop 

 between a center of distribution and the lamps is generally 

 limited "to a certain percentage of the voltage at the lamps. A 

 total drop of about 5 per cent, 2 per cent, in the mains and 3 

 per cent, in the service wires, is frequently allowed ; although a 

 greater or less drop may be advisable if the distance between the 

 center of distribution and the lamps is very great or very small 

 respectively. 



Dependence of total weight of wire upon the voltage at the lamps 

 and upon the distance of the lamps from the center of distribution. 

 If a given amount of power, P, is to be supplied to lamps at 

 voltage, E, with a given percentage drop, />, by a separate pair of 

 wires leading from the center of distribution to the lamps, then 

 the weight of the copper required is proportional to I^JE 2 , where 

 / is the distance between the center of distribution and the lamps. 

 This is evident when we consider that the current is P\E, that 

 the drop pEj 100 is equal to R x P\E, that R= IO.8 x 2ljd 2 , 

 and that W= 0.00000303 x 2ld 2 , so that 



PI 2 



2 , (43) 



