A 120-V source is to be connected to a load by two copper wires. The load is loc

Question

A 120-V source is to be connected to a load by two copper wires. The load is located 500 feet from the source. The maximum voltage drop due to the wire resistance is 10 V, which makes the minimum load voltage 110 V. The maximum current drawn by the load is 50 A. The figure below is the circuit at maximum current and minimum load voltage. Each wire must be 500 feet long. What is the smallest copper wire that can be used? Calculate the wire resistance using the copper wire table.
With the wire size you selected, calculate the load voltage for 50 A when the source voltage is 120 V.

Explanation / Answer

The maximum current through the wire = 50A The total resistance of both the wires = 10/50 = 0.2
Resistance of each wire will be = 0.2/2 = 0.1
R = 0.1 l =500ft = (500)(0.3048)m = 152.4m for copper = 1.68*10-8 m
We have R = l/A Where A is the area of cross section of the copper wire A = l/R A = (1.68*10-8)(152.4)/0.1 A = 0.0000256m2 is the smallest copper wire required.
With R=0.1, I=50A The voltage drop across both the wires = 2RI = 2(0.1)(50)=10V Load voltage = 120-10 = 110V

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