Answer:
The required cross-sectional area of the copper wires used to connect the source to the mine lights is 0.029mm²
Explanation:
Given.
Copper Resistivity = 1.69 *10^-8Ωm
The mine lights use a total of 5 kW and operate at 120 V dc.
So,
Power = 5kW
Power Required = 5% of 5kW
Power Required = 0.05 of 5000W
Power Required = 250W
Calculating the Resistance
The power lost in the wires is given by 120² / R, where R is the resistance.
250 = 120²/R
R = 120²/250
R = 57.6 Ω
This is small amount over 100 m.
Calculating the Cross-sectional area.
The resistance of the wires is given by:
R = 1.69 * 10^(-8) *100 / A
R = 1.69 * 10^-6/A, where A is the cross-sectional area.
1.69 * 10^(-6)/A = 57.6
A = 1.69 * 10^-6/57.6
A = 2.9340277777777E−8m²
A = 0.029mm²
