The portable lighting equipment for a mine is located 100 meters from its dc supply source. The mine lights use a total of 5 kW and operate at 120 V dc. Determine the required cross-sectional area of the copper wires used to connect the source to the mine lights if we require that the power lost in the copper wires be less than or equal to 5 percent of the power required by the mine lights.

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

answer:

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