Answer:
a) 43.98 V
b) E = 21.37 MJ
Explanation:
Parameters given:
Length of cable = 180 km = 180000 m
Diameter of cable = 11 cm = 0.11 m
Radius = 0.11 / 2 = 0.055 m
Current, I = 135 A
a) To find the potential drop, we have to find the voltage across the wire:
V = IR
=> V = IρL / A
where R = resistance
L = length of cable
A = cross-sectional area
ρ = resistivity of the copper wire = 1.72 * 10^(-8) Ωm
Therefore:
V = (135 * 1.72 * 10^(-8) * 180000) / (π * 0.055^2)
V = 43.98 V
The potential drop across the cable is 43.98 V
b) Electrical energy is given as:
E = IVt
where t = time taken = 1 hour = 3600 s
Therefore, the energy dissipated per hour is:
E = 135 * 43.98 * 3600
E = 21.37 MJ (mega joules, 10^6)
