Answer:
The answer is below
Explanation:
The resistivity of copper is ρ = 1.72 * 10⁻⁸ Ωm, diameter d = 10 cm = 0.1 m
The resistance (R) of transmission line is given as:
Rtl = ρL / A; where ρ = resistivity of copper = 1.72 * 10⁻⁸ Ωm, L = length of transmission line = 2000 km = 2000000 m, A is the area of the wire = πd²/4 = π(0.1)²/4
[tex]R_{tl}=\frac{\rho L}{A}=\frac{1.72*10^{-8}*2000000}{\pi*0.1^2/4}=4.4 \ ohm[/tex]
Power = [tex]\frac{V_L^2}{R_L}[/tex]
Power = 10 MW = 10 * 10⁶ W
[tex]10*10^6=\frac{(250*10^3)^2}{R_L} \\\\R_L=\frac{(250*10^3)^2}{10*10^6} \\\\R_L=6250\ ohm[/tex]
[tex]I_L=\frac{V_L}{R_L} \\\\I_L=\frac{250*10^3}{6250} =40\ A[/tex]
a) Since there are two tranmission lines, the power consumed by the lines is:
[tex]P_{TL}=2*I_L^2*R_{TL}=2*40^2*4.4=14080\ W[/tex]
b) The energy generated by the source = 10 * 10⁶ W + 14080 W = 10014080 W
Fraction used = 10 * 10⁶ / 10014080 * 100% = 99.86%
