Answer
See answers.
Work Step by Step
As the problem demands, set the power loss equal to 2.5 percent of the total power.
Use equation 18–6a to write the power loss in terms of the current (I = P/V; that is, current is power divided by the voltage drop) and the resistance.
We also use equation 18–3 to relate the cross-sectional area of each wire to aluminum’s resistivity and the wire length.
$$P_{loss}=0.025P=I^2R=(\frac{P}{V})^2(\frac{\rho\mathcal{l}}{A})$$
Solve for the area in terms of the wire’s radius.
$$A=\frac{P\rho \mathcal{l}}{0.025V^2}=\pi r^2$$
Assume that we need only one wire transmitting the power over 185 km, because the Earth completes the circuit.
$$r=\sqrt{\frac{P\rho \mathcal{l}}{0.025\pi V^2}}$$
$$r=\sqrt{\frac{(925\times10^6W)(2.65\times10^{-8}\Omega \cdot m) (185000m)}{0.025\pi (660000V)^2}}=0.012m$$
The transmission line must have a radius greater than 1.2 cm.