Answer
a. 12 W.
b. 4.6 W.
Work Step by Step
a. The wasted power is due to heating losses in the wire. Calculate the current from equation 18–5.
$$P_{lost}=I^2R=\frac{P^2}{V^2}R=\frac{P^2}{V^2}\frac{\rho L}{A}=\frac{P^2}{V^2}\frac{\rho L}{\pi r^2}$$
$$P_{lost} =\frac{P^2}{V^2}\frac{4 \rho L}{\pi d^2}$$
$$P_{lost} =\frac{(1450W)^2}{(120V)^2}\frac{4 (1.68\times10^{-8}\Omega \cdot m)(25.0m)}{\pi (0.00259m)^2}\approx 12W$$
b. Repeat, using a larger wire.
$$P_{lost}=I^2R=\frac{P^2}{V^2}R=\frac{P^2}{V^2}\frac{\rho L}{A}=\frac{P^2}{V^2}\frac{\rho L}{\pi r^2}$$
$$P_{lost} =\frac{P^2}{V^2}\frac{4 \rho L}{\pi d^2}$$
$$P_{lost} =\frac{(1450W)^2}{(120V)^2}\frac{4 (1.68\times10^{-8}\Omega \cdot m)(25.0m)}{\pi (0.00412m)^2}\approx 4.6W$$
