Answer
22 W
Work Step by Step
Use equation 18–3 to find the resistance. The cross-sectional area $A=\pi r^2=\frac{\pi d^2}{4}$.
$$R=\rho\frac{\mathcal{l}}{A}=\rho\frac{4 \mathcal{l}}{\pi d^2}$$
Find the power dissipated in the cord by using equation 18–6a. The length $\mathcal{l}$ is two times 2.7 meters.
$$P=I^2R=I^2\rho\frac{4 \mathcal{l}}{\pi d^2}$$
$$=(18.0A)^2(1.68\times10^{-8}\Omega \cdot m)\frac{(4)(2\times2.7m)}{\pi (0.129\times10^{-2}m)^2}$$
$$=22\;watts$$
