Answer
$R = 6.70 \times 10^{-3} Ω$
Work Step by Step
Required:
Using the data in Table 18.1, compute the resistance of an aluminum wire 5 mm (0.20 in.) in diameter and 5 m (200 in.) in length
Solution:
Combining Equations 18.2 and 18.4 and using that $σ = 3.8 \times 10^{7} (Ω−m)^{-1}$ (from table 18.1), it follows:
$R = \frac{ρl}{A} = \frac{l}{σA} = \frac{l}{σπ(\frac{d}{2})^{2}}$
$R = \frac{5 m}{[3.8 \times 10^{7} (Ω−m)^{-1}]π(\frac{5 \times 10^{-3} m)}{2})^{2}} = 6.70 \times 10^{-3} Ω$
