Answer
$d = 2.59 \times 10^{-3} m = 2.59 mm $
Work Step by Step
Given:
An aluminum wire 10 m long must experience a voltage drop of less than 1.0 V when a current of 5 A passes through it
Required:
The minimum diameter of the wire, considering the data in Table 18.1.
Solution:
Combining Equations 18.3 and 18.4:
$A = \frac{Ilρ}{V} = \frac{Il}{Vσ} $
From Table 18.1, $σ = 3.8 \times 10^{7} (Ω-m)^{-1}$, and for cylindrical wire, $A = π(\frac{d}{2})^{2}$, so:
$π(\frac{d}{2})^{2} = \frac{Il}{Vσ} $
Solving for $d$:
$d = \sqrt \frac{4Il}{πVσ} = \sqrt \frac{(4)(5 A)(10 m)}{π(1.0 V) [3.8 \times 10^{7} (Ω-m)^{-1}]} = 2.59 \times 10^{-3} m = 2.59 mm $
