#### Answer

Copper and Silver

#### Work Step by Step

Given:
3 mm (0.12 in) in diameter cylindrical wire
Current = 12 A
minimum 0.01 V drop per foot (300 mm)
Required:
Metals and alloys that are possible candidates to use
Solution:
Using Equation 18.3, p = $\frac{VA}{Il}$, and Equation 18.4, σ = $\frac{1}{p}$, determine the minimum conductivity required and then select from Table 18.1 those metals with conductivities greater than the computed value of minimum conductivity.
By combining these two equations,
σ = $\frac{1}{p}$ = $\frac{Il}{VA}$ = $\frac{Il}{(V)(π)(\frac{d}{2})^{2} }$
σ = $\frac{(12 A)(300 \times 10^{-3} m)}{(0.01 V) (3.1416) (\frac{3 \times 10^{-3}m}{2})^{2}}$
σ = $5.1 \times 10^{7} (Ω-m)^{-1}$
From Table 18.1, the only candidates are Copper ($6.0 \times 10^{7}(Ω-m)^{-1}$)
and Silver ($6.8 \times 10^{7}(Ω-m)^{-1}$).