Materials Science and Engineering: An Introduction

Published by Wiley
ISBN 10: 1118324579
ISBN 13: 978-1-11832-457-8

Chapter 18 - Electrical Properties - Questions and Problems - Page 779: 18.17


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 ($6.8 \times 10^{7}(Ω-m)^{-1}$).
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