Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 18 - Electric Currents - General Problems - Page 524: 69

Answer

4.2 mm.

Work Step by Step

Assume that we have exactly one meter of wire, carrying 35 A of current, dissipating 1.5 W of heat. Solve for the diameter. The power dissipated is $I^2R$ and the resistance is $R=\frac{\rho\mathcal{l}}{A}$. $$P=I^2\frac{\rho\mathcal{l}}{\pi(d/2)^2}$$ $$d=\sqrt{I^2\frac{4\rho\mathcal{l}}{P\pi}}=2I\sqrt{ \frac{\rho\mathcal{l}}{P\pi}}$$ $$d=2(35A)\sqrt{ \frac{(1.68\times10^{-8}\Omega\cdot m)(1.0m)}{(1.5W)\pi}}=4.2mm$$
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