Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 27 - Current and Resistance - Exercises and Problems - Page 763: 33

Answer

$I = 1.65~A$

Work Step by Step

We can find the resistance: $R = \frac{\rho~L}{A}$ $R = \frac{\rho~L}{\pi~r^2}$ $R = \frac{(3.5\times 10^{-5}~\Omega~m)(6.0\times 10^{-2}~m)}{(\pi)~(0.35\times 10^{-3}~m)^2}$ $R = 5.46~\Omega$ We can find the current: $I = \frac{V}{R}$ $I = \frac{9.0~V}{5.46~\Omega}$ $I = 1.65~A$
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