Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 20 - Electric Circuits - Problems - Page 573: 15



Work Step by Step

$R_{1}=R_{2}$ $\implies \rho_{Al}\times\frac{l_{1}}{A_{1}}=\rho_{Cu}\times\frac{l_{2}}{A_{2}}$ Given that $l_{1}=l_{2}=l$ Therefore, $\rho_{Al}\times\frac{l}{A_{1}}=\rho_{Cu}\times\frac{l}{A_{2}}$ $\implies \frac{A_{1}}{A_{2}}=\frac{\rho_{Al}}{\rho_{Cu}}=\frac{2.82\times10^{-8}\,\Omega\cdot m}{1.72\times10^{-8}\,\Omega\cdot m}=1.64$
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