Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 26 - Current and Resistance - Problems - Page 766: 15

Answer

$R=2.4{\Omega}$

Work Step by Step

The resistance $R$ can be determined as: $R=\frac{{\rho}L}{A}$ $R=\frac{{\rho}L}{\pi r^2}$...........................eq(1) Also, $L=N\times2{\pi}R=250\times2\times3.1416\times0.12=188.496m$ Now putting values in eq(1), we get $R=\frac{1.69\times10^{-8}(188.496)}{3.1416(\frac{1.3\times10^{-3}}{2})^2}$ $R=2.4{\Omega}$
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