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 28 - Fundamentals of Circuits - Exercises and Problems - Page 790: 13

Answer

The diameter of the wire is approximately $24\mu m$

Work Step by Step

Power loss across the tungsten wire can be taken by the equation, $$P=VI$$ The voltage drop across the tungsten wire can be taken by the equation, $$V=IR$$ Taking $I$ as a common value these two equations can be equalized as $$\frac{P}{V}=\frac{V}{R}$$ Substitute R in above for $\frac{\rho\times l}{A}$ $\frac{\rho\times l}{A}=\frac{V^{2}}{P}$ $π\times\frac{d^{2}}{4}=\frac{P\times\rho\times l}{V^{2}}$ $d=\sqrt \frac{4P\times\rho\times l}{πV^{2}} = \sqrt \frac{400W\times\ 9.0\times 10^{-7}Ωm\times 0.07m}{π\times120^{2}}$ $d = 2.36 \times 10^{-5}m \approx 24\mu m$
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