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 - Problems - Page 522: 20

Answer

The new resistance is one-fourth of the original resistance.

Work Step by Step

Use equation 18–3 to find the resistance. $$R_{original}=\rho\frac{\mathcal{l}}{A}$$ The length of the new wire is half that of the original wire, and the cross-sectional area of the new wire is twice that of the original wire. $$R_{new}=\rho\frac{0.5\mathcal{l}}{2A}=\frac{1}{4}\rho\frac{\mathcal{l}}{A}=\frac{1}{4}R_{ original }$$
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