College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 17 - Electric Current and Resistance - Learning Path Questions and Exercises - Conceptual Questions - Page 618: 13


Diameter must be equal to $\frac{1}{\sqrt 2}$ times the original value.

Work Step by Step

$R=\frac{\rho l}{A}$ Length is halved, i.e., $l'=\frac{l}{2}$. If the new area $A'=\frac{A}{2}$, then $\frac{1}{2}$ in the numerator and $\frac{1}{2}$ in the denominator will get canceled and the resistance will be constant. $A'=\frac{A}{2} \implies \pi r_{2}^{2}=\frac{1}{2}\pi r_{1}^{2}$ $\implies r_{2}=\frac{r_{1}}{\sqrt 2}$ or $d_{2}=\frac{d_{1}}{\sqrt 2}$
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