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 - General Problems - Page 524: 86

Answer

Power is reduced by a factor of 4.

Work Step by Step

First find the change in resistance. The amount of metal is unchanged, so the volume stays the same. If the length of the wire increases by a factor of 2, the cross-sectional area decreases by the same factor of 2. $$R_f=\rho\frac{\mathcal{l}_f}{A_f}=\rho\frac{2\mathcal{l}_i}{A_i/2}=4\rho\frac{\mathcal{l}_i}{A_i}=4R_i$$ The final resistance is 4 times the old resistance. The power dissipated is $P=\frac{V^2}{R}$. Since the resistance went up by a factor of 4, the power is reduced by a factor of 4, i.e., the new power dissipated is one-fourth of the old power.
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