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: 25

Answer

a) $R_{total}=14.1$ b) $I=0.00674A$ c) $V_{Al}=0.0579V$ $V_{Cu}=0.0367V$

Work Step by Step

a) We can find the total resistance by adding the resistance of the copper wire and the resistance of the aluminum wire. We can use the formula $R=\frac{pl}{A}$, where $A=\pi(\frac{1.4\times10^-4}{2})^2=1.54\times10^{-8}m^{2}$ $R_{Cu}=\frac{1.68\times10^{-8}\times5.0m}{1.54\times10^{-8}}=5.45$ $R_{Al}=\frac{2.65\times10^{-8}\times5.0m}{1.54\times10^{-8}}=8.60$ $R_{total}=5.45+8.60=14.1$ b) We can use Ohm's Law $V=IR$ $I=\frac{V}{R}=\frac{0.095V}{14.1}=0.00674A$ c) $V_{Al}=IR_{Al}=0.00674A\times8.60=0.0579V$ $V_{Cu}=IR_{Cu}=0.00674A\times5.45=0.0367V$
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