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 - Search and Learn - Page 525: 6

Answer

a) $0.634$ b) $1.01\;\rm \Omega$ c) $1.59\;\rm \Omega$ d) $327\;\rm W, \;515\;W$ e) $2.04\;\rm mm$ f) Yes

Work Step by Step

a) We know that the resistance of a wire is given by $R=\dfrac{\rho l}{A}$ So the ratio of the resistance of a copper wire to that of an aluminum wire is $$\dfrac{R_{Cu}}{R_{Al}}=\dfrac{\dfrac{\rho_{Cu} l}{A}}{\dfrac{\rho_{Al} l}{A}}=\dfrac{\rho_{Cu}}{\rho_{Al}}$$ since the two wires are having the same length and the same. Plugging from table 18-1; $$\dfrac{R_{Cu}}{R_{Al}} =\dfrac{1.68\times10^{-8}}{2.65\times10^{-8}}=\color{red}{\bf 0.634}$$ b) $$R_{Cu}=\dfrac{\rho_{Cu}l_{Cu}}{A_{cu}}=\dfrac{\rho_{Cu}l_{Cu}}{\pi r_{Cu}^2}=\dfrac{\rho_{Cu}l_{Cu}}{\pi \left(\dfrac{D_{Cu}}{2}\right)^2}$$ whereas $r=0.5D$ and $D$ is the diameter. Plugging the known; $$R_{Cu}= \dfrac{1.68\times10^{-8}\cdot 125}{\pi \left(\dfrac{1.63\times10^{-3}}{2}\right)^2}=\color{red}{\bf1.01}\;\rm \Omega$$ c) Using the last formula above; $$R_{Al} =\dfrac{\rho_{Al}l_{Al}}{\pi \left(\dfrac{D_{Al}}{2}\right)^2}=\dfrac{2.65\times10^{-8}\cdot 125}{\pi \left(\dfrac{1.63\times10^{-3}}{2}\right)^2}=\color{red}{\bf 1.59}\;\rm \Omega$$ d) We know that the power is given by $$P=I^2R$$ Thus, $$P_{Cu}=I^2R_{Cu}=18^2\cdot 1.01=\color{red}{\bf 327}\;\rm W$$ $$P_{Al}=I^2R_{Al}=18^2\cdot 1.59=\color{red}{\bf 515}\;\rm W$$ e) $$R_{Al}=R_{Cu}=\dfrac{\rho_{Al}l_{Al}}{\pi \left(\dfrac{D_{Al}}{2}\right)^2}=1.01$$ and solving for $D_{Al}$; $$ \left(\dfrac{D_{Al}}{2}\right)^2=\dfrac{\rho_{Al}l_{Al}}{1.01\pi }$$ $$ D_{Al} =\sqrt{\dfrac{4\rho_{Al}l_{Al}}{1.01\pi }}=\sqrt{\dfrac{4\cdot 2.65\times10^{-8}\cdot 125}{1.01\pi }}=2.04\times10^{-3}\;\rm m$$ $$ D_{Al} =\color{red}{\bf 2.04}\;\rm mm$$ f) The assumption in section (18-4) is that an aluminum wire of the same mass and length as the copper wire would have a smaller resistance. So, first, we need to find the mass of our copper wire. We need to use the mass density law; $$\rho_{Cu}=\dfrac{m_{Cu}}{V_{Cu}}$$ Thus, $$m_{Cu}=\rho_{Cu}V_{Cu}=\rho_{Cu}A_{Cu}l_{Cu}=\pi \rho_{Cu}l_{Cu}\left(\dfrac{D_{Cu}}{2}\right)^2$$ $$m_{Cu}= \pi\cdot 8900\cdot 125\cdot \left(\dfrac{ 1.63\times10^{-3} }{2}\right)^2=\bf2.32 \;\rm kg$$ Now we need to find the cross-sectional area of an aluminum wire that has the same length as the copper wire and the same mass. $$m_{Al}=m_{Cu}=\rho_{Al}V_{Al}=\rho_{Al}A_{Al}l_{Al} $$ Solving for $A_{Al}$; $$A_{Al}=\dfrac{m_{Al}}{\rho_{Al} l_{Al} }$$ Plugging the known; $$A_{Al}=\dfrac{2.32}{2700\cdot 125 }=\bf 6.87\times 10^{-6}\;\rm m^2$$ Thus, $$R_{Al}=\dfrac{\rho_{Al} l_{Al}}{A_{Al}}=\dfrac{2.65\times10^{-8}\cdot 125}{6.87\times 10^{-6}}=\color{red}{\bf 0.482}\;\rm \Omega$$ It is obvious now that an aluminum wire with the same mass and length as a copper wire will have less resistance. Actually, its resistance is less than half of copper's wire resistance which its resistance, in this case, is 1.01 $\Omega$
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