## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 18 - Problems - Page 699: 25

#### Answer

The ratio of the diameter of the copper wire to that of the aluminum wire is 0.80

#### Work Step by Step

We can find an expression for the resistance of the copper wire: $R = \frac{\rho_c~L}{A_c}$ $R = \frac{\rho_c~L}{\pi~(\frac{d_c}{2})^2}$ $R = \frac{4~\rho_c~L}{\pi~d_c^2}$ We can find an expression for the resistance of the aluminum wire: $R = \frac{\rho_a~L}{A_a}$ $R = \frac{\rho_a~L}{\pi~(\frac{d_a}{2})^2}$ $R = \frac{4~\rho_a~L}{\pi~d_a^2}$ Since the resistance is equal in both wires, we can equate the two expressions: $\frac{4~\rho_a~L}{\pi~d_a^2} = \frac{4~\rho_c~L}{\pi~d_c^2}$ $\frac{d_c^2}{d_a^2} = \frac{\rho_c}{\rho_a}$ $\frac{d_c}{d_a} = \sqrt{\frac{\rho_c}{\rho_a}}$ $\frac{d_c}{d_a} = \sqrt{\frac{1.68\times 10^{-8}~\Omega~m}{2.65\times 10^{-8}~\Omega~m}}$ $\frac{d_c}{d_a} = 0.80$ The ratio of the diameter of the copper wire to that of the aluminum wire is 0.80

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