## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

The required diameter of the aluminum wire is $0.64~mm$
We can write a general expression for the resistance: $R = \frac{\rho~L}{A} = \frac{\rho~L}{\pi~r^2}$ We can write an expression for the resistance of the copper wire: $R_c = \frac{\rho_c~L}{\pi~r_c^2}$ We can write an expression for the resistance of the aluminum wire: $R_a = \frac{\rho_a~L}{\pi~r_a^2}$ Since the resistance must be equal, we can equate both expressions for the resistance: $R_c = R_a$ $\frac{\rho_c~L}{\pi~r_c^2} = \frac{\rho_a~L}{\pi~r_a^2}$ $r_a^2 = \frac{\rho_a}{\rho_c}~r_c^2$ $r_a = \sqrt{\frac{\rho_a}{\rho_c}}~r_c$ $r_a = \sqrt{\frac{2.8\times 10^{-8}~\Omega~m}{1.7\times 10^{-8}~\Omega~m}}~(0.25~mm)$ $r_a = 0.32~mm$ Since the diameter is twice the radius, the required diameter of the aluminum wire is $0.64~mm$