(a) What are the ratios of the resistances of (a) silver and (b) aluminum wire to the resistance of copper wire $\left(R_{\mathrm{Ag}} / R_{\mathrm{Cu}} \text { and } R_{\mathrm{A}} / R_{\mathrm{Ca}}\right)$ for wires of the same length and the same diameter? (c) Which material is the best conductor, for wires of equal length and diameter?

The resistance formula can be written as: \(R = \frac{\rho * L}{A}\), where \(R\) is resistance, \(\rho\) is resistivity, \(L\) is the length of the wire, \(A\) is cross-sectional area. For our given problem, we know that the length and diameter of the wires are equal, meaning that the length \(L\) and the cross-sectional area \(A\) will remain constant in our calculations.

We need to find \(R_{\mathrm{Ag}} / R_{\mathrm{Cu}}\) and \(R_{\mathrm{Al}} / R_{\mathrm{Cu}}\). We know that \(\frac{R_{1}}{R_{2}} = \frac{\rho_{1}}{\rho_{2}}\) because \(L\) and \(A\) are equal. Given the resistivities of Silver (\(\rho_{\mathrm{Ag}} = 1.59 \times 10^{-8} \,Ω \cdot m\)), Aluminum (\(\rho_{\mathrm{Al}} = 2.82 \times 10^{-8} \,Ω \cdot m\)), and Copper (\(\rho_{\mathrm{Cu}} = 1.72 \times 10^{-8} \,Ω \cdot m\)): \(R_{\mathrm{Ag}} / R_{\mathrm{Cu}} = \frac{\rho_{\mathrm{Ag}}}{\rho_{\mathrm{Cu}}} = \frac{1.59 \times 10^{-8}}{1.72 \times 10^{-8}} \approx 0.92\) \(R_{\mathrm{Al}} / R_{\mathrm{Cu}} = \frac{\rho_{\mathrm{Al}}}{\rho_{\mathrm{Cu}}} = \frac{2.82 \times 10^{-8}}{1.72 \times 10^{-8}} \approx 1.64\)

We now have the resistance ratios of Silver and Aluminum compared to Copper. The best conductor is the one with the lowest resistance for the same length and diameter wires. From our calculations, we can see that Silver has the lowest resistance, with a ratio of approximately \(0.92\). Therefore, Silver is the best conductor among the three materials, followed by Copper, and finally, Aluminum.