Question

Two wires have the same resistance and radius. If the wires are made of copper and platinum with resistivities respectively of 1.70 ✕ 10−8 Ω · m and 11.0 ✕ 10−8 Ω · m, determine the ratio of their lengths.

Answer 1

Answer:

L1/L2 = 6.47

Explanation:

In order to calculate the ratio of the lengths of the wires you use the following formula for the resistivity of a wire:

$\rho=\frac{\pi r R}{L}$        (1)

r: radius of the cross-sectional area of the wire

R: resistance of the wire

L: length of the wire

Then, you have for each wire:

$\rho_1=\frac{\pi r_1^2R_1}{L_1}=1.70*10^{-8}\Omega.m\\\\\rho_2=\frac{\pi r_2^2R_2}{L_2}=11.0*10^{10^{-8}}\Omega.m$

The resistance and radius of the wires are the same, that is, R1 = R2 = R and r1 = r2 = r. By taking into account this last and dive the equation for the wire 2 into the wire 1, you obtain:

$\frac{\rho_2}{\rho_1}=\frac{11.0*10^{-8}\Omega.m}{1.70*10^{-8}\Omega.m}=\frac{L_1}{L_2}\\\\\frac{L_1}{L_2}=6.47$

The ratio of the lengthd of the wires is L1/L2 = 6.47