What is the ratio of resistances of two copper conductors whose l

SOLUTION

Concept:

The resistance of a conductor is given by:

\(R = \rho\frac{{ ~l}}{A}\)

ρ = resistivity

R = Resistance

l = length of wire

A = cross-sectional area of the wire

Application:

Let R₁ = Resistance of 1^st copper conductor

R₂ = Resistance of 2^nd copper conductor

\(\frac{R_1}{R_2}=\frac{l_1/A_1}{l_2/A_2}\)

Since A = πr², the above can be written as:

\(\frac{R_1}{R_2}=\frac{l_1/\pi r_1^2}{l_2/\pi r_2^2}=\frac{l_1r_2^2}{l_2r_1^2}\)

Given \(\frac{l_1}{l_2}=\frac{1}{4}\)and \(\frac{r_1}{r_2}=\frac{1}{2}\)

\(\frac{R_1}{R_2}=\frac{1}{4}\times (\frac{2}{1})^2\)

\(\frac{R_1}{R_2}=\frac{1}{1}\)