Question

Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1.5 mm. Conductor B is a hollow tube of outside diameter 7.4 mm and inside diameter 6.4 mm. What is the resistance ratio RA/RB , measured between their ends?

Answer 1

Concepts and reason

The concepts used to solve this problem are resistance of a solid wire and hollow tube.

First find the resistance for the solid and then find the area of the hollow tube. Later find the resistance for the hollow tube and finally take the ratio of the resistance of these conductors.

Fundamentals

Resistance is defined as the amount of electric current a component can resist.

The expression for resistance is,

$R = \frac{{\rho L}}{A}$

Here, $\rho$is the resistivity, $L$is the length, and $A$ is the area.

Express the area of the wire.

$A = \frac{{\pi {d^2}}}{4}$

Here, $d$ is the diameter of the wire.

Rewrite the expression in terms of area.

$R = \frac{{4\rho L}}{{\pi {d^2}}}$

The two conductors are made up of the same material, thus, the resistivity is same for both the materials.

The expression for resistance ratio is,

${\rm{Resistance}}\,{\rm{ratio}} = \frac{{{R_A}}}{{{R_B}}}$

Here, ${R_A}$ is the resistance of a solid wire and ${R_B}$ is the resistance of a hollow tube.

The expression for resistance of the solid wire is,

${R_A} = \frac{{4\rho L}}{{\pi {d^2}}}$

Substitute $1.5\,{\rm{mm}}$ for the diameter $d$.

$\begin{array}{c}\\{R_A} = \frac{{4\rho L}}{{\pi {{\left( {{\rm{1}}{\rm{.5}}\,{\rm{mm }} \times \frac{{{{10}^{ - 3}}\,{\rm{m}}}}{{1\,{\rm{mm}}}}} \right)}^2}}}\\\\ = \frac{{\rho L}}{{1.76 \times {{10}^{ - 6}}{\rm{ }}{{\rm{m}}^2}}}\\\end{array}$

The expression to calculate the resistance of the hollow tube is,

${R_B} = \frac{{4\rho L}}{{\pi {d^2}}}$

The expression to calculate the area of the hollow tube is,

$\frac{{\pi {d^2}}}{4} = \frac{\pi }{4}\left( {d_{{\rm{outer}}}^2 - d_{{\rm{inner}}}^2} \right)$

Here, ${d_{{\rm{outer}}}}$ is the outer diameter of the hollow tube and ${d_{{\rm{inner}}}}$ is the inner diameter of the hollow tube.

Use $\frac{\pi }{4}\left( {d_{{\rm{outer}}}^2 - d_{{\rm{inner}}}^2} \right)$ for $\frac{{\pi {d^2}}}{4}$ in ${R_B} = \frac{{4\rho L}}{{\pi {d^2}}}$. The resistance of the hollow tube is,

${R_B} = \frac{{4\rho L}}{{\pi \left( {d_{{\rm{outer}}}^2 - d_{{\rm{inner}}}^2} \right)}}$

Substitute 7.4 mm for ${d_{{\rm{outer}}}}$ and 6.4 mm for ${d_{{\rm{inner}}}}$ in the above expression.

$\begin{array}{c}\\{R_B} = \frac{{4\rho L}}{{\pi \left( {{{\left( {\left( {7.4\,{\rm{mm}}} \right)\left( {\frac{{1\,{\rm{m}}}}{{{{10}^3}\,{\rm{mm}}}}} \right)} \right)}^2} - {{\left( {\left( {6.4\,{\rm{mm}}} \right)\left( {\frac{{1\,{\rm{m}}}}{{{{10}^3}\,{\rm{mm}}}}} \right)} \right)}^2}} \right)}}\\\\ = \frac{{\rho L}}{{1.0833 \times {{10}^{ - 5}}\,{{\rm{m}}^{\rm{2}}}}}\\\end{array}$

${\rm{Resistance}}\,{\rm{ratio}} = \frac{{{R_A}}}{{{R_B}}}$

Substitute $\frac{{\rho L}}{{1.76 \times {{10}^{ - 6}}{\rm{ }}{{\rm{m}}^2}}}$ for ${R_A}$ and $\left[ {\frac{{\rho L}}{{1.08 \times {{10}^{ - 5}}{\rm{ }}{{\rm{m}}^2}}}} \right]$ for ${R_B}$ in$\frac{{{R_A}}}{{{R_B}}}$.

$\begin{array}{c}\\\frac{{{R_A}}}{{{R_B}}} = \frac{{\left( {\frac{{\rho L}}{{1.76 \times {{10}^{ - 6}}{\rm{ }}{{\rm{m}}^2}}}} \right)}}{{\left( {\frac{{\rho L}}{{1.08 \times {{10}^{ - 5}}{\rm{ }}{{\rm{m}}^2}}}} \right)}}\\\\ = \left[ {\frac{{\rho L}}{{1.76 \times {{10}^{ - 6}}{\rm{ }}{{\rm{m}}^2}}}} \right]\left[ {\frac{{1.08 \times {{10}^{ - 5}}{\rm{ }}{{\rm{m}}^2}}}{{\rho L}}} \right]\\\\ = \frac{{1.08 \times {{10}^{ - 5}}{\rm{ }}{{\rm{m}}^2}}}{{1.76 \times {{10}^{ - 6}}{\rm{ }}{{\rm{m}}^2}}}\\\\ = 6.136\\\end{array}$

Ans:

The resistance ratio $\frac{{{R_A}}}{{{R_B}}}$ measured between their ends is 6.136.