Question

What diameter must a copper wire have if its resistance is to be the same as that of an equal length of aluminum wire with diameter 3.48 mm ?

d = ____________ mm

please show steps and reasoning.

Answer 1

Concepts and reason

Use the formula for the resistance of the wire in terms of resistivity, length and cross-sectional are of the wire.

Initially, write the formula for the resistance of the copper and aluminium wires. Compare the resistance formulas of the copper and aluminium wires to obtain the expression for the diameter of the copper. Finally, substitute the values and find the diameter of the copper wire.

Fundamentals

Resistivity is the resistance of the wire when the wire is having unit length and unit cross-sectional area.

The expression for the resistance in terms of resistivity is expressed as follows:

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

Here, $\rho$ is the resistivity of the wire, l is the length of the wire and A is the cross-sectional area of the wire.

The cross-sectional area of the wire in terms of diameter d is expressed as follows:

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

The resistance of the copper wire is expressed as follows:

${R_{Cu}} = \frac{{{\rho _{Cu}}{l_{Cu}}}}{{{A_{Cu}}}}$ …… (1)

Here, ${\rho _{Cu}}$ is the resistivity of the copper wire, ${l_{Cu}}$ is the length of the copper wire and ${A_{Cu}}$ is the cross sectional area of the copper wire.

The cross-sectional area of the copper wire is expressed as follows:

${A_{Cu}} = \frac{{\pi d_{Cu}^2}}{4}$

Here, ${d_{Cu}}$ is the diameter of the copper wire.

The resistance of the aluminium wire is expressed as follows:

${R_{Al}} = \frac{{{\rho _{Al}}{l_{Al}}}}{{{A_{Al}}}}$ …… (2)

Here, ${\rho _{Al}}$ is the resistivity of the aluminium wire, ${l_{Al}}$ is the length of the aluminium wire and ${A_{Al}}$ is the cross sectional area of the aluminium wire.

The cross-sectional area of the aluminium wire is expressed as follows:

${A_{Al}} = \frac{{\pi d_{Al}^2}}{4}$

Here, ${d_{Al}}$ is the diameter of the copper wire.

Divide equation (1) and (2) are as follows:

$\frac{{{R_{Cu}}}}{{{R_{Al}}}} = \frac{{{\rho _{Cu}}{l_{Cu}}}}{{{\rho _{Al}}{l_{Al}}}}\frac{{{A_{Al}}}}{{{A_{Cu}}}}$

Substitute, $\frac{{\pi d_{Cu}^2}}{4}$ for ${A_{Cu}}$ and $\frac{{\pi d_{Al}^2}}{4}$ for ${A_{Al}}$ and rearrange the equation for ${d_{Cu}}$ .

$\begin{array}{l}\\\frac{{{R_{Cu}}}}{{{R_{Al}}}} = \frac{{{\rho _{Cu}}{l_{Cu}}}}{{{\rho _{Al}}{l_{Al}}}}\frac{{\left( {\frac{{\pi d_{Al}^2}}{4}} \right)}}{{\left( {\frac{{\pi d_{Cu}^2}}{4}} \right)}}\\\\{d_{Cu}} = d_{Al}^{}\sqrt {\left( {\frac{{{\rho _{Cu}}{l_{Cu}}{R_{Al}}}}{{{\rho _{Al}}{l_{Al}}{R_{Cu}}}}} \right)} \\\end{array}$

According to the data given in the question, both copper and aluminium wires have the same resistance and length.

Substitute, ${l_{Cu}}$ for ${l_{Al}}$ , ${R_{Cu}}$ for ${R_{Al}}$ , 3.48 mm for ${d_{Al}}$ , $1.7 \times {10^{ - 8}}{\rm{ }}\Omega {\rm{m}}$ for ${\rho _{Cu}}$ , and $2.7 \times {10^{ - 8}}{\rm{ }}\Omega {\rm{m}}$ for ${\rho _{Al}}$ .

$\begin{array}{c}\\{d_{Cu}} = \left( {3.48{\rm{ mm}}} \right)\sqrt {\left( {\frac{{\left( {1.7 \times {{10}^{ - 8}}{\rm{ }}\Omega {\rm{m}}} \right){l_{Cu}}{R_{Cu}}}}{{\left( {2.7 \times {{10}^{ - 8}}{\rm{ }}\Omega {\rm{m}}} \right){l_{Cu}}{R_{Cu}}}}} \right)} \\\\ = 2.76{\rm{ mm}}\\\end{array}$

Ans:

The diameter of the copper wire is 2.76 mm.