Question

Find the ratio of the diameter of aluminum to copper wire, if they have the same resistance per unit length (as they might in household wiring).

Short answer

The ratio of the diameter of the aluminum to the copper wire is obtained as:$1.219$.

Step-by-step solution

Define Resistance

In an electrical circuit, resistance is a measure of the resistance to current flow. The Greek letter omega $\left(\Omega \right)$ is used to represent resistance in ohms.

Concepts and Principles

The electrical resistance of the value $R$of a resistive circuit element depends on its geometric structure (its length symbolized as $L$ and cross-sectional area symbolized as $A$) and on the resistivity data-custom-editor="chemistry" $p$ of the material of which it is made:

$R=p\frac{\mathrm{L}}{\mathrm{A}}$

The given data

• Two wires are made of aluminum and copper.
• Two wires have the same resistance per unit length.
• The resistivity of aluminum is: $p_{AI}=2.56\times {10}^{-8}\mathrm{N}.\mathrm{m}.$.
• The resistivity of copper is: $p_{Cu}=1.72\times {10}^{-8}\mathrm{N}.\mathrm{m}.$.

Equation to find the length of aluminum and copper wire

Let the values $R_{AI}$,$D_{AI}$and$L_{AI}$ be the resistance, the diameter, and the length of the aluminum wire, respectively.

Let the values $R_{Cu}$,$D_{Cu}$ and$L_{Cu}$ be the resistance, the diameter, and the length of the copper wire, respectively.

The resistance per unit length of the aluminum wire is found by solving the above equation for the values of $\frac{R_{AI}}{L_{AI}}$;

$\frac{R_{AI}}{L_{AI}}=\frac{p_{AI}}{A_{AI}}$

The value of $A_{AI}=\pi D_{AI}^2/4$, which is the cross-sectional area of the aluminum wire:

$\frac{R_{AI}}{L_{AI}}=\frac{p_{AI}}{\pi D_{AI}^2\mathit{/}\mathit{4}}$

The resistance per unit length of the copper wire is:

$\frac{R_{Cu}}{L_{Cu}}=\frac{p_{Cu}}{\pi D_{Cu}^2\mathit{/}\mathit{4}}$

Evaluation of the diameter of the aluminum to the copper wire

Both the wires have the same resistance per unit length, so we get:

$\frac{p_{AI}}{\pi D_{AI}^2\mathit{/}\mathit{4}}=\frac{p_{Cu}}{\pi D_{Cu}^2\mathit{/}\mathit{4}}$

Rearranging and then solving for$D_{AI}/D_{Cu}$, we get:

$$\begin{gathered} {\left(\frac{D_{AI}}{D_{Cu}}\right)}^2=\frac{p_{AI}}{p_{Cu}} \\ \left(\frac{D_{AI}}{D_{Cu}}\right)=\sqrt{\frac{p_{AI}}{p_{Cu}}} \end{gathered}$$

Putting the values, we get:

$$\begin{gathered} \frac{D_{AI}}{D_{Cu}}=\sqrt{\frac{2.56\times {10}^{-8}\mathrm{\Omega }.\mathrm{m}}{1.72\times {10}^{-8}\mathrm{\Omega }.\mathrm{m}}} \\ =1.219 \end{gathered}$$

Therefore, the ratio of the diameter of the aluminum to the copper wire is:$1.219$.