Question

A copper wire has cross sectional area $4\times {10}^{-6}{m}^2$ and resistivity of $1.6\times {10}^{-8}ohmmeter$. Calculate the length of the wire to make its resistance$10ohms$. By how much does the resistance change if the diameter is doubled?

Answer 1

Step 1 - Given data and formula

Area of cross-section $A=4\times {10}^{-6}{m}^2$

Resistivity $\rho =1.6\times {10}^{-8}\Omega m$

Resistance $R=10\Omega$

Step 2 - Finding the formula for the length of the wire

Consider the length of the wire is $l$.

The formula for resistivity is given by the expression

$\rho =\frac{R\times A}{l}...\left(1\right)$

Rearrange the above equation.

$$\begin{gathered} \rho =\frac{R\times A}{l} \\ l=\frac{R\times A}{\rho } \end{gathered}$$

Step 3 - Finding the length of the wire

Substitute the values into the above equation to calculate the length of the wire.

$$\begin{gathered} l=\frac{(10\Omega )\times (4\times {10}^{-6}{m}^2)}{(1.6\times {10}^{-8}\Omega m)} \\ l=2500m \end{gathered}$$

Step 4 - Finding the new area if the diameter is doubled

The formula for area is given by the expression

$A=\frac{\pi d^2}{4}...\left(2\right)$

Where $d$ is the diameter of the wire.

The formula for the area after the diameter is changed to $d\text{'}$ is given by the expression

$A\text{'}=\frac{\pi d{\text{'}}^2}{4}...\left(3\right)$

Where $A\text{'}$ is the new area of the cross section.

If the diameter is doubled then $d\text{'}=2d$

Substitute the value of $d\text{'}$ into equation$\left(3\right)$.

$$\begin{gathered} A\text{'}=\frac{\pi {\left(2d\right)}^2}{4} \\ A\text{'}=4\times \frac{\pi d^2}{4} \\ A\text{'}=4\times A \end{gathered}$$

Step 5 - Finding the change in resistance if the diameter is doubled

Consider that the new resistance is $R\text{'}$.

The formula for resistivity can be reduced as

$\rho =\frac{R\text{'}\times A\text{'}}{l}$

Substitute $4A$ for $A\text{'}$ into the above formula.

$$\begin{gathered} \rho =\frac{R\text{'}\times \left(4A\right)}{l} \\ \rho =4\left(\frac{R\text{'}\times A}{l}\right)...\left(4\right) \end{gathered}$$

Divide equation $\left(4\right)$ with equation $\left(1\right)$.

$$\begin{gathered} \frac{\rho }{\rho }=\frac{4\left(\frac{R\text{'}\times A}{l}\right)}{\frac{R\times A}{l}} \\ 1=\frac{4\times R\text{'}}{R} \\ R\text{'}=\frac{1}{4}\times R \end{gathered}$$

So, resistance reduces by factor$\frac{1}{4}$.

Final answer: The length of the wire is $2500m$ and if the diameter is doubled then the resistance becomes one-fourth.