Question

Two copper wires, one of length $1\mathrm{m}$ and the other of length $9\mathrm{m}$, are found to have the same resistance. Their diameters are in the ratio?

Answer 1

Step 1: Given data

Two copper wires having the same resistance i.e. $R$ and length

$$\begin{gathered} L_1=1m \\ {L}_2=9m \end{gathered}$$

As they are made up of the same material then they have the same resistivity($\rho$).

Step 2: Formula used

Let they have diameter$D_1$ and $D_2$.

So two wires have the area,

$$\begin{gathered} A_1=\pi \frac{{D_1}^2}{4} \\ {A}_2=\pi \frac{{D^2}_2}{4} \end{gathered}$$

The resistance of a current-carrying wire is given by,

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

Where,

$L$ is the length of the wire

$A$ is the crossectional area of the wire

$\rho$ is the resistivity of the material

Step 3: Finding the ratio of diameters of two wires

Now for the two given wires, we have the same resistance($R$) and the same resistivity($\rho$).

So from equation (1), for two wires we can write,

$\begin{array}{rcl}& \Rightarrow & \frac{L_1}{A_1}=\frac{L_2}{A_2}\\ & \Rightarrow & \frac{L_1}{L_2}=\frac{A_1}{A_2}\\ & \Rightarrow & \frac{L_1}{L_2}=\frac{\pi {\displaystyle \frac{{D^2}_1}{4}}}{\pi {\displaystyle \frac{{D^2}_2}{4}}}\\ & =& \frac{{D^2}_1}{{D^2}_2}\end{array}$

So, we can write

$\begin{array}{rcl}& \Rightarrow & \frac{D_1}{D_2}=\sqrt{\frac{L_1}{L_2}}\\ & =& \sqrt{\frac{1}{9}}\\ & =& \frac{1}{3}\end{array}$

Hence, their diameters are in the ratio $1:3$.