Question

A transmission-line cable with a length of $\mathbf{2}\mathbf{km}$consists of 19 strandsof identical copper conductors, each $\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{mm}$ in diameter. Because of thetwist of the strands, the actual length of each conductor is increased by $\mathbf{5}\mathbf{\%}$. Determine the resistance of the cable if the resistivity of copper is$\mathbf{1}\mathbf{.}\mathbf{72}\mathbf{\mu \Omega }\mathbf{-}\mathbf{cm}$ at data-custom-editor="chemistry" $\mathbf{20}\mathbf{°}\mathbf{C}$.

Short answer

Therefore, the resistance of cable is$1.536\Omega$.

Step-by-step solution

Given data.

The resistivity of the cable is,

$\begin{array}{c}\mathrm{\rho }=1.72\mathrm{\mu \Omega }-\mathrm{cm}\\ =0.0172\mathrm{\mu \Omega }-\mathrm{m}\end{array}$

The diameter of each strand is,

$\begin{array}{c}\mathrm{d}=1.5\mathrm{mm}\\ =1.5\times {10}^{-3}\mathrm{m}\end{array}$

The length of the cable is,

$\begin{array}{c}\mathrm{l}=2\mathrm{km}\\ =2000\mathrm{m}\end{array}$.

The length is increased by $5\%$.

The number of strands is 19.

Determine the formula of diameter of strand and resistance of cable.

Write the formula of cross-sectional area because of 19 strands.

$\mathrm{A}=\mathrm{n}\times \frac{\mathrm{\pi }}{4}{\mathrm{d}}^2$ ……. (1)

Here,$\mathrm{n}$is number of strands, $\mathrm{d}$is original diameter of cable and ${\mathrm{d}}_1$ is diameter because of strands.

$\mathrm{R}=\frac{\mathrm{\rho l}}{\mathrm{A}}$ ……. (2)

Write the formula of resistance.

Here, $\rho$is resistivity, $l$is length and $\mathrm{A}$ is area.

Determine the resistance of cable.

Determine the area of cable.

Substitute $19$ for $\mathrm{n}$ and $1500\mathrm{m}$ for $\mathrm{d}$ in equation (1).

$\begin{array}{c}\mathrm{A}=19\times \frac{\mathrm{\pi }}{4}{(1.5\times {10}^{-3})}^2\\ =33.576\times {10}^{-6}{\mathrm{m}}^2\end{array}$

The length is increased by 5 %, so new length is,

$\begin{array}{c}\mathrm{l}=1.5\times 2000\mathrm{m}\\ =3000\mathrm{m}\end{array}$

Determine the resistance of cable.

Substitute role="math" localid="1655298130170" $3000\mathrm{m}$ for role="math" localid="1655298154910" $\mathbf{I}$, $33.576\times {10}^{-6}{\mathrm{m}}^2$ for $\mathrm{A}$ and $0.0172\mathrm{\mu \Omega }-\mathrm{m}$ for $\rho$ in equation (2).

$\begin{array}{c}\mathrm{R}=\frac{0.0172\mathrm{\mu \Omega }-\mathrm{m}\times 3000\mathrm{m}}{33.576\times {10}^{-6}{\mathrm{m}}^2}\\ =1.536\mathrm{\Omega }\end{array}$

Therefore, the resistance of cable is $1.536\Omega$.