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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# The diameter of 0-gauge copper wire is 8.252 m m. Find the resistance of a 1.00 km length of such wire used for power transmission.

The resistance of the copper wire is obtained as: $\mathrm{R}=321\mathrm{m\Omega }.$

See the step by step solution

## Step 1: 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.

## Step 2: 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 p of the material of which it is made:

$R=P=\frac{L}{A}$

## Step 3: The given data

$•\mathrm{The}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{wire}\mathrm{is}:L=1.00\left[1\mathrm{km}×\frac{1000\mathrm{m}}{1\mathrm{km}}\right]=1000\mathrm{m}.\phantom{\rule{0ex}{0ex}}•\mathrm{The}\mathrm{diameter}\mathrm{of}\mathrm{the}\mathrm{wire}\mathrm{is}:D=\left(8.252\right).\left[1\mathrm{mm}×\frac{1\mathrm{m}}{1000\mathrm{mm}}\right]=8.252×{10}^{-3}\mathrm{m}.\phantom{\rule{0ex}{0ex}}•\mathrm{The}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{wire}\mathrm{is}:r=D/2=4.126×{10}^{-3}\mathrm{m}.\phantom{\rule{0ex}{0ex}}•\mathrm{The}\mathrm{resistivity}\mathrm{of}\mathrm{the}\mathrm{copper}\mathrm{is}:p=1.72×{10}^{-8}\mathrm{\Omega }.\mathrm{m}$

## Step 4: Evaluation of resistance of the copper wire

The resistance of the copper wire is found from the above equation as:

$R=p\frac{L}{A}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{value}\mathrm{of}A={\mathrm{\pi r}}^{2}\mathrm{which}\mathrm{is}\mathrm{the}\mathrm{cross}-\mathrm{sectional}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{wire}:\phantom{\rule{0ex}{0ex}}R=p\frac{L}{\pi {r}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Entering}\mathrm{the}\mathrm{values},\mathrm{we}\mathrm{obtain}:\phantom{\rule{0ex}{0ex}}\mathrm{R}=1.72×{10}^{-8}\mathrm{\Omega }.\mathrm{m}×\frac{1000\mathrm{m}}{\mathrm{\pi }×{\left(4.126×{10}^{-3}\mathrm{m}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=0.321\mathrm{\Omega }\phantom{\rule{0ex}{0ex}}=\left(0.321\right).\left(1\mathrm{\Omega }×\frac{1000\mathrm{m\Omega }}{1\mathrm{\Omega }}\right)\phantom{\rule{0ex}{0ex}}=321\mathrm{m\Omega }\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{the}\mathrm{resistance}\mathrm{of}\mathrm{the}\mathrm{copper}\mathrm{wire}\mathrm{is}:R=321\mathrm{m\Omega }$