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Question: Find the resistance of the copper conductor where length is \( 1\text{ cm} \) long and area of cross...

Find the resistance of the copper conductor where length is 1 cm1\text{ cm} long and area of cross-section is 05×106 m20\cdot 5\times {{10}^{-6}}\text{ }{{\text{m}}^{-2}} and resistivity =0176×106Ω m1=0\cdot 176\times {{10}^{-6}}\Omega \text{ }{{\text{m}}^{-1}} .

Explanation

Solution

Resistance is a measure of opposition to current flow in an electrical circuit. It is measured in ohms. It is an electrical quantity that measures how the device or material reduces the electric current. How through it, it can be calculated by
R= !!ρ!! lA\text{R}=\text{ }\\!\\!\rho\\!\\!\text{ }\dfrac{\text{l}}{\text{A}} ,
Where l = length
 !!ρ!! =\text{ }\\!\\!\rho\\!\\!\text{ }= resistivity
A = Area.

Complete step by step solution
Given
l=1 cm  =1100  =001 m A=05×106 m2  !!ρ!! =00176×106Ω m1 \begin{aligned} & \text{l}=1\text{ cm} \\\ & \text{ }=\dfrac{1}{100} \\\ & \text{ }=0\cdot 01\text{ m} \\\ & \text{A}=0\cdot 5\times {{10}^{-6}}\text{ }{{\text{m}}^{2}} \\\ & \text{ }\\!\\!\rho\\!\\!\text{ }=0\cdot 0176\times {{10}^{-6}}\Omega \text{ }{{\text{m}}^{-1}} \\\ \end{aligned}
Using R= !!ρ!! lA\text{R}=\text{ }\\!\\!\rho\\!\\!\text{ }\dfrac{\text{l}}{\text{A}}
We can find out the values of resistance putting all values
R=00176×106×00105×106  =0000176×10605×106  =0000352Ω  =352×105Ω \begin{aligned} & \text{R}=\dfrac{0\cdot 0176\times {{10}^{-6}}\times 0\cdot 01}{0\cdot 5\times {{10}^{-6}}} \\\ & \text{ }=\dfrac{0\cdot 000176\times {{10}^{-6}}}{0\cdot 5\times {{10}^{-6}}} \\\ & \text{ }=0\cdot 000352\Omega \\\ & \text{ }=35\cdot 2\times {{10}^{-5}}\Omega \\\ \end{aligned}
That is resistance of copper conductor is 352×105Ω35\cdot 2\times {{10}^{-5}}\Omega

Note
Resistance causes some of the electrical energy to turn heat so some electrical energy is lost along the way. Therefore, it is sometimes useful to add components called resistors into an electrical circuit to restrict the flow of electricity and protect the components in the circuit.