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Difficulty: EasyElectric Current and Resistance

A uniform conductor of length 4.0 m4.0\,\text{m} and cross-sectional area 2.0×10−7 m22.0 \times 10^{-7}\,\text{m}^2 has an electrical resistance of 0.50 Ω0.50\,\Omega. What is the electrical resistivity of the material of the conductor in units of 10−8 Ω⋅m10^{-8}\,\Omega\cdot\text{m}?

Answer: 2.5 10^-8 \Omega\cdot m

Answer

The electrical resistivity of the material is 2.5×10−8 Ω⋅m2.5 \times 10^{-8}\,\Omega\cdot\text{m}, which gives a numerical value of 2.52.5 in units of 10−8 Ω⋅m10^{-8}\,\Omega\cdot\text{m}.
The electrical resistance RR of a conductor is given by R=ρLAR = \frac{\rho L}{A}, where ρ\rho is the resistivity, LL is the length, and AA is the cross-sectional area. Rearranging to solve for resistivity yields ρ=R⋅AL\rho = \frac{R \cdot A}{L}. Substituting R=0.50 ΩR = 0.50\,\Omega, A=2.0×10−7 m2A = 2.0 \times 10^{-7}\,\text{m}^2, and L=4.0 mL = 4.0\,\text{m} gives ρ=0.50×2.0×10−74.0=2.5×10−8 Ω⋅m\rho = \frac{0.50 \times 2.0 \times 10^{-7}}{4.0} = 2.5 \times 10^{-8}\,\Omega\cdot\text{m}. Expressed in units of 10−8 Ω⋅m10^{-8}\,\Omega\cdot\text{m}, the answer is 2.52.5.

Step-by-Step Solution

1
Identify the relevant formula linking resistance, resistivity, length, and area.
R=ρLAR = \frac{\rho L}{A}
The resistance of a uniform conductor is directly proportional to its length and inversely proportional to its cross-sectional area.
2
Rearrange the equation to solve for resistivity ρ\rho.
ρ=R⋅AL\rho = \frac{R \cdot A}{L}
Isolating ρ\rho allows direct evaluation using the given quantitative values.
3
Substitute the known numerical values into the equation.
ρ=0.50×(2.0×10−7)4.0=2.5×10−8 Ω⋅m\rho = \frac{0.50 \times (2.0 \times 10^{-7})}{4.0} = 2.5 \times 10^{-8}\,\Omega\cdot\text{m}
Performing the algebraic calculation gives the resistivity in SI units.

Key Concept

Electrical Resistivity and Conductor Dimensions
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Topics

Electric Current and Resistance

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