To determine the electrical resistivity (\( \rho \)) when resistance (\( R \)), cross-sectional area (\( A \)), and length (\( L \)) are known, use the formula:
\[ \rho = \dfrac{R \cdot A}{L} \]
where:
- \( \rho \) is the resistivity (in ohm meters, \( \Omega \cdot \text{m} \)),
- \( R \) is the resistance (in ohms, \( \Omega \)),
- \( A \) is the cross-sectional area (in square meters, \( \text{m}^2 \)),
- \( L \) is the length (in meters, \( \text{m} \)).
Problem 1: Resistivity of a Copper Wire
Scenario: A copper wire has a resistance of \( 0.5 \, \Omega \), a cross-sectional area of \( 1.5 \times 10^{-6} \, \text{m}^2 \), and a length of \( 10 \, \text{m} \). What is its resistivity?
Calculation:
1. Given:
\[ R = 0.5 \, \Omega \]
\[ A = 1.5 \times 10^{-6} \, \text{m}^2 \]
\[ L = 10 \, \text{m} \]
2. Substitute into the Resistivity Formula:
\[ \rho = \dfrac{R \cdot A}{L} \]
\[ \rho = \dfrac{0.5 \cdot 1.5 \times 10^{-6}}{10} \]
3. Calculate:
\[ \rho = \dfrac{7.5 \times 10^{-7}}{10} = 7.5 \times 10^{-8} \, \Omega \cdot \text{m} \]
Answer: The resistivity of the copper wire is \( 7.5 \times 10^{-8} \, \Omega \cdot \text{m} \).
Problem 2: Resistivity of an Aluminum Rod
Scenario: An aluminum rod has a resistance of \( 2 \, \Omega \), a cross-sectional area of \( 3 \times 10^{-4} \, \text{m}^2 \), and a length of \( 1 \, \text{m} \). Find its resistivity.
Calculation:
1. Given:
\[ R = 2 \, \Omega \]
\[ A = 3 \times 10^{-4} \, \text{m}^2 \]
\[ L = 1 \, \text{m} \]
2. Substitute into the Resistivity Formula:
\[ \rho = \dfrac{R \cdot A}{L} \]
\[ \rho = \dfrac{2 \cdot 3 \times 10^{-4}}{1} \]
3. Calculate:
\[ \rho = 6 \times 10^{-4} \, \Omega \cdot \text{m} \]
Answer: The resistivity of the aluminum rod is \( 6 \times 10^{-4} \, \Omega \cdot \text{m} \).
Problem 3: Resistivity of a Graphite Pencil
Scenario: A graphite pencil lead has a resistance of \( 0.1 \, \Omega \), a cross-sectional area of \( 5 \times 10^{-8} \, \text{m}^2 \), and a length of \( 0.05 \, \text{m} \). Determine its resistivity.
Calculation:
1. Given:
\[ R = 0.1 \, \Omega \]
\[ A = 5 \times 10^{-8} \, \text{m}^2 \]
\[ L = 0.05 \, \text{m} \]
2. Substitute into the Resistivity Formula:
\[ \rho = \dfrac{R \cdot A}{L} \]
\[ \rho = \dfrac{0.1 \cdot 5 \times 10^{-8}}{0.05} \]
3. Calculate:
\[ \rho = \dfrac{5 \times 10^{-9}}{0.05} = 1 \times 10^{-7} \, \Omega \cdot \text{m} \]
Answer: The resistivity of the graphite pencil lead is \( 1 \times 10^{-7} \, \Omega \cdot \text{m} \).
