To determine the cross-sectional area (\( A \)) when resistivity (\( \rho \)), resistance (\( R \)), and length (\( L \)) are known, use the formula:
\[ A = \dfrac{\rho \cdot L}{R} \]
where:
- \( A \) is the cross-sectional area (in square meters, \( \text{m}^2 \)),
- \( \rho \) is the resistivity (in ohm meters, \( \Omega \cdot \text{m} \)),
- \( R \) is the resistance (in ohms, \( \Omega \)),
- \( L \) is the length (in meters, \( \text{m} \)).
Problem 1: Area of a Copper Conductor
Scenario: A copper conductor has a resistivity of \( 1.68 \times 10^{-8} \, \Omega \cdot \text{m} \), a resistance of \( 0.2 \, \Omega \), and a length of \( 5 \, \text{m} \). What is its cross-sectional area?
Calculation:
1. Given:
\[ \rho = 1.68 \times 10^{-8} \, \Omega \cdot \text{m} \]
\[ R = 0.2 \, \Omega \]
\[ L = 5 \, \text{m} \]
2. Substitute into the Area Formula:
\[ A = \dfrac{\rho \cdot L}{R} \]
\[ A = \dfrac{1.68 \times 10^{-8} \cdot 5}{0.2} \]
3. Calculate:
\[ A = \dfrac{8.4 \times 10^{-8}}{0.2} = 4.2 \times 10^{-7} \, \text{m}^2 \]
Answer: The cross-sectional area of the copper conductor is \( 4.2 \times 10^{-7} \, \text{m}^2 \).
Problem 2: Area of a Gold Wire
Scenario: A gold wire has a resistivity of \( 2.44 \times 10^{-8} \, \Omega \cdot \text{m} \), a resistance of \( 0.5 \, \Omega \), and a length of \( 3 \, \text{m} \). Determine its cross-sectional area.
Calculation:
1. Given:
\[ \rho = 2.44 \times 10^{-8} \, \Omega \cdot \text{m} \]
\[ R = 0.5 \, \Omega \]
\[ L = 3 \, \text{m} \]
2. Substitute into the Area Formula:
\[ A = \dfrac{\rho \cdot L}{R} \]
\[ A = \dfrac{2.44 \times 10^{-8} \cdot 3}{0.5} \]
3. Calculate:
\[ A = \dfrac{7.32 \times 10^{-8}}{0.5} = 1.464 \times 10^{-7} \, \text{m}^2 \]
Answer: The cross-sectional area of the gold wire is \( 1.464 \times 10^{-7} \, \text{m}^2 \).
Problem 3: Area of a Tungsten Filament
Scenario: A tungsten filament has a resistivity of \( 5.6 \times 10^{-8} \, \Omega \cdot \text{m} \), a resistance of \( 1 \, \Omega \), and a length of \( 0.1 \, \text{m} \). Find its cross-sectional area.
Calculation:
1. Given:
\[ \rho = 5.6 \times 10^{-8} \, \Omega \cdot \text{m} \]
\[ R = 1 \, \Omega \]
\[ L = 0.1 \, \text{m} \]
2. Substitute into the Area Formula:
\[ A = \dfrac{\rho \cdot L}{R} \]
\[ A = \dfrac{5.6 \times 10^{-8} \cdot 0.1}{1} \]
3. Calculate:
\[ A = \dfrac{5.6 \times 10^{-9}}{1} = 5.6 \times 10^{-9} \, \text{m}^2 \]
Answer: The cross-sectional area of the tungsten filament is \( 5.6 \times 10^{-9} \, \text{m}^2 \).
