How To Find Resistance Using Length And Cross-Sectional Area

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Resistance (\( R \)) is a measure of how much a material opposes the flow of electric current. It can be determined using the formula:

\[ R = \dfrac{\rho \cdot L}{A} \]


where:

  • \( R \) is the resistance (in ohms, \( \Omega \)),
  • \( \rho \) is the resistivity (in ohm meters, \( \Omega \cdot \text{m} \)),
  • \( L \) is the length of the material (in meters, m),
  • \( A \) is the cross-sectional area (in square meters, \( \text{m}^2 \)).


We’ll illustrate how to find resistance with five practical examples.


Example 1: Resistance of a Copper Wire

Scenario: A copper wire has a length of \( 2 \, \text{m} \), a cross-sectional area of \( 1 \, \text{mm}^2 \) (\( 1 \times 10^{-6} \, \text{m}^2 \)), and a resistivity of \( 1.68 \times 10^{-8} \, \Omega \cdot \text{m} \). What is the resistance?


Step-by-Step Calculation:

1. Given:

  \[ \rho = 1.68 \times 10^{-8} \, \Omega \cdot \text{m} \]

  \[ L = 2 \, \text{m} \]

  \[ A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \]


2. Substitute Values into the Resistance Formula:

  \[ R = \dfrac{\rho \cdot L}{A} \]

  \[ R = \dfrac{1.68 \times 10^{-8} \cdot 2}{1 \times 10^{-6}} \]


3. Perform the Calculation:

  \[ R = 3.36 \times 10^{-2} \, \Omega \]


Final Value

The resistance of the copper wire is:


\[ R = 0.0336 \, \Omega \]


Example 2: Resistance of an Aluminum Rod

Scenario: An aluminum rod has a length of \( 5 \, \text{m} \), a cross-sectional area of \( 2 \, \text{mm}^2 \) (\( 2 \times 10^{-6} \, \text{m}^2 \)), and a resistivity of \( 2.65 \times 10^{-8} \, \Omega \cdot \text{m} \). Calculate the resistance.


Step-by-Step Calculation:

1. Given:

  \[ \rho = 2.65 \times 10^{-8} \, \Omega \cdot \text{m} \]

  \[ L = 5 \, \text{m} \]

  \[ A = 2 \times 10^{-6} \, \text{m}^2 \]


2. Substitute Values into the Resistance Formula:

  \[ R = \dfrac{\rho \cdot L}{A} \]

  \[ R = \dfrac{2.65 \times 10^{-8} \cdot 5}{2 \times 10^{-6}} \]


3. Perform the Calculation:

  \[ R = 6.625 \times 10^{-2} \, \Omega \]


Final Value

The resistance of the aluminum rod is:

\[ R = 0.06625 \, \Omega \]


Example 3: Resistance of a Silver Cable

Scenario: A silver cable has a length of \( 10 \, \text{m} \), a cross-sectional area of \( 0.5 \, \text{mm}^2 \) (\( 0.5 \times 10^{-6} \, \text{m}^2 \)), and a resistivity of \( 1.59 \times 10^{-8} \, \Omega \cdot \text{m} \). What is the resistance?


Step-by-Step Calculation:

1. Given:

  \[ \rho = 1.59 \times 10^{-8} \, \Omega \cdot \text{m} \]

  \[ L = 10 \, \text{m} \]

  \[ A = 0.5 \times 10^{-6} \, \text{m}^2 \]


2. Substitute Values into the Resistance Formula:

  \[ R = \dfrac{\rho \cdot L}{A} \]

  \[ R = \dfrac{1.59 \times 10^{-8} \cdot 10}{0.5 \times 10^{-6}} \]


3. Perform the Calculation:

  \[ R = 3.18 \times 10^{-1} \, \Omega \]


Final Value

The resistance of the silver cable is:

\[ R = 0.318 \, \Omega \]


Example 4: Resistance of an Iron Bar

Scenario: An iron bar has a length of \( 3 \, \text{m} \), a cross-sectional area of \( 3 \, \text{mm}^2 \) (\( 3 \times 10^{-6} \, \text{m}^2 \)), and a resistivity of \( 9.71 \times 10^{-8} \, \Omega \cdot \text{m} \). Calculate the resistance.


Step-by-Step Calculation:

1. Given:

  \[ \rho = 9.71 \times 10^{-8} \, \Omega \cdot \text{m} \]

  \[ L = 3 \, \text{m} \]

  \[ A = 3 \times 10^{-6} \, \text{m}^2 \]


2. Substitute Values into the Resistance Formula:

  \[ R = \dfrac{\rho \cdot L}{A} \]

  \[ R = \dfrac{9.71 \times 10^{-8} \cdot 3}{3 \times 10^{-6}} \]


3. Perform the Calculation:

  \[ R = 9.71 \times 10^{-2} \, \Omega \]


Final Value

The resistance of the iron bar is:

\[ R = 0.0971 \, \Omega \]


Example 5: Resistance of a Gold Wire

Scenario: A gold wire has a length of \( 0.5 \, \text{m} \), a cross-sectional area of \( 0.1 \, \text{mm}^2 \) (\( 0.1 \times 10^{-6} \, \text{m}^2 \)), and a resistivity of \( 2.44 \times 10^{-8} \, \Omega \cdot \text{m} \). What is the resistance?


Step-by-Step Calculation:

1. Given:

  \[ \rho = 2.44 \times 10^{-8} \, \Omega \cdot \text{m} \]

  \[ L = 0.5 \, \text{m} \]

  \[ A = 0.1 \times 10^{-6} \, \text{m}^2 \]


2. Substitute Values into the Resistance Formula:

  \[ R = \dfrac{\rho \cdot L}{A} \]

  \[ R = \dfrac{2.44 \times 10^{-8} \cdot 0.5}{0.1 \times 10^{-6}} \]


3. Perform the Calculation:

  \[ R = 1.22 \times 10^{-1} \, \Omega \]


Final Value

The resistance of the gold wire is:

\[ R = 0.122 \, \Omega \]


Summary

To find the resistance (\( R \)) given the resistivity (\( \rho \)), length (\( L \)), and cross-sectional area (\( A \)), use the formula:

\[ R = \dfrac{\rho \cdot L}{A} \]


In the examples provided:

1. Copper wire: \( R = 0.0336 \, \Omega \)

2. Aluminum rod: \( R = 0.06625 \, \Omega \)

3. Silver cable: \( R = 0.318 \, \Omega \)

4. Iron bar: \( R = 0.0971 \, \Omega \)

5. Gold wire: \( R = 0.122 \, \Omega \)

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