Wire Resistance Principle

Wire resistance is the opposition to the flow of electric current through a conductor, caused by collisions between electrons and the atomic structure of the material.

Formula: \( R = \rho \cdot \frac{L}{A} \)

Where:

• R is the resistance of the wire (measured in Ohms, Ω).
• ρ (rho) is the resistivity of the material (measured in Ω·m).
• L is the length of the wire (measured in meters).
• A is the cross-sectional area of the wire (measured in square meters).

Key Points:

• Resistance increases with the length of the wire.
• Resistance decreases as the cross-sectional area increases.
• The material's resistivity depends on its type and temperature.
• Wires with higher resistivity materials, such as carbon, exhibit greater resistance compared to metals like copper or aluminum.

Example:

Input:

• Material: Annealed Copper \( \rho = 1.68 \times 10^{-8} \, \Omega \cdot \text{m} \)
• Length: 10 \( \text{m} \)
• Diameter: 2 \( \text{mm} \)

Steps:

1. Convert diameter to meters: \( 2 \, \text{mm} = 0.002 \, \text{m} \)
2. Calculate radius: \( r = \frac{\text{Diameter}}{2} = \frac{0.002}{2} = 0.001 \, \text{m} \)
3. Calculate cross-sectional area: \( A = \pi \cdot r^2 = \pi \cdot (0.001)^2 = 3.14 \times 10^{-6} \, \text{m}^2 \)
4. Calculate resistance: \( R = \rho \cdot \frac{L}{A} = (1.68 \times 10^{-8}) \cdot \frac{10}{3.14 \times 10^{-6}} = 0.0535 \, \Omega \)
5. Calculate conductance: \( G = \frac{1}{R} = \frac{1}{0.0535} \approx 18.69 \, \text{S} \)

Output: Cross-sectional Area = \( 3.14 \, \text{mm}^2 \), Conductance = \( 18.69 \, \text{S} \)