Lecture on Resistivity and Resistance of Wires

Resistance Calculation

• Formula: ( R = \frac{\rho \cdot L}{A} )
  • ( \rho ) (rho) is the resistivity
  • ( L ) is the length of the wire
  • ( A ) is the cross-sectional area

Factors Affecting Resistance

Length Effect

• Longer wires have more resistance.
• Reason: As length ( L ) increases, ( R ) increases because ( L ) is in the numerator.

Thickness Effect

• Thinner wires have more resistance if length is constant.
• Reason: As cross-sectional area ( A ) increases, ( R ) decreases because ( A ) is in the denominator.
• Analogy: A thin wire is like a one-lane highway (less flow, more resistance), a thick wire is like a seven-lane highway (more flow, less resistance).

Resistivity ( \rho )

• Property of the material.
• Metals: Low resistivity ( \approx 10^{-8} )
  • Good conductors: Silver (( 1.59 \times 10^{-8} )), Copper (( 1.68 \times 10^{-8} ))
  • Silver is a better conductor than copper due to lower resistivity.

Conductivity vs. Resistivity

• Inversely related. Lower resistivity means better conductivity.

Semiconductors and Insulators

• Semiconductors: Higher resistivity (e.g., carbon graphite: ( 3-60 \times 10^{-5} ))
• Insulators: Very high resistivity (e.g., glass: ( 10^9 - 10^{12} ))

Temperature Dependence

• Equation: ( \rho_T = \rho_0 \times (1 + \alpha (T - T_0)) )
• Metals:
  • Higher temperature -> higher resistivity.
  • Conductivity decreases with rising temperature.
  • Can become superconductors at very low temperatures.
• Semiconductors:
  • Higher temperature -> lower resistivity.
  • Conductivity increases with rising temperature.

Applications

• Used in digital thermometers to measure temperature based on resistance changes.

Problem Examples

Example 1

• Calculate resistance of a 15m copper wire with a 3mm radius.
• Given ( \rho = 1.68 \times 10^{-8} ), ( \alpha = 0.0068 ).

Calculation Steps:

1. Resistance at 20°C: ( R = \frac{\rho \times L}{\pi r^2} \approx 0.008913 \Omega )

2. Resistance at 50°C:

   • ( R = R_0 (1 + \alpha (T - T_0)) \approx 0.01073 \Omega )

3. Voltage Drop for 200 mA Current at 20°C:

   • Converted to amps: 0.2 A
   • Voltage drop ( = I \times R \approx 0.0017826 V ) or 1.783 mV

4. Voltage Drop per Meter:

   • ( \approx 0.119 \text{mV/m} )

Example 2

• Copper wire connected to 12V battery, initial current 0.45 A at 20°C, new current 0.41 A.
• Determine if the new temperature is higher or lower.

Reasoning and Calculation:

• Decreased current implies increased resistance, thus higher temperature.
• Calculated new temperature: 34.34°C

---

These notes summarize the core concepts and example problems discussed in the lecture on resistivity and resistance of wires.