The resistivity of material is 10 ohm m at 30 C when the temperature drops to 28 then the resistivity drops by 2. The temperature coefficient of resistivity of material is

To determine the temperature coefficient of resistivity for the material, we can use the formula that relates resistivity, temperature change, and the temperature coefficient. The temperature coefficient of resistivity (\(\alpha\)) quantifies how much the resistivity of a material changes with temperature. The formula is given by:

Understanding the Relationship

The resistivity (\(\rho\)) of a material at a specific temperature can be expressed as:

\(\rho(T) = \rho_0 [1 + \alpha (T - T_0)]\)

Where:

• \(\rho(T)\) = resistivity at temperature \(T\)
• \(\rho_0\) = resistivity at reference temperature \(T_0\)
• \(\alpha\) = temperature coefficient of resistivity
• \(T\) = new temperature
• \(T_0\) = reference temperature

Given Data

From your question, we have:

• Initial resistivity at 30°C, \(\rho_0 = 10 \, \Omega \cdot m\)
• Resistivity at 28°C, \(\rho(28) = 10 \, \Omega \cdot m - 2 \, \Omega \cdot m = 8 \, \Omega \cdot m\)
• Temperature change, \(\Delta T = 28°C - 30°C = -2°C\)

Calculating the Temperature Coefficient

We can rearrange the resistivity formula to solve for \(\alpha\):

\(\alpha = \frac{\rho(T) - \rho_0}{\rho_0 \cdot (T - T_0)}\)

Substituting the known values into the equation:

\(\alpha = \frac{8 \, \Omega \cdot m - 10 \, \Omega \cdot m}{10 \, \Omega \cdot m \cdot (-2°C)}\)

This simplifies to:

\(\alpha = \frac{-2 \, \Omega \cdot m}{10 \, \Omega \cdot m \cdot (-2)}\)

\(\alpha = \frac{-2}{-20} = \frac{1}{10} = 0.1 \, °C^{-1}\)

Final Result

The temperature coefficient of resistivity for the material is \(0.1 \, °C^{-1}\). This means that for every degree Celsius increase in temperature, the resistivity of the material increases by 10% of its value at 30°C. Understanding this relationship is crucial in applications where temperature variations can significantly affect the performance of electrical materials.