Question

The resistance of a wire at $30^{\circ}\text{C}$ and $40^{\circ}\text{C}$ are respectively $5\ \Omega$ and $6\ \Omega$. The temperature coefficient of resistance is:

• A. 0.04
• B. 0.05
• C. 0.02
• D. 0.03
• E. 0.01

Hint:

$\alpha = \frac{R_2 - R_1}{R_1 \Delta T}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The resistance of most conducting materials changes with temperature. For many materials, over a certain range, this change is approximately linear. The temperature coefficient of resistance (\( \alpha \)) quantifies this change.
Step 2: Key Formula or Approach:
The resistance \( R_T \) at a temperature T is related to the resistance \( R_0 \) at a reference temperature \( T_0 \) by the formula: \[ R_T = R_0[1 + \alpha(T - T_0)] \] We are given resistances at two different temperatures, so we can set up a system of two equations. Let \( T_0 \) be a reference temperature, say 0°C, and \( R_0 \) be the resistance at 0°C. (1) \( R_{30} = 5 = R_0[1 + \alpha(30 - 0)] = R_0(1 + 30\alpha) \) (2) \( R_{40} = 6 = R_0[1 + \alpha(40 - 0)] = R_0(1 + 40\alpha) \) We can solve this system for \( \alpha \). A common method is to divide the two equations. Step 3: Detailed Explanation:
Divide equation (2) by equation (1): \[ \frac{6}{5} = \frac{R_0(1 + 40\alpha)}{R_0(1 + 30\alpha)} \] Cancel \( R_0 \): \[ \frac{6}{5} = \frac{1 + 40\alpha}{1 + 30\alpha} \] Now, cross-multiply to solve for \( \alpha \): \[ 6(1 + 30\alpha) = 5(1 + 40\alpha) \] \[ 6 + 180\alpha = 5 + 200\alpha \] Rearrange the terms to isolate \( \alpha \): \[ 6 - 5 = 200\alpha - 180\alpha \] \[ 1 = 20\alpha \] \[ \alpha = \frac{1}{20} = 0.05 \] The units are per degree Celsius (\( /^\circ\text{C} \) or \( (^\circ\text{C})^{-1} \)). This calculation matches option (B). My initial check was incorrect. Step 4: Final Answer:
The temperature coefficient of resistance is 0.05 per degree Celcius.