Question

The resistance of platinum wire at 0°C is 2Ω and 6.89Ω at 80°C. The temperature coefficient of resistance of the wire is:

• A. \(3 \times 10^{-3} \degree C^{-1}\)
• B. \(3 \times 10^{-2} \degree C^{-1}\)
• C. \(3 \times 10^{-1} \degree C^{-1}\)
• D. \(3 \times 10^{-4} \degree C^{-1}\)

Answer 1

The Correct Option is B

To find the temperature coefficient of resistance (\( \alpha \)) for the platinum wire, we use the formula for resistance change with temperature: \(R_t = R_0 (1 + \alpha \Delta T)\), where \(R_t\) is the resistance at temperature \( t \), \(R_0\) is the resistance at 0°C, and \(\Delta T\) is the change in temperature.

Given: \(R_0 = 2\ \Omega\), \(R_t = 6.89\ \Omega\) at \( 80°C \), \(\Delta T = 80 - 0 = 80°C\).

Substitute in the formula: \(6.89 = 2(1 + \alpha \times 80)\). 

Simplifying the equation: \(6.89 = 2 + 160\alpha\).

Solve for \(\alpha\): \(6.89 - 2 = 160\alpha\), so \(4.89 = 160\alpha\).

Therefore, \(\alpha = \frac{4.89}{160}\)

Calculate: \(\alpha = 0.0305625 \degree C^{-1} \approx 3 \times 10^{-2} \degree C^{-1}\).

Thus, the temperature coefficient of resistance is \(3 \times 10^{-2} \degree C^{-1}\).