Question

The resistance of a wire at 25°C is 10.0 \( \Omega \). When heated to 125°C, its resistance becomes 10.5 \( \Omega \). Find (i) the temperature coefficient of resistance of the wire, and (ii) the resistance of the wire at 425°C.

Hint:

To calculate the resistance at a different temperature, use the formula \( R_t = R_0 [1 + \alpha (t - t_0)] \), where \( \alpha \) is the temperature coefficient of resistance. This formula assumes the material's temperature coefficient is constant over the given temperature range.

Answer 1

The temperature dependence of the resistance of a conductor is given by the formula: \[ R_t = R_0 [1 + \alpha (t
- t_0)] \] Where:
- \( R_t \) is the resistance at temperature \( t \),
- \( R_0 \) is the resistance at a reference temperature \( t_0 \),
- \( \alpha \) is the temperature coefficient of resistance,
- \( t \) is the temperature at which the resistance is measured,
- \( t_0 \) is the reference temperature. Given:
- \( R_0 = 10.0 \, \Omega \) (resistance at 25°C),
- \( R_{125} = 10.5 \, \Omega \) (resistance at 125°C),
- \( t_0 = 25°C \),
- \( t = 125°C \). (i) To find the temperature coefficient of resistance \( \alpha \): Using the formula: \[ R_{125} = R_0 [1 + \alpha (125
- 25)] \] Substitute the known values: \[ 10.5 = 10.0 [1 + \alpha (100)] \] Simplifying: \[ 1.05 = 1 + 100 \alpha \] Solving for \( \alpha \): \[ 100 \alpha = 0.05 \] \[ \alpha = \frac{0.05}{100} = 0.0005 \, \text{per°C} \] Thus, the temperature coefficient of resistance is \( \alpha = 0.0005 \, \text{per°C} \). (ii) To find the resistance of the wire at 425°C, use the formula again: \[ R_{425} = R_0 [1 + \alpha (425
- 25)] \] Substitute the known values: \[ R_{425} = 10.0 [1 + 0.0005 (425
- 25)] \] \[ R_{425} = 10.0 [1 + 0.0005 \times 400] \] \[ R_{425} = 10.0 [1 + 0.2] \] \[ R_{425} = 10.0 \times 1.2 = 12.0 \, \Omega \] Thus, the resistance of the wire at 425°C is \( 12.0 \, \Omega \).