Question:

Find the temperature at which the resistance of a conductor increases by 25% of its value at \( 27^\circ \text{C} \). The temperature coefficient of resistance of the conductor is \( 2.0 \times 10^{-4} \, \text{C}^{-1} \).

Show Hint

Use the formula \( R_T = R_0 \left( 1 + \alpha (T - T_0) \right) \) to calculate changes in resistance with temperature.
Updated On: Feb 13, 2025
Hide Solution
collegedunia
Verified By Collegedunia

Solution and Explanation

The resistance of a conductor at a temperature \( T \) is given by: \[ R_T = R_0 \left( 1 + \alpha (T - T_0) \right), \] where:   \( R_T = 1.25R_0 \) (final resistance is 25% greater than the initial resistance),  \( R_0 \) is the resistance at \( T_0 = 27^\circ \text{C} \), \( \alpha = 2.0 \times 10^{-4} \, \text{C}^{-1} \) (temperature coefficient of resistance). Substitute \( R_T = 1.25R_0 \) into the equation: \[ 1.25R_0 = R_0 \left( 1 + \alpha (T - T_0) \right). \] Simplify: \[ 1.25 = 1 + \alpha (T - 27). \] Rearrange: \[ \alpha (T - 27) = 0.25. \] Substitute \( \alpha = 2.0 \times 10^{-4} \): \[ (2.0 \times 10^{-4}) (T - 27) = 0.25. \] Solve for \( T \): \[ T - 27 = \frac{0.25}{2.0 \times 10^{-4}} = 1250. \] \[ T = 27 + 1250 = 1277^\circ \text{C}. \] Answer: The temperature is \( 1277^\circ \text{C} \).

Was this answer helpful?
0
0

Questions Asked in CBSE CLASS XII exam

View More Questions