Question

A heating element using nichrome is connected to a 220 V supply. Initially it draws a current of 2.9 A. After some time, the current attains a steady value of 2.5 A. Find the steady temperature of the heating element if the room temperature is 27 \(^{\circ}\)C. The temperature coefficient of resistance of nichrome is \(1.7 \times 10^{-4}\ ^{\circ}\text{C}^{-1}\).

Hint:

For heating elements: - Use \(R = \frac{V}{I}\) - Then apply \(R_T = R_0(1 + \alpha \Delta T)\) - Always compute resistance ratio first for faster solving.

Answer 1

Concept: The resistance of a conductor increases with temperature according to: \[ R_T = R_0 \left(1 + \alpha \Delta T \right) \] where:
• \(R_0\) = resistance at initial temperature \(T_0\)
• \(R_T\) = resistance at temperature \(T\)
• \(\alpha\) = temperature coefficient of resistance
• \(\Delta T = T - T_0\) Since voltage is constant: \[ R = \frac{V}{I} \]

Step 1: Initial resistance
\[ R_0 = \frac{V}{I_1} = \frac{220}{2.9} \] \[ R_0 \approx 75.86\ \Omega \]

Step 2: Final (steady) resistance
\[ R_T = \frac{V}{I_2} = \frac{220}{2.5} \] \[ R_T = 88\ \Omega \]

Step 3: Apply temperature dependence of resistance
\[ \frac{R_T}{R_0} = 1 + \alpha (T - T_0) \] Substitute values: \[ \frac{88}{75.86} = 1 + (1.7 \times 10^{-4})(T - 27) \]

Step 4: Simplify left-hand side
\[ \frac{88}{75.86} \approx 1.160 \] So: \[ 1.160 = 1 + (1.7 \times 10^{-4})(T - 27) \]

Step 5: Solve for temperature
\[ 0.160 = (1.7 \times 10^{-4})(T - 27) \] \[ T - 27 = \frac{0.160}{1.7 \times 10^{-4}} \] \[ T - 27 \approx 941.18 \] \[ T \approx 968^{\circ}\text{C} \] Final Answer: \[ \boxed{T \approx 9.7 \times 10^2\ ^{\circ}\text{C}} \]