Question

The resistance of a heating element is found to be \(120\Omega\) at room temperature which is \(20^\circ C\). If the temperature coefficient of the material of the resistor is \(1.6 \times 10^{-4}\,^\circ C^{-1}\) and the resistance is found to be \(160\Omega\), the temperature of the element is:

• A. \(1203^\circ C\)
• B. \(2083^\circ C\)
• C. \(2310^\circ C\)
• D. \(2013^\circ C\)

Hint:

Always use \(R = R_0[1 + \alpha (T - T_0)]\) for temperature problems and carefully handle small values of \(\alpha\).

Answer 1

The Correct Option is B

Step 1: Use resistance-temperature relation.
\[ R = R_0 \left[1 + \alpha (T - T_0)\right] \]

Step 2: Substitute given values.
\[ R = 160\Omega,\quad R_0 = 120\Omega \]
\[ \alpha = 1.6 \times 10^{-4},\quad T_0 = 20^\circ C \]
\[ 160 = 120 \left[1 + 1.6 \times 10^{-4}(T - 20)\right] \]

Step 3: Divide both sides by 120.
\[ \frac{160}{120} = 1 + 1.6 \times 10^{-4}(T - 20) \]
\[ \frac{4}{3} = 1 + 1.6 \times 10^{-4}(T - 20) \]

Step 4: Simplify.
\[ \frac{1}{3} = 1.6 \times 10^{-4}(T - 20) \]

Step 5: Solve for temperature.
\[ T - 20 = \frac{1/3}{1.6 \times 10^{-4}} \]
\[ T - 20 = \frac{1}{3 \times 1.6 \times 10^{-4}} \]
\[ T - 20 = \frac{1}{4.8 \times 10^{-4}} \]
\[ T - 20 \approx 2083 \]
\[ T \approx 2103^\circ C \]

Step 6: Match with closest option.
Closest value is \(2083^\circ C\).

Step 7: Final answer.
\[ \boxed{2083^\circ C} \]