Question

The resistance of a conductor at $15^\circ\text{C}$ is 16 $\Omega$ and at $100^\circ\text{C}$ is 20 $\Omega$. What will be the temperature coefficient of resistance of the conductor?

• A. $0.003^\circ\text{C}^{-1}$
• B. $0.010^\circ\text{C}^{-1}$
• C. $0.033^\circ\text{C}^{-1}$
• D. $0.042^\circ\text{C}^{-1}$

Hint:

The formula $R_2 = R_1(1 + \alpha \Delta T)$ assumes $\alpha$ is constant over the temperature range, which is a good approximation for many conductors over moderate temperature changes. Always use the initial resistance $R_1$ in the denominator.

Answer 1

The Correct Option is A

The relationship between resistance and temperature for a conductor is given by the formula:
$R_2 = R_1 (1 + \alpha (T_2 - T_1))$
Where:
$R_1$ is the resistance at temperature $T_1$.
$R_2$ is the resistance at temperature $T_2$.
$\alpha$ is the temperature coefficient of resistance.
We are given:
$T_1 = 15^\circ\text{C}$, $R_1 = 16 \Omega$.
$T_2 = 100^\circ\text{C}$, $R_2 = 20 \Omega$.
We need to find $\alpha$. Let's rearrange the formula to solve for $\alpha$.
$R_2 - R_1 = R_1 \alpha (T_2 - T_1)$.
$\alpha = \frac{R_2 - R_1}{R_1 (T_2 - T_1)}$.
Now, substitute the given values into the equation.
$\alpha = \frac{20 - 16}{16 (100 - 15)}$.
$\alpha = \frac{4}{16 \times 85}$.
$\alpha = \frac{1}{4 \times 85} = \frac{1}{340}$.
$\alpha \approx 0.00294 \ ^\circ\text{C}^{-1}$.
This value is approximately $0.003 \ ^\circ\text{C}^{-1}$.