Question

a)What is the resistance of a nichrome wire at 0.0 degrees C if its resistance 100.00 ohms at 11.5 degrees C?

 

(b)What is the resistance of a carbon rod at 25.8 degrees C if its resistance is 0.0160 ohms at 0.0 degrees C?

Answer 1

Concepts and reasons

The concept required to solve this question is the variation of resistance with temperature. So, first, find the temperature difference and then find the resistance at 0 degrees celsius by using the expression for the variation of the resistance with temperature.

Fundamentals

The expression for the variation of resistance with temperature is given by \(R=R_{\mathrm{o}}\left(1+\alpha\left(T-T_{\mathrm{o}}\right)\right)\)

Here, \(\mathrm{R}\) is the material's resistance at \(T\) degrees Celsius, \(R_{\mathrm{o}}\) is the resistance of the same material at \(T_{\mathrm{o}}\) degrees Celsius, and \(\alpha\) is the temperature coefficient of the material. The expression for the change in the temperature is given by, \(\Delta T=T-T_{\mathrm{o}}\)

Here, \(T_{\mathrm{o}}\) is the initial temperature, and \(\mathrm{T}\) is the final temperature.

(a) The change in the temperature is given by the following expression:

\(\Delta T=T-T_{\mathrm{o}}\)

Here, the nichrome wire was initially at temperature \(T_{\mathrm{o}}\) when its resistance was recorded; the final temperature is \(\mathrm{T}\). Substitute \(11.50 \mathrm{C}\) for \(T_{\mathrm{o}}\) and \(0 \circ \mathrm{C}\) for \(\mathrm{T}\) in the above expression.

$$ \begin{aligned} \Delta T=&\left(0^{\circ} \mathrm{C}\right)-\left(11.5^{\circ} \mathrm{C}\right) \\ &=-11.5^{\circ} \mathrm{C} \end{aligned} $$

The negative change in the temperature means that the temperature of the nichrome wire is decreased.

The expression for the variation of resistance with temperature is given as follows:

\(R=R_{0}\left(1+\alpha\left(T-T_{\mathrm{o}}\right)\right)\)

Every material has a different temperature coefficient depending on its properties. Substitute \(4.0 \times 10^{-4} /{ }^{\circ} \mathrm{C}\) for \(\alpha,-11.5^{\circ} \mathrm{C}\) for \(\left(T-T_{\mathrm{o}}\right)\), and \(100 \Omega\) for \(R_{\mathrm{o}}\) in the above expression.

$$ \begin{array}{c} R=(100 \Omega)\left(1+\left(4.0 \times 10^{-4} /{ }^{\circ} \mathrm{C}\right)\left(-11.5^{\circ} \mathrm{C}\right)\right) \\ =99.54 \Omega \end{array} $$

The resistance of the nichrome wire at a higher temperature was higher than the resistance at a lower temperature. This is because the resistance of the nichrome wire increases with increasing temperature.

(b) The change in the temperature is given by the following expression:

\(\Delta T=T-T_{\mathrm{o}}\)

Here, the carbon rod was initially at temperature. When its resistance was recorded, the final temperature is \(\mathrm{T}\). Substitute 00 C for \(T_{\mathrm{o}}\) and \(25.80 \mathrm{C}\) for \(\mathrm{T}\) in the above expression.

$$ \begin{aligned} \Delta T=&\left(25.8^{\circ} \mathrm{C}\right)-\left(0^{\circ} \mathrm{C}\right) \\ &=25.8^{\circ} \mathrm{C} \end{aligned} $$

The positive change in the temperature means that the temperature of the carbon rod is increased.

The expression for the temperature coefficient \((\alpha)\) of resistance of any material is given as follows:

\(R=R_{\mathrm{o}}\left(1+\alpha\left(T-T_{\mathrm{o}}\right)\right)\)

Substitute \(-5.0 \times 10^{-4} /{ }^{\circ} \mathrm{C}\) for \(\alpha, 25.8^{\circ} \mathrm{C}\) for \(\left(T-T_{\mathrm{o}}\right)\), and \(0.016 \Omega\) for \(R_{\mathrm{o}}\) in the above expression.

$$ \begin{array}{c} R=(0.016 \Omega)\left(1+\left(-5.0 \times 10^{-4} /{ }^{\circ} \mathrm{C}\right)\left(25.8^{\circ} \mathrm{C}\right)\right) \\ =1.57 \times 10^{-2} \Omega \end{array} $$

The resistance of the carbon rod at a higher temperature was lower than the resistance at a lower temperature. The resistance of the carbon rod decreases with increasing temperature.

Part a The resistance of nichrome wire at \(0^{\circ} \mathrm{C}\) is \(99.54 \Omega\).

Part b The resistance of the carbon rod at \(25.8^{\circ} \mathrm{C}\) is \(1.57 \times 10^{-2} \Omega\).