Chapter 27, Problem 27.75AP

Review. When a straight wire is warmed, its resistance is given by R = R₀,[1 + a(T – T₀)] according to Equation 27.20, where a is the temperature coefficient of resistivity. This expression needs to be modified if we include the change in dimensions of the wire due to thermal expansion. For a copper wire of radius 0.100 0 mm and length 2.000 m, find its resistance at 100.0°C, including the effects of both thermal expansion and temperature variation of resistivity. Assume the coefficients are known to four significant figures.

To determine

The resistance of wire at $100°\text{C}$.

Answer to Problem 27.75AP

The resistance of wire at $100°\text{C}$ is $1.418\Omega$.

Explanation of Solution

Given info: The radius of copper wire is $0.100\text{mm}$ and length is $2.000\text{m}$. The temperature coefficient of copper is $3.9\times {10}^{-3}{\left(°\text{C}\right)}^{-1}$. The resistance of a straight wire is $R=R_0\left[1+\alpha \left(T-T_0\right)\right]$.

Write the expression of the resistance of the wire.

$R=R_0\left[1+\alpha \left(T-T_0\right)\right]$

Here,

$R$ is the resistance of the wire.

$R_0$ is the resistance of initial temperature.

$\alpha$ is the temperature coefficient of resistivity.

$T$ is the increased temperature

$T_0$ is the initial temperature.

The resistance of wire is increased with the increase of temperature because increase in temperature causes to collision between free electrons in metal with crystal lattice ions.

The length of wire is changed with the increase of temperature. So the final length of the wire is,

$l=l_0\left(1+{\alpha }^{\prime }\left(T-T_0\right)\right)$

Here,

$l$ is the length of the wire at temperature $T$.

$l_0$ is the length of wire at temperature $T_0$.

${\alpha }^{\prime }$ is the coefficient of linear expansion.

The area of the cross section of the wire at temperature $T$ is,

$A=A_0\left(1+2{\alpha }^{\prime }\left(T-T_0\right)\right)$

Here,

$A$ is the area of the cross section of wire at temperature $T$.

$A_0$ is the area of the cross section of wire at temperature $T_0$.

Resistivity of wire with change in temperature is,

$\rho ={\rho }_0\left(1+\alpha \left(T-T_0\right)\right)$

Here,

$\rho$ is the resistivity of wire at temperature $T$.

${\rho }_0$ is the resistivity of wire at temperature $T_0$.

The resistivity of copper is $1.7\times {10}^{-8}\Omega \cdot \text{m}$.

Write the expression for the resistance of wire at $100°\text{C}$.

$R_0={\rho }_0\frac{l_0}{A_0}$ (1)

The area of wire is,

$A_0=\pi r^2$

Here,

$r$ is the radius of wire.

Substitute $0.100\text{mm}$ for $r$ in above equation.

$\begin{array}{c}{A}_0=\pi {\left(0.100\text{mm}\left(\frac{1\times {10}^{-3}\text{m}}{1\text{mm}}\right)\right)}^2\\ =\pi {\left(0.100\times {10}^{-3}\text{m}\right)}^2\\ =0.0314\times {10}^{-6}{\text{m}}^2\end{array}$

Thus, the area of the wire is $0.0314\times {10}^{-6}{\text{m}}^2$.

Substitute $1.7\times {10}^{-8}\Omega \cdot \text{m}$ for ${\rho }_0$, $2.000\text{m}$ for $l_0$ and $0.0314\times {10}^{-6}{\text{m}}^2$ for $A_0$ in equation (1).

$\begin{array}{c}{R}_0=1.7\times {10}^{-8}\Omega \cdot \text{m}\times \left(\frac{2.000\text{m}}{0.0314\times {10}^{-6}{\text{m}}^2}\right)\\ =\left(1.7\times {10}^{-8}\Omega \cdot \text{m}\right)\left(63.69\times {10}^6{\text{m}}^{-1}\right)\\ =108.273\times {10}^{-2}\Omega \end{array}$

The relation between resistance of wire, the length of wire and the cross sectional area at final temperature is,

$R=\rho \frac{l}{A}$

Substitute ${\rho }_0\left(1+\alpha \left(T-T_0\right)\right)$ for $\rho$, $l_0\left(1+{\alpha }^{\prime }\left(T-T_0\right)\right)$ for $l$ and $A_0\left(1+{\alpha }^{\prime }\left(T-T_0\right)\right)$ for $A$ in above equation.

$\begin{array}{c}R={\rho }_0\left(1+\alpha \left(T-T_0\right)\right)\frac{l_0\left(1+{\alpha }^{\prime }\left(T-T_0\right)\right)}{A_0\left(1+2{\alpha }^{\prime }\left(T-T_0\right)\right)}\\ =\frac{{\rho }_0{l}_0}{A_0}\times \frac{\left[1+\alpha \left(T-T_0\right)\right]\left(1+{\alpha }^{\prime }\left(T-T_0\right)\right)}{\left(1+2{\alpha }^{\prime }\left(T-T_0\right)\right)}\end{array}$

Substitute $R_0$ for $\frac{{\rho }_0{l}_0}{A_0}$ in above equation.

$R=R_0\times \frac{\left[1+\alpha \left(T-T_0\right)\right]\left(1+{\alpha }^{\prime }\left(T-T_0\right)\right)}{\left(1+2{\alpha }^{\prime }\left(T-T_0\right)\right)}$

The coefficient of linear expansion is $17\times {10}^{-6}{\left(\text{°C}\right)}^{-1}$.

Substitute $108.273\times {10}^{-2}\Omega$ for $R_0$, $17\times {10}^{-6}{\left(\text{°C}\right)}^{-1}$ for ${\alpha }^{\prime }$, $3.9\times {10}^{-3}{\left(°\text{C}\right)}^{-1}$ for $\alpha$, $20°\text{C}$ for $T_0$ and $100°\text{C}$ for $T$ in above equation.

$\begin{array}{c}R=\frac{\left[\begin{array}{l}108.273\times {10}^{-2}\Omega \times \left[1+\left(3.9\times {10}^{-3}{\left(°\text{C}\right)}^{-1}\right)\left(100-20\right)°\text{C}\right]\\ \left(1+\left(17\times {10}^{-6}{\left(\text{°C}\right)}^{-1}\right)\left(100-20\right)°\text{C}\right)\end{array}\right]}{\left(1+2\left(17\times {10}^{-6}{\left(\text{°C}\right)}^{-1}\right)\left(100-20\right)°\text{C}\right)}\\ =108.273\times {10}^{-2}\Omega \times \frac{\left[1.312\right]\left(1.00136\right)}{\left(1.00272\right)}\\ =1.418\Omega \end{array}$

Conclusion:

Therefore, the resistance of wire at $100°\text{C}$ is $1.418\Omega$.