Chapter 17, Problem 66AP

When a straight wire is heated, its resistance changes according to the equation

R = R₀ [1 + α(T − T₀)]

(Eq. 17.7), where α is the temperature coefficient of resistivity. (a) Show that a more precise result, which includes the length and area of a wire change when it is heated, is

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

where α′ is the coefficient of linear expansion. (See Topic 10.) (b) Compare the two results for a 2.00-m-long copper wire of radius 0.100 mm, starting at 20.0°C and heated to 100.0°C.

To determine

The formula to calculate the variation of resistance with change in temperature on taking into account the variation of length and area of cross section of resistor with respect to temperature.

Answer to Problem 66AP

It is proved that the required formula is $R=R_0\frac{\left[\left(1+\alpha \left(T-T_0\right)\right)\right]\left[\left(1+{\alpha }^{\prime }\left(T-T_0\right)\right)\right]}{\left[\left(1+2{\alpha }^{\prime }\left(T-T_0\right)\right)\right]}$

Explanation of Solution

Given Info: Variation of resistance with respect to temperature is given by $R=R_0\left(1+\alpha \left(T-T_0\right)\right)$.

Explanation:

Formula to calculate the resistance at temperature $T_0$ is,

$R_0=\frac{{\rho }_0{L}_0}{A_0}$ (I)

• $R_0$ is the resistance at temperature $T_0$
• ${\rho }_0$ is the resistivity of the material of resistor at temperature $T_0$
• $L_0$ is the length of the resistor at temperature $T_0$
• $A_0$ is the area of cross section of the resistor at temperature $T_0$

Formula to calculate the resistance at temperature $T$ is,

$R=\frac{\rho L}{A}$ (II)

• $R$ is the resistance at temperature $T$
• $\rho$ is the resistivity of the material of resistor at temperature $T$
• $L$ is the length of the resistor at temperature $T$
• $A$ is the area of cross section of the resistor at temperature $T$

Formula to calculate the variation of resistivity with respect to temperature is,

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

• $\rho$ is the resistivity at temperature $T$,
• ${\rho }_0$ is the resistivity at temperature $T_0$,
• $\alpha$ is the temperature coefficient of resistivity,

Formula to calculate the variation of $L$ with respect to temperature is,

$L=L_0\left(1+{\alpha }^{\prime }\left(T-T_0\right)\right)$ (IV)

• $L_0$ is the length of the resistor at temperature $T_0$,
• ${\alpha }^{\prime }$ is the coefficient of linear expansion,

Formula to calculate the variation of $L$ with respect to temperature is,

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

• $A_0$ is the area of cross section of the resistor at temperature $T_0$

Substitute equation (II), equation (III) and equation (IV) in equation (I) and rewrite $R$.

$R=\left(\frac{{\rho }_0{L}_0}{A_0}\right)\frac{\left[\left(1+\alpha \left(T-T_0\right)\right)\right]\left[\left(1+{\alpha }^{\prime }\left(T-T_0\right)\right)\right]}{\left[\left(1+2{\alpha }^{\prime }\left(T-T_0\right)\right)\right]}$.(VI)

Substitute $R_0$ for $\frac{{\rho }_0{L}_0}{A_0}$ from eq.(I) in the above formula and rewrite $R$

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

is the required formula.

Conclusion: It is proved that the formula to calculate the variation of resistance with change in temperature on taking into account the variation of length and area of cross section of resistor with respect to temperature is $R=R_0\frac{\left[\left(1+\alpha \left(T-T_0\right)\right)\right]\left[\left(1+{\alpha }^{\prime }\left(T-T_0\right)\right)\right]}{\left[\left(1+2{\alpha }^{\prime }\left(T-T_0\right)\right)\right]}$

To determine

The resistance of copper wire at $100.0\text{°C}$ using formulas with and without considering the temperature dependence of length and area. Compare the results.

Answer to Problem 66AP

The resistance of copper wire at $100.0\text{°C}$ without considering the temperature dependence of length and area of cross section is $1.42\Omega$. The resistance of copper wire at $100.0\text{°C}$ on considering the temperature dependence of length and area of cross section is $1.418\Omega$.

Explanation of Solution

Given Info: The length of copper wire is $2.00\text{m}$ and radius is $0.100\text{mm}$. The copper wire is heated from $20.0\text{°C}$ to $100.0\text{°C}$

Explanation:

Formula to calculate the variation of resistance with respect to temperature without considering the temperature dependence of length and area of cross section is

$R=\left(\frac{{\rho }_0{L}_0}{\pi r_0{}^2}\right)\left(1+\alpha \left(T-T_0\right)\right)$

• $r_0$ is the radius of copper wire at $20.0\text{°C}$

Substitute $1.7\times {10}^{-8}\text{Ω}\cdot \text{m}$ for ${\rho }_0$, $2.00\text{m}$ for $L_0$, $3.14$ for $\pi$, $0.100\text{mm}$ for $r_0$, $3.9\times {10}^{-3}\text{°C}$ for $\alpha$, $100.0\text{°C}$ for $T$ and $20.0\text{°C}$ for $T_0$ in the above equation to find $R$.

$\begin{array}{c}R=\frac{\left(1.7\times {10}^{-8}\text{Ω}\cdot \text{m}\right)\left(2.00\text{m}\right)}{\left(3.14\right)\left(0.100\text{mm}\right)\left(\frac{{\text{10}}^{\text{-6}}{\text{m}}^{\text{2}}}{\text{1}{\text{mm}}^{\text{2}}}\right)}\left(1+\left[\left(3.9\times {10}^{-3}\text{°C}\right)\left(100.0\text{°C}-20.0\text{°C}\right)\right]\right)\\ =1.420\Omega \end{array}$

The resistance is $1.42\Omega$

Formula to calculate the variation of resistance with respect to temperature on considering the temperature dependence of length and area of cross section is

$R=\left(\frac{{\rho }_0{L}_0}{\pi r_0{}^2}\right)\frac{\left[\left(1+\alpha \left(T-T_0\right)\right)\right]\left[\left(1+{\alpha }^{\prime }\left(T-T_0\right)\right)\right]}{\left[\left(1+2{\alpha }^{\prime }\left(T-T_0\right)\right)\right]}$

Substitute $1.7\times {10}^{-8}\text{Ω}\cdot \text{m}$ for ${\rho }_0$, $2.00\text{m}$ for $L_0$, $3.14$ for $\pi$, $0.100\text{mm}$ for $r_0$, $3.9\times {10}^{-3}\text{°C}$ for $\alpha$, $17\times {10}^{-6}{\left(\text{°C}\right)}^{-1}$ for ${\alpha }^{\prime }$, $100.0\text{°C}$ for $T$ and $20.0\text{°C}$ for $T_0$ in the above equation to find $R$.

$\begin{array}{c}R=\left\{\begin{array}{l}\left[\frac{\left(1.7\times {10}^{-8}\text{Ω}\cdot \text{m}\right)\left(2.00\text{m}\right)}{\left(3.14\right)\left(0.100\text{mm}\right)\left(\frac{{\text{10}}^{\text{-6}}{\text{m}}^{\text{2}}}{\text{1}{\text{mm}}^{\text{2}}}\right)}\right]\cdot \\ \left[\frac{\left[\left(1+\left(3.9\times {10}^{-3}\text{°C}\right)\left(100.0\text{°C}-20.0\text{°C}\right)\right)\right]\left[\left(1+\left(17\times {10}^{-6}{\left(\text{°C}\right)}^{-1}\right)\left(100.0\text{°C}-20.0\text{°C}\right)\right)\right]}{\left[\left(1+2\left(\left(17\times {10}^{-6}{\left(\text{°C}\right)}^{-1}\right)\right)\left(100.0\text{°C}-20.0\text{°C}\right)\right)\right]}\right]\end{array}\right\}\\ =1.418\Omega \end{array}$

The resistance is $1.418\Omega$

Conclusion: The resistance of copper wire at $100.0\text{°C}$ without considering the temperature dependence of length and area of cross section is $1.42\Omega$. The resistance of copper wire at $100.0\text{°C}$ on considering the temperature dependence of length and area of cross section is $1.418\Omega$.