Chapter 27, Problem 32P

(a)

To determine

Weather the design goal is achieved with this method or not.

Answer to Problem 32P

The design goal is achieved with this method.

Explanation of Solution

Write the expression to obtain the resistance in the wire.

    $R=\frac{\rho l}{A}$

Here, $R$ is the resistance in the wire, $\rho$ is the resistivity, $l$ is the length of the wire and $A$ is the area of cross section of the wire.

Substitute $\pi r^2$ for $A$ in the above equation.

    $R=\frac{\rho l}{\pi r^2}$                                                                                                                     (I)

Here, $r$ is the radius of the wire.

Write the expression to obtain the carbon resistance in the wire.

    $R_1=\frac{{\rho }_1{l}_1}{\pi r^2}$                                                                                                                  (II)

Here, $R_1$ is the carbon resistance in the wire, ${\rho }_1$ is the carbon resistivity, $l_1$ is the length of the carbon wire and $r$ is the radius of the wire.

Write the expression to obtain the nichrome resistance in the wire.

    $R_2=\frac{{\rho }_2{l}_2}{\pi r^2}$                                                                                                                 (III)

Here, $R_2$ is the nichrome resistance in the wire, ${\rho }_2$ is the nichrome resistivity, $l_2$ is the length of the nichrome wire and $r$ is the radius of the wire.

Write the expression to obtain the variation in resistance with change in temperature.

    $R_{\text{T}}=R\left(1+\alpha \Delta T\right)$                                                                                                    (IV)

Here, $R_{\text{T}}$ is the resistance with temperature change, $R$ is the initial resistance, $\alpha$ is the coefficient of thermal expansion and $\Delta T$ is the change in temperature.

Write the expression to obtain the final resistance.

    $R_{\text{1T}}+R_{\text{2T}}=R_1\left(1+{\alpha }_1\Delta T\right)+R_2\left(1+{\alpha }_2\Delta T\right)$                                                               (V)

Here, $R_{\text{1T}}$ is carbon resistance variation with temperature, $R_{\text{2T}}$ is nichrome resistance variation with temperature, $R_1$ is the initial carbon resistance, $R_2$ is the initial nichrome resistance, ${\alpha }_1$ is the coefficient of thermal resistance for carbon resistance, ${\alpha }_2$ is the coefficient of thermal resistance for nichrome resistance and $\Delta T$ is the change in temperature.

Resistance is independent of temperature variation, thus,

    $R_{\text{1T}}+R_{\text{2T}}=R_{\text{1}}+R_{\text{2}}$                                                                                                 (VI)

Compare equation (III) and (IV).

    $\begin{array}{l}{R}_1{\alpha }_1\Delta T+R_2{\alpha }_2\Delta T=0\\ \frac{R_1}{R_2}=-\frac{{\alpha }_2}{{\alpha }_1}\end{array}$

Substitute $-0.5\times {10}^{-3}{\text{°C}}^{-1}$ for ${\alpha }_1$ and $0.4\times {10}^{-3}{\text{°C}}^{-1}$ for ${\alpha }_2$ in the above equation.

    $\begin{array}{c}\frac{R_1}{R_2}=-\frac{0.4\times {10}^{-3}{\text{°C}}^{-1}}{\left(-0.5\times {10}^{-3}{\text{°C}}^{-1}\right)}\\ \frac{R_1}{R_2}=0.8\\ R_1=0.8R_2\end{array}$                                                                                      (VII)

The sum of the resistances is given as $10.0\text{Ω}$.

    $R_1+R_2=10.0\text{Ω}$                                                                                                    (VII)

Conclusion:

Substitute $0.8R_2$ for $R_1$ in the above equation to calculate $R_2$.

    $\begin{array}{c}0.8R_2+R_2=10.0\text{Ω}\\ 1.8R_2=10.0\text{Ω}\\ R_2=\frac{10.0\text{Ω}}{1.8}\\ R_2=5.56\text{Ω}\end{array}$

Substitute $5.56\text{Ω}$ for $R_2$ in equation (VII) to calculate $R_1$.

    $\begin{array}{c}{R}_1=0.8\left(5.56\text{Ω}\right)\\ =4.44\text{Ω}\end{array}$

As the equation (VII) and (VIII) provides the suitable values of the resistances,

Therefore, the design goal is achieved with this method.

(b)

To determine

The length of the carbon and the nichrome cylinder.

Answer to Problem 32P

The length of the carbon cylinder is $0.8967\text{m}$ and length of the nichrome cylinder is $26.2\text{m}$.

Explanation of Solution

Conclusion:

Substitute $4.44\text{Ω}$ for $R_1$ and $1.50\text{mm}$ for $r$ and $3.50\times {10}^{-5}\text{Ω-m}$ for ${\rho }_1$ in equation (II) to calculate $l_1$.

    $\begin{array}{c}4.44\text{Ω}=\frac{\left(3.50\times {10}^{-5}\text{Ω-m}\right)l_1}{\pi {\left(1.50\text{mm}\right)}^2}\\ l_1=\frac{\left(4.44\text{Ω}\right)\left(\pi {\left(1.50\text{mm}\right)}^2\right)}{\left(3.50\times {10}^{-5}\text{Ω-m}\right)}\\ l_1=\frac{\left(4.44\text{Ω}\right)\left(\pi {\left(1.50\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\right)}{\left(3.50\times {10}^{-5}\text{Ω-m}\right)}\\ l_1=0.8967\text{m}\end{array}$

Substitute $5.56\text{Ω}$ for $R_2$ and $1.50\text{mm}$ for $r$ and $1.50\times {10}^{-6}\text{Ω-m}$ for ${\rho }_2$  in equation (III) to calculate $l_2$.

    $\begin{array}{c}5.56\text{Ω}=\frac{\left(1.50\times {10}^{-6}\text{Ω-m}\right)l_2}{\pi {\left(1.50\text{mm}\right)}^2}\\ l_2=\frac{\left(5.56\text{Ω}\right)\left(\pi {\left(1.50\text{mm}\right)}^2\right)}{\left(1.50\times {10}^{-6}\text{Ω-m}\right)}\\ l_2=\frac{\left(5.56\text{Ω}\right)\left(\pi {\left(1.50\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\right)}{\left(1.50\times {10}^{-6}\text{Ω-m}\right)}\\ l_2=26.2\text{m}\end{array}$

Therefore, the length of the carbon cylinder is $0.8967\text{m}$ and length of the nichrome cylinder is $26.2\text{m}$.