Chapter 18, Problem 143P

(a)

To determine

Find the resistance of aluminum wire.

Answer to Problem 143P

The resistance of aluminum wire is $29\Omega$.

Explanation of Solution

Given, $L_{Al}=3L_{Cu}$ and $r_{Al}=2r_{Cu}$

Write the equation for resistance.

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

Here, $\rho$ is the resistivity, $R$ is the resistance, $L$ is the length and $A$ is the area.

Rewrite the above equation for aluminum and copper wire and divide them.

$\frac{R_{Al}}{R_{Cu}}=\frac{{\rho }_{Al}\frac{L_{Al}}{A_{Al}}}{{\rho }_{Cu}\frac{L_{Cu}}{A_{Cu}}}$ (II)

Rewrite the above equation and substitute the area values to get resistance of aluminum.

$R_{Al}=\frac{{\rho }_{Al}{L}_{Al}{A}_{Cu}}{{\rho }_{Cu}{L}_{Cu}{A}_{Al}}{R}_{Cu}=\frac{{\rho }_{Al}{L}_{Al}\pi r_{Cu}^2}{{\rho }_{Cu}{L}_{Cu}\pi r_{Al}^2}{R}_{Cu}$ (III)

Conclusion:

Substitute $24\Omega$ for $R_{Cu}$, $2.65\times {10}^{-8}\Omega \text{m}$ for ${\rho }_{Al}$, $1.67\times {10}^{-8}\Omega \text{m}$ for ${\rho }_{Cu}$, $L_{Al}=3L_{Cu}$ and $r_{Al}=2r_{Cu}$ in equation III.

$\begin{array}{c}{R}_{Al}=\frac{\left(2.65\times {10}^{-8}\Omega \text{m}\right)\left(3L_{Cu}\right)\pi r_{Cu}^2}{\left(1.67\times {10}^{-8}\Omega \text{m}\right)L_{Cu}\pi {\left(2r_{Cu}\right)}^2}\left(24\Omega \right)\\ =29\Omega \end{array}$

Therefore, the resistance of aluminum wire is $29\Omega$.

(b)

To determine

Find the resistance of the wire when operating steadily.

Answer to Problem 143P

The resistance of the wire when operating steadily is $\text{30 }\Omega$. 

Explanation of Solution

Write the equation for resistance.

$R=\frac{V}{I}$ (IV)

Here, $R$ is the resistance, $I$ is the current and $V$ is potential difference.

Conclusion:

Substitute $300\text{ V}$ for $V$ and $10\text{ A}$ for $I$ in equation IV.

$\begin{array}{c}R=\frac{300\text{ V}}{10\text{ A}}\\ =30\Omega \end{array}$

Therefore, the resistance of the wire when operating steadily is $\text{30 }\Omega$.

(c)

To determine

Find the temperature of the copper wire.

Answer to Problem 143P

The temperature of the copper wire is $82°\text{C}$. 

Explanation of Solution

Write the equation for resistance.

$R=R_0\left(1+\alpha \Delta T\right)$

Here, $R$ is the resistance, $R_0$ is the initial resistance, $\alpha$ is the temperature coefficient of resistivity and $\Delta T$ is change in temperature.

Substitute $\Delta T=T-T_0$ in above equation and rewrite to get final temperature $T$.

$T=\frac{1}{\alpha }\left(\frac{R}{R_0}-1\right)+T_0$ (V)

Conclusion:

Substitute $20°\text{C}$ for $T_0$, $4.05\times {10}^{-3}°{\text{C}}^{-1}$ for $\alpha$, $\text{30 }\Omega$ for $R$ and $\text{24 }\Omega$ for $R_0$ in equation V.

$\begin{array}{c}T=\frac{1}{4.05\times {10}^{-3}°{\text{C}}^{-1}}\left(\frac{\text{30 }\Omega }{\text{24 }\Omega }-1\right)+20°\text{C}\\ =82°\text{C}\end{array}$

Therefore, the temperature of the copper wire is $82°\text{C}$.

(d)

To determine

Does the thermal expansion of wire changes the answer in part c.

Answer to Problem 143P

The thermal expansion of wire will not changes the answer in part c, since the expansion in wire is less than $0.1\%$.

Explanation of Solution

The thermal expansion in the wire for the temperature of $62°\text{C}$ is less than $0.1\%$. This change in the wire dimensions will not affect the answer in part c.

Conclusion:

Therefore, the thermal expansion of wire will not changes the answer in part c, since the expansion in wire is less than $0.1\%$.