Chapter 17, Problem 63E

(a)

To determine

The room temperature resistance of the aluminum wire.

Answer to Problem 63E

The room temperature resistance of the aluminum wire is $\underset{\_}{2.09\times {10}^{-3}\text{ohm}}$.

Explanation of Solution

Given that the length of the wire is $1\text{m}$, the radius is $0.002\text{m}$. The resistivity of aluminum is $2.63\times {10}^{-8}\text{ohm}\cdot \text{m}$.

Write the expression for the resistance in terms of dimensions of the conductor.

  $R=\frac{\rho l}{A}$        (I)

Here, $R$ is the resistance, $\rho$ is the resistivity, $l$ is the length, and $A$ is the cross sectional area.

Write the expression for the cross sectional area of the wire.

  $A=\pi r^2$        (II)

Here, $r$ is the radius.

Use equation (II) in (I).

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

Conclusion:

Substitute $2.63\times {10}^{-8}\text{ohm}\cdot \text{m}$ for $\rho$, $1\text{m}$ for $l$, and $0.002\text{m}$ for $r$ in equation (III) to find $R$.

  $\begin{array}{c}R=\frac{\left(2.63\times {10}^{-8}\text{ohm}\cdot \text{m}\right)\left(1\text{m}\right)}{\pi {\left(0.002\text{m}\right)}^2}\\ =2.09\times {10}^{-3}\text{ohm}\end{array}$

Therefore, the room temperature resistance of the aluminum wire is $\underset{\_}{2.09\times {10}^{-3}\text{ohm}}$.

(b)

To determine

The radius of the copper wire with same resistance as of aluminum.

Answer to Problem 63E

The radius of the copper wire with same resistance as of aluminum is $\underset{\_}{0.00162\text{m}}$.

Explanation of Solution

Solve equation (III) for $r$.

  $r=\sqrt{\frac{\rho l}{\pi R}}$        (IV)

It is obtained that the resistance is $2.09\times {10}^{-3}\text{ohm}$, the length is $1\text{m}$, and the resistivity of copper is $1.72\times {10}^{-8}\text{ohm}\cdot \text{m}$.

Conclusion:

Substitute $1.72\times {10}^{-8}\text{ohm}\cdot \text{m}$ for $\rho$, $1\text{m}$ for $l$, and $2.09\times {10}^{-3}\text{ohm}$ for $R$ in equation (IV) to find $r$.

  $\begin{array}{c}r=\sqrt{\frac{\left(1.72\times {10}^{-8}\text{ohm}\cdot \text{m}\right)\left(1\text{m}\right)}{\pi \left(2.09\times {10}^{-3}\text{ohm}\right)}}\\ =0.00162\text{m}\end{array}$

Therefore, the radius of the copper wire with same resistance as of aluminum is $\underset{\_}{0.00162\text{m}}$.

(c)

To determine

To compare the weight of aluminum wire and copper wire.

Answer to Problem 63E

The weight of aluminum is $\underset{\_}{0.462\text{times}}$ of that of copper.

Explanation of Solution

Given that the length of both wires is $1\text{m}$, the density of copper is $8900\text{kg}\cdot {\text{m}}^{-3}$, the density of aluminum is $2700\text{kg}\cdot {\text{m}}^{-3}$. The radius of copper wire is $0.00162\text{m}$, the radius of aluminum wire is $0.002\text{m}$.

Write the expression for the mass in terms of the density.

  $m=DV$        (V)

Here, $m$ is the mass, $D$ is the density, and $V$ is the volume.

Write the expression for the volume of the wire.

  $V=\pi r^{2}l$        (VI)

The ratio of weights will be equal to the ratio of masses of the two wires.

  $\frac{w_{\text{Al}}}{w_{\text{Cu}}}=\frac{D_{\text{Al}}{V}_{\text{Al}}}{D_{\text{Cu}}{V}_{\text{Cu}}}$        (VII)

Here, $w_{\text{Al}}$ is the weight of aluminum, and $w_{\text{Cu}}$ is the weight of copper.

Use the expression for the volume in equation (VII) (consider the equal lengths of the wire).

  $\begin{array}{c}\frac{w_{\text{Al}}}{w_{\text{Cu}}}=\frac{D_{\text{Al}}\left(\pi r_{\text{Al}}^{2}l\right)}{D_{\text{Cu}}\left(\pi r_{\text{Cu}}^{2}l\right)}\\ =\frac{D_{\text{Al}}{r}_{\text{Al}}^2}{D_{\text{Cu}}{r}_{\text{Cu}}^2}\end{array}$        (VIII)

Conclusion:

Substitute $8900\text{kg}\cdot {\text{m}}^{-3}$ for $D_{\text{Cu}}$, $2700\text{kg}\cdot {\text{m}}^{-3}$ for $D_{\text{Al}}$, $0.00162\text{m}$ for $r_{\text{Cu}}$, and $0.002\text{m}$ for $r_{\text{Al}}$ in equation (VIII) to find $w_{\text{Al}}/w_{\text{Cu}}$.

  $\begin{array}{c}\frac{w_{\text{Al}}}{w_{\text{Cu}}}=\frac{\left(2700\text{kg}\cdot {\text{m}}^{-3}\right){\left(0.002\text{m}\right)}^2}{\left(8900\text{kg}\cdot {\text{m}}^{-3}\right){\left(0.00162\text{m}\right)}^2}\\ =0.462\\ w_{\text{Al}}=0.462w_{\text{Cu}}\end{array}$

Therefore, the weight of aluminum is $\underset{\_}{0.462\text{times}}$ of that of copper.