Chapter 28, Problem 28PQ

(a)

To determine

The estimation of the resistivity and conductivity of a roll of the pure copper pennies.

Answer to Problem 28PQ

The estimation of the resistivity and conductivity of a roll of the pure copper pennies are $1.678\times {10}^{-8}\text{Ω}\cdot \text{m}$ and $5.959\times {10}^7{\text{Ω}}^{-1}\cdot {\text{m}}^{-1}$.

Explanation of Solution

The estimated values of pure substances are given in Table 28.2.

For the pure copper, the resistivity $\left(\rho \right)$ and the conductivity $\left(\sigma \right)$ are $1.678\times {10}^{-8}\text{Ω}\cdot \text{m}$ and $5.959\times {10}^7{\text{Ω}}^{-1}\cdot {\text{m}}^{-1}$.

Conclusion:

Therefore, the estimation of the resistivity and conductivity of a roll of the pure copper pennies are $1.678\times {10}^{-8}\text{Ω}\cdot \text{m}$ and $5.959\times {10}^7{\text{Ω}}^{-1}\cdot {\text{m}}^{-1}$.

(b)

To determine

The estimation of the resistance of the roll of pure copper pennies.

Answer to Problem 28PQ

The estimation of the resistance of the roll of pure copper pennies is $3.7\times {10}^{-6}\text{Ω}$.

Explanation of Solution

The radius and the length of a stack of 50 pennies are $10\text{mm}$ and $70\text{mm}$.

Write the relation for the resistance of the role of pure copper pennies.

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

Here, $R$ is the resistance of the role of pure copper pennies, $\rho$ is resistivity of the pure copper, $l$ is the length of the copper role, and $A$ is cross-section area of a copper penny.

Write the relation for the cross-sectional area of the copper penny.

   $A=\pi r^2$                                                                                                                  (II)

Here, $r$ is radius of the copper penny.

Use equation (II) in equation (I) to get $R$.

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

Conclusion:

Substitute $1.678\times {10}^{-8}\text{Ω}\cdot \text{m}$ for $\rho$, $70\text{mm}$ for $l$, and $10\text{mm}$ for $r$ to find $R$.

   $\begin{array}{c}R=\left(1.678\times {10}^{-8}\text{Ω}\cdot \text{m}\right)\frac{\left(70\text{mm}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}{\pi {\left[\left(10\text{mm}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right]}^2}\\ =3.7\times {10}^{-6}\text{Ω}\end{array}$

Therefore, the estimation of the resistance of the roll of pure copper pennies is $3.7\times {10}^{-6}\text{Ω}$.

(b)

To determine

The estimation of the resistance of the roll of pure aluminum pennies.

Answer to Problem 28PQ

The estimation of the resistance of the roll of pure aluminum pennies is $5.9\times {10}^{-6}\text{Ω}$.

Explanation of Solution

Rewrite the equation (III) for the resistance of the aluminum pennies $R$.

   $R=\rho \frac{l}{\pi r^2}$

Conclusion:

Substitute $2.655\times {10}^{-8}\text{Ω}\cdot \text{m}$ for $\rho$, $70\text{mm}$ for $l$, and $10\text{mm}$ for $r$ to find $R$.

   $\begin{array}{c}R=\left(2.655\times {10}^{-8}\text{Ω}\cdot \text{m}\right)\frac{\left(70\text{mm}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}{\pi {\left[\left(10\text{mm}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right]}^2}\\ =5.9\times {10}^{-6}\text{Ω}\end{array}$

Therefore, the estimation of the resistance of the roll of pure aluminum pennies is $5.9\times {10}^{-6}\text{Ω}$.