Chapter 27, Problem 62AP

(a)

To determine

The resistance and resistivity of the wire in each case.

Answer to Problem 62AP

The resistance and resistivity of the wire in each case is given in the table below.

$L\left(\text{m}\right)$	$\Delta V(\text{V})$	$I(A)$	$R\left(\text{Ω}\right)$	$\rho \left(\text{Ω}\cdot \text{m}\right)$
$0.540$	$5.22$	$0.72$	$7.25$	$9.8\times {10}^{-7}$
$1.028$	$5.82$	$0.414$	$14.1$	$1.0\times {10}^{-6}$
$1.543$	$5.94$	$0.281$	$21.1$	$1.0\times {10}^{-6}$

Explanation of Solution

Write the expression to obtain the resistance of the wire.

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

Here, $R$ is the resistance, $\Delta V$ is the voltage and $I$ is the current.

Write the expression to obtain the resistivity of the wire.

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

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

Conclusion:

Case (i):

Substitute $5.22\text{V}$ for $\Delta V$ and $0.72\text{A}$ for $I$ in equation (I) to calculate $R$.

    $\begin{array}{c}R=\frac{5.22\text{V}}{0.72\text{A}}\\ =7.25\text{Ω}\end{array}$

Substitute $7.25\text{Ω}$ for $R$, $0.540\text{m}$ for $L$ and $7.30\times {10}^{-8}{\text{m}}^2$ for $A$ in equation (II) to calculate $\rho$.

    $\begin{array}{c}\rho =\left(7.25\text{Ω}\right)\frac{7.30\times {10}^{-8}{\text{m}}^2}{0.540\text{m}}\\ =9.8\times {10}^{-7}\text{Ω}\cdot \text{m}\end{array}$

Case (ii):

Substitute $5.82\text{V}$ for $\Delta V$ and $0.414\text{A}$ for $I$ in equation (I) to calculate $R$.

    $\begin{array}{c}R=\frac{5.82\text{V}}{0.414\text{A}}\\ =14.1\text{Ω}\end{array}$

Substitute $14.1\text{Ω}$ for $R$, $1.028\text{m}$ for $L$ and $7.30\times {10}^{-8}{\text{m}}^2$ for $A$ in equation (II) to calculate $\rho$.

    $\begin{array}{c}\rho =\left(14.1\text{Ω}\right)\frac{7.30\times {10}^{-8}{\text{m}}^2}{1.028\text{m}}\\ =1.0\times {10}^{-6}\text{Ω}\cdot \text{m}\end{array}$

Case (iii):

Substitute $5.94\text{V}$ for $\Delta V$ and $0.281\text{A}$ for $I$ in equation (I) to calculate $R$.

    $\begin{array}{c}R=\frac{5.94\text{V}}{0.281\text{A}}\\ =21.1\text{Ω}\end{array}$

Substitute $21.1\text{Ω}$ for $R$, $1.543\text{m}$ for $L$ and $7.30\times {10}^{-8}{\text{m}}^2$ for $A$ in equation (II) to calculate $\rho$.

    $\begin{array}{c}\rho =\left(21.1\text{Ω}\right)\frac{7.30\times {10}^{-8}{\text{m}}^2}{1.543\text{m}}\\ =1.0\times {10}^{-6}\text{Ω}\cdot \text{m}\end{array}$

Therefore, the resistance and resistivity of the wire in each case is given in the table below.

$L\left(\text{m}\right)$	$\Delta V(\text{V})$	$I(A)$	$R\left(\text{Ω}\right)$	$\rho \left(\text{Ω}\cdot \text{m}\right)$
$0.540$	$5.22$	$0.72$	$7.25$	$9.8\times {10}^{-7}$
$1.028$	$5.82$	$0.414$	$14.1$	$1.0\times {10}^{-6}$
$1.543$	$5.94$	$0.281$	$21.1$	$1.0\times {10}^{-6}$

(b)

To determine

The average resistivity.

Answer to Problem 62AP

The average resistivity is $0.993\times {10}^{-6}\text{Ω}\cdot \text{m}$.

Explanation of Solution

Write the expression to obtain the average resistivity.

    ${\rho }_{\text{avg}}\frac{{\rho }_1+{\rho }_2+{\rho }_3}{3}$

Here, ${\rho }_{\text{avg}}$ is the average resistivity, ${\rho }_1$ is the resistivity of material $1$, ${\rho }_2$ is the resistivity of material $2$ and ${\rho }_3$ is the resistivity of material $3$.

Conclusion:

Substitute $9.8\times {10}^{-7}\text{Ω}\cdot \text{m}$ for ${\rho }_1$. $1.0\times {10}^{-6}\text{Ω}\cdot \text{m}$ for ${\rho }_2$ and $1.0\times {10}^{-6}\text{Ω}\cdot \text{m}$ for ${\rho }_3$ in the above equation to calculate ${\rho }_{\text{avg}}$.

    $\begin{array}{c}{\rho }_{\text{avg}}=\frac{\left(9.8\times {10}^{-7}\text{Ω}\cdot \text{m}\right)+\left(1.0\times {10}^{-6}\text{Ω}\cdot \text{m}\right)+\left(1.0\times {10}^{-6}\text{Ω}\cdot \text{m}\right)}{3}\\ =\frac{2.98\times {10}^{-6}\text{Ω}\cdot \text{m}}{3}\\ =0.993\times {10}^{-6}\text{Ω}\cdot \text{m}\end{array}$

Therefore, the average resistivity is $0.993\times {10}^{-6}\text{Ω}\cdot \text{m}$.

(c)

To determine

The comparison between the average resistivity and the resistivity values for different length as shown in table in part (a).

Answer to Problem 62AP

There is not much variation in the resistivity value for different lengths for the same material.

Explanation of Solution

The values of the resistivity with different length and cross section is as below.

$L\left(\text{m}\right)$	$\Delta V(\text{V})$	$I(A)$	$R\left(\text{Ω}\right)$	$\rho \left(\text{Ω-m}\right)$
$0.540$	$5.22$	$0.72$	$7.25$	$9.8\times {10}^{-7}$
$1.028$	$5.82$	$0.414$	$14.1$	$1.0\times {10}^{-6}$
$1.543$	$5.94$	$0.281$	$21.1$	$1.0\times {10}^{-6}$

The average value of resistivity is $0.993\times {10}^{-6}\text{Ω-m}$.

The values of the resistivity in the above table and the average value are almost similar.

Conclusion:

Therefore, there is not much variation in the resistivity value for different length for the same material.