Chapter 26, Problem 38AP

(a)

To determine

The resistance of the wires and the corresponding values of the resistivity.

Answer to Problem 38AP

The resistance of wire 1 is $7.25\text{Ω}$, the resistance of wire 2 is $14.06\text{Ω}$, the resistance of wire 3 is $21.14\text{Ω}$, the resistivity of wire 1 is $9.80\times {10}^{-7}\text{Ωm}$, the resistivity of wire 2 is $9.98\times {10}^{-7}\text{Ωm}$, the resistivity of wire 3 is $1.00\times {10}^{-6}\text{Ωm}$.

Explanation of Solution

Given information: Area of cross section of gauge wire is $7.30\times {10}^{-8}{\text{m}}^2$, length of gauge wire 1 is $0.540\text{m}$, length of gauge wire 2 is $1.028\text{m}$, length of gauge wire 3 is $1.543\text{m}$, voltage across the wire 1 is $5.22\text{V}$, voltage across the wire 2 is $5.82\text{V}$, voltage across the wire 3 is $5.94\text{V}$, current across the wire 1 is $0.72\text{A}$, current across the wire 2 is $0.414\text{A}$, current across the wire 3 is $0.281\text{A}$.

Formula to calculate the resistance of wire 1.

$\begin{array}{c}{V}_1=I_1{R}_1\\ R_1=\frac{V_1}{I_1}\end{array}$ (1)

Here,

$R_1$ is the resistance of wire 1.

$I_1$ is the current across the wire 1.

$V_1$ is the voltage across the wire 1.

Substitute $5.22\text{V}$ for $V_1$, $0.72\text{A}$ for $I_1$ in equation (1) to find $R_1$,

$\begin{array}{c}{R}_1=\frac{5.22\text{V}}{0.72\text{A}}\\ =7.25\text{Ω}\end{array}$

Thus, the resistance of wire 1 is $7.25\text{Ω}$.

Formula to calculate the resistivity of wire 1.

$\begin{array}{c}{R}_1={\rho }_1\frac{l_1}{A}\\ {\rho }_1=\frac{R_1\times A}{l_1}\end{array}$ (2)

Here,

$l_1$ is the the length of the wire 1.

${\rho }_1$ is the resistivity of wire 1.

$A$ is the cross section of the gauge wire.

Substitute $7.25\text{Ω}$ for $R_1$, $0.540\text{m}$ for $l_1$, $7.30\times {10}^{-8}{\text{m}}^2$ for $A$ in equation (2) to find ${\rho }_1$.

$\begin{array}{c}{\rho }_1=\frac{7.25\text{Ω}\times 7.30\times {10}^{-8}{\text{m}}^2}{0.540\text{m}}\\ =9.80\times {10}^{-7}\text{Ωm}\end{array}$

Thus, the resistivity of wire 1 is $9.80\times {10}^{-7}\text{Ωm}$.

Formula to calculate the resistance of wire 2.

$\begin{array}{c}{V}_2=I_2{R}_2\\ R_2=\frac{V_2}{I_2}\end{array}$ (3)

Here,

$R_2$ is the resistance of wire 2.

$I_2$ is the current across the wire 2.

$V_2$ is the voltage across the wire 2.

Substitute $5.82\text{V}$ for $V_2$, $0.414\text{A}$ for $I_2$ in equation (3) to find $R_2$,

$\begin{array}{c}{R}_1=\frac{5.82\text{V}}{0.414\text{A}}\\ =14.057\text{Ω}\\ \approx 14.06\text{Ω}\end{array}$

Thus, the resistance of wire 2 is $14.06\text{Ω}$.

Formula to calculate the resistivity of wire 2.

$\begin{array}{c}{R}_2={\rho }_2\frac{l_2}{A}\\ {\rho }_2=\frac{R_2\times A}{l_2}\end{array}$ (4)

Here,

$l_2$ is the the length of the wire 2.

${\rho }_2$ is the resistivity of wire 2.

$A$ is the cross section of the gauge wire.

Substitute $14.06\text{Ω}$ for $R_2$, $1.028\text{m}$ for $l_2$, $7.30\times {10}^{-8}{\text{m}}^2$ for $A$ in equation (4) to find ${\rho }_2$,

$\begin{array}{c}{\rho }_2=\frac{14.06\text{Ω}\times 7.30\times {10}^{-8}{\text{m}}^2}{1.028\text{m}}\\ =9.98\times {10}^{-7}\text{Ωm}\end{array}$

Thus, the resistivity of wire 2 is $9.98\times {10}^{-7}\text{Ωm}$.

Formula to calculate the resistance of wire 3.

$\begin{array}{c}{V}_3=I_3{R}_3\\ R_3=\frac{V_3}{I_3}\end{array}$ (5)

Here,

$R_3$ is the resistance of wire 3.

$I_3$ is the current across the wire 3.

$V_3$ is the voltage across the wire 3.

Substitute $5.94\text{V}$ for $V_3$, $0.281\text{A}$ for $I_3$ in equation (5) to find $R_3$,

$\begin{array}{c}{R}_3=\frac{5.94\text{V}}{0.281\text{A}}\\ =21.138\text{Ω}\\ \approx 21.14\text{Ω}\end{array}$

Thus, the resistance of wire 3 is $21.14\text{Ω}$.

Formula to calculate the resistivity of wire 3.

$\begin{array}{c}{R}_3={\rho }_3\frac{l_3}{A}\\ {\rho }_3=\frac{R_3\times A}{l_3}\end{array}$ (6)

Here,

$l_3$ is the the length of the wire 3.

${\rho }_3$ is the resistivity of wire 3.

$A$ is the cross section of the gauge wire.

Substitute $21.14\text{Ω}$ for $R_3$, $1.543\text{m}$ for $l_3$, $7.30\times {10}^{-8}{\text{m}}^2$ for $A$ in equation (6) to find ${\rho }_3$,

$\begin{array}{c}{\rho }_3=\frac{21.14\text{Ω}\times 7.30\times {10}^{-8}{\text{m}}^2}{1.543\text{m}}\\ =1.00\times {10}^{-6}\text{Ωm}\end{array}$

Thus, the resistivity of wire 3 is $1.00\times {10}^{-6}\text{Ωm}$.

Conclusion:

Therefore, the resistance of wire 1 is $7.25\text{Ω}$, the resistance of wire 2 is $14.06\text{Ω}$, the resistance of wire 3 is $21.14\text{Ω}$, the resistivity of wire 1 is $9.80\times {10}^{-7}\text{Ωm}$, the resistivity of wire 2 is $9.98\times {10}^{-7}\text{Ωm}$, the resistivity of wire 3 is $1.00\times {10}^{-6}\text{Ωm}$.

(b)

To determine

The average value of resistivity.

Answer to Problem 38AP

The average value of resistivity is $1.0\times {10}^{-6}\text{Ωm}$.

Explanation of Solution

Given information: Area of cross section of gauge wire is $7.30\times {10}^{-8}{\text{m}}^2$, length of gauge wire 1 is $0.540\text{m}$, length of gauge wire 2 is $1.028\text{m}$, length of gauge wire 3 is $1.543\text{m}$, voltage across the wire 1 is $5.22\text{V}$, voltage across the wire 2 is $5.82\text{V}$, voltage across the wire 3 is $5.94\text{V}$, current across the wire 1 is $0.72\text{A}$, current across the wire 2 is $0.414\text{A}$, current across the wire 3 is $0.281\text{A}$.

Formula to calculate the average value of resistivity.

${\rho }_{\text{avg}}=\frac{{\rho }_1+{\rho }_2+{\rho }_3}{3}$ (7)

Here,

${\rho }_{\text{avg}}$ is the average value of resistivity.

Substitute $9.80\times {10}^{-7}\text{Ωm}$ for ${\rho }_1$, $9.98\times {10}^{-7}\text{Ωm}$ for ${\rho }_2$, $1.00\times {10}^{-6}\text{Ωm}$ for ${\rho }_3$ in equation (7) to find ${\rho }_{\text{avg}}$,

$\begin{array}{c}{\rho }_{\text{avg}}=\frac{\left(9.80\times {10}^{-7}\text{Ωm}\right)+\left(9.98\times {10}^{-7}\text{Ωm}\right)+\left(1.00\times {10}^{-6}\text{Ωm}\right)}{3}\\ =0.9926\times {10}^{-6}\text{Ωm}\\ \approx 1.0\times {10}^{-6}\text{Ωm}\end{array}$

Thus, the average value of resistivity is $1.0\times {10}^{-6}\text{Ωm}$.

Conclusion:

Therefore, the average value of resistivity is $1.0\times {10}^{-6}\text{Ωm}$.

(c)

To determine

The way in which this average value of resistivity compares with the value given in the table 26.2.

Answer to Problem 38AP

This average value of resistivity compares with the value given in the table 26.2. It is observed that the average value of the resistivity is equal to the resistivity of the nichrome. So, a experiment is done by the student to measure the electrical resistance of Nichrome in the form of wires of different lengths and cross sectional areas is correct.

Explanation of Solution

Given information: Area of cross section of gauge wire is $7.30\times {10}^{-8}{\text{m}}^2$, length of gauge wire 1 is $0.540\text{m}$, length of gauge wire 2 is $1.028\text{m}$, length of gauge wire 3 is $1.543\text{m}$, voltage across the wire 1 is $5.22\text{V}$, voltage across the wire 2 is $5.82\text{V}$, voltage across the wire 3 is $5.94\text{V}$, current across the wire 1 is $0.72\text{A}$, current across the wire 2 is $0.414\text{A}$, current across the wire 3 is $0.281\text{A}$.

The average value of resistivity is $1.0\times {10}^{-6}\text{Ωm}$.

The value of resistivity of Nichrome is ranging from $1.00\times {10}^{-6}\text{Ωm}\text{to}1.50\times {10}^{-6}\text{Ωm}$.

It is observed that the average value of the resistivity is equal to the resistivity of the nichrome. So, a experiment is done by the student to measure the electrical resistance of Nichrome in the form of wires of different lengths and cross sectional areas is correct.

Thus, this average value of resistivity compares with the value given in the table 26.2. It is observed that the average value of the resistivity is equal to the resistivity of the nichrome. So, a experiment is done by the student to measure the electrical resistance of Nichrome in the form of wires of different lengths and cross sectional areas is correct.

Conclusion:

Therefore, this average value of resistivity compares with the value given in the table 26.2. It is observed that the average value of the resistivity is equal to the resistivity of the nichrome. So, a experiment is done by the student to measure the electrical resistance of Nichrome in the form of wires of different lengths and cross sectional areas is correct.