Question

Copper and aluminum are being considered for a high-voltagetransmission line that must carry a current of 50.0 A. The resistance per unit length is to be0.200 /km. The densities of copper and aluminum are 8960 and 2700kg/m³, respectively.

Compute:

(a) the magnitude J of the averagecurrent density for a copper cable
1 A/m²

(b) the mass per unit length λ for a coppercable.
2 kg/m

(c) the magnitude J of the average current density for analuminum cable
3 A/m²

(d) the mass per unit length λ for an alumniumcable.
4 kg/m

Answer 1

Concepts and reason

The required concepts to solve these questions are current density, resistivity of conductor and electric field of conductor.

In part (a) calculate the current density for copper cable with the help of electric field equation and then in part (b) calculate the mass per unit length for the copper cable by using resistivity equation. In part (c), calculate the current density for aluminum cable and in part (d), calculate the mass per unit length for the aluminum cable by using mass density of aluminum equation.

Fundamentals

The expression for current density in term of electric field is,

$J = \frac{E}{\rho }$

Here, $E$ is the electric field and $\rho$ is the resistivity of the conductor.

The expression for electric field of conductor is,

$E = \frac{V}{L}$

Here, $V$ is the potential difference and $L$ is the length of the conductor.

The expression for resistivity of the conductor is,

$\rho = \frac{{RA}}{L}$

Here, $A$ is the area and $R$ is the resistance of the conductor.

The general equation of voltage is,

$V = iR$

Here, $i$ is the current and $R$ is the resistance.

(a)

The expression for electric field of conductor is,

$E = \frac{V}{L}$

The general equation of voltage is,

$V = iR$

The expression for current density in term of electric field is,

$J = \frac{E}{\rho }$

Substitute $\left( {\frac{V}{L}} \right)$ for $E$ in the above equation of current density.

$J = \frac{V}{{\rho L}}$ …… (1)

Substitute $iR$ for $V$ in equation (1).

$J = \frac{{iR}}{{\rho L}}$ …… (2)

Substitute $1.69 \times {10^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}$ for $\rho$ , $50{\rm{ A}}$ for $i$ and $0.200{\rm{ }}\Omega \cdot {\rm{k}}{{\rm{m}}^{ - 1}}$ for $\left( {\frac{R}{L}} \right)$ in the equation (2).

$\begin{array}{c}\\J = \frac{{\left( {50{\rm{ A}}} \right)\left( {0.200{\rm{ }}\frac{\Omega }{{{\rm{km}}}}\left( {\frac{{1{\rm{ km}}}}{{{{10}^3}{\rm{ m}}}}} \right)} \right)}}{{\left( {1.69 \times {{10}^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}} \right)}}\\\\ = 5.91 \times {10^5}{\rm{ A}} \cdot {{\rm{m}}^{ - 2}}\\\end{array}$

(b)

The expression for resistivity of the conductor is,

$\rho = \frac{{RA}}{L}$

Rearrange the equation in term of $A$ .

$A = \frac{{\rho L}}{R}$

The expression for mass density of copper is,

$m = dV$

Here, $V$ is the density of the copper.

Substitute $AL$ for $V$ in equation of the mass density of copper.

$\begin{array}{l}\\m = d\left( {AL} \right)\\\\\frac{m}{L} = dA\\\end{array}$

Here, $\lambda$ is used for $\left( {\frac{m}{L}} \right)$ mass per unit length.

Substitute $\frac{{\rho L}}{R}$ for $A$ and $\lambda$ for $\frac{m}{L}$ in above equation.

$\lambda = d\frac{\rho }{{\left( {\frac{R}{L}} \right)}}$ …… (3)

Substitute $0.200{\rm{ }}\Omega \cdot {\rm{k}}{{\rm{m}}^{ - 1}}$ for $\left( {\frac{R}{L}} \right)$ , $8960{\rm{ kg}} \cdot {{\rm{m}}^{ - 3}}$ for $d$ and $1.69 \times {10^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}$ in above equation.

$\begin{array}{c}\\\lambda = \left( {8960{\rm{ kg}} \cdot {{\rm{m}}^{ - 3}}} \right)\left( {\frac{{1.69 \times {{10}^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}}}{{0.200{\rm{ }}\frac{\Omega }{{{\rm{km}}}}\left( {\frac{{1{\rm{ km}}}}{{{{10}^3}{\rm{ m}}}}} \right)}}} \right)\\\\ = \left( {8960{\rm{ kg}} \cdot {{\rm{m}}^{ - 3}}} \right)\left( {\frac{{1.69 \times {{10}^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}}}{{0.200 \times {{10}^{ - 3}}{\rm{ }}\Omega \cdot {{\rm{m}}^{ - 1}}}}} \right)\\\\ = 0.757{\rm{ kg}} \cdot {{\rm{m}}^{ - 1}}\\\end{array}$

(c)

Using the equation (2) current density for aluminum cable is,

$J = \frac{{iR}}{{\rho L}}$

Here, $\rho$ is the resistivity of the aluminum cable.

Substitute $2.82 \times {10^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}$ for $\rho$ , $50{\rm{ A}}$ for $i$ and $0.200{\rm{ }}\Omega \cdot {\rm{k}}{{\rm{m}}^{ - 1}}$ for $\left( {\frac{R}{L}} \right)$ in the above equation.

$\begin{array}{c}\\J = \frac{{\left( {50{\rm{ A}}} \right)\left( {0.200{\rm{ }}\frac{\Omega }{{{\rm{km}}}}\left( {\frac{{1{\rm{ km}}}}{{{{10}^3}{\rm{ m}}}}} \right)} \right)}}{{\left( {2.82 \times {{10}^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}} \right)}}\\\\ = 3.5 \times {10^5}{\rm{ A}} \cdot {{\rm{m}}^{ - 2}}\\\end{array}$

(d)

Using the equation (3) mass per unit length for aluminum cable is,

$\lambda = d\frac{\rho }{{\left( {\frac{R}{L}} \right)}}$

Here, $\rho$ is the resistivity and is the density of the aluminum cable.

Substitute $0.200{\rm{ }}\Omega \cdot {\rm{k}}{{\rm{m}}^{ - 1}}$ for $\left( {\frac{R}{L}} \right)$ , $2700{\rm{ kg}} \cdot {{\rm{m}}^{ - 3}}$ for $d$ and $2.82 \times {10^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}$ in above equation.

$\begin{array}{c}\\\lambda = \left( {2700{\rm{ kg}} \cdot {{\rm{m}}^{ - 3}}} \right)\left( {\frac{{2.82 \times {{10}^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}}}{{0.200{\rm{ }}\frac{\Omega }{{{\rm{km}}}}\left( {\frac{{1{\rm{ km}}}}{{{{10}^3}{\rm{ m}}}}} \right)}}} \right)\\\\ = \left( {2700{\rm{ kg}} \cdot {{\rm{m}}^{ - 3}}} \right)\left( {\frac{{2.82 \times {{10}^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}}}{{0.200 \times {{10}^{ - 3}}{\rm{ }}\Omega \cdot {{\rm{m}}^{ - 1}}}}} \right)\\\\ = 0.380{\rm{ kg}} \cdot {{\rm{m}}^{ - 1}}\\\end{array}$

Ans: Part a

The magnitude $J$ of the current density for a copper cable is $5.91 \times {10^5}{\rm{ A}} \cdot {{\rm{m}}^{ - 2}}$ .

Part b

The mass per unit length $\lambda$ for a copper cable is $0.757{\rm{ kg}} \cdot {{\rm{m}}^{ - 1}}$ .

Part c

The magnitude $J$ of the current density for an aluminum cable is $3.5 \times {10^5}{\rm{ A}} \cdot {{\rm{m}}^{ - 2}}$ .

Part d

The mass per unit length $\lambda$ for an aluminum cable is $0.380{\rm{ kg}} \cdot {{\rm{m}}^{ - 1}}$ .