The (United States) National Electric Code, which sets maximum safe currents for insulated copper wires of various diameters, is given (in part) in the table. Plot the safe current density as a function of diameter. Which wire gauge has the maximum safe current density? (“Gauge” is a way of identifying wire diameters, and$\mathbf{1}\mathbf{mil}\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$in.)

A table showing the values of diameters and safe currents for different gauges is given.

The current density is the current across the unit area at a given point in the conductor. We can find the wire gauge that has the maximum safe current density by substituting the given values of safe currents and diameters in the formula for current density.

Formulae:

The current density of the flowing current through an area,$J=\frac{i}{A}$ ...(i)

The cross-sectional area of a wire gauge,$A={\mathrm{\pi R}}^2$ …(ii)

Wherei is the current flowing through the wire,A is the cross sectional area of wire andR is the radius of wire

Substituting the value of area of the wire gauge from equation (ii) in equation (i) for radius$R=\frac{D}{2}$, we can get the current density formula of the wire gauge as follows

$J=\frac{4i}{\pi D^2}.......................\left(iii\right)$

By using this formula, we can calculate the current density for different wire gauges.

a) Gauge 4:

The given values of diameter and current:

$D=204\mathrm{miles}\mathrm{and}i=70A$

Since,

$$\begin{gathered} \mathrm{imil}={10}^{-3}\mathrm{inch} \\ ={10}^{-3}\times 0.0254\mathrm{m}...................\left(\mathrm{iv}\right) \end{gathered}$$

Thus, the value of diameter becomes,

$$\begin{gathered} D=204\times {10}^{-3}\times 0.0254\mathrm{m} \\ =5.1816\times {10}^{-3}\mathrm{m} \end{gathered}$$

The current density is then given by suing the given data in equation (iii) as follows:

role="math" localid="1661837200987" $$\begin{gathered} J=\frac{4\left(70\mathrm{A}\right)}{\left(\mathrm{\pi }\right){\left(5.1816\times {10}^{-3}\mathrm{m}\right)}^2} \\ =3.3195\times {10}^6/{\mathrm{m}}^2 \\ \approx 3.3\times {10}^6\mathrm{A}/{\mathrm{m}}^2 \end{gathered}$$

b)Gauge 6:

The values of diameter using equation (iv) and current from table:

$$\begin{gathered} D=162milsor4.1148\times {10}^{-3}m \\ i=50A \end{gathered}$$

The current density is then given by using the given data in equation (iii) as follows:

role="math" localid="1661837324296" $$\begin{gathered} J=\frac{4\left(50\mathrm{A}\right)}{\left(\mathrm{\pi }\right){\left(4.1148\times {10}^{-3}\mathrm{m}\right)}^2} \\ =3.759955\times {10}^{6}A/{\mathrm{m}}^2 \\ \approx 3.8\times {10}^6\mathrm{A}/{\mathrm{m}}^2 \end{gathered}$$

c)Gauge 8:

The values of diameter using equation (iv) and current from table:

role="math" localid="1661837376268" $$\begin{gathered} D=129milsor3.2766\times {10}^{-3}m \\ i=35A \end{gathered}$$

The current density is then given by using the given data in equation (iii) as follows:

$$\begin{gathered} J=\frac{4\left(35\mathrm{A}\right)}{\left(\mathrm{\pi }\right){\left(3.2766\times {10}^{-3}\mathrm{m}\right)}^2} \\ =4.150795\times {10}^{6}A/{\mathrm{m}}^2 \\ \approx 4.2\times {10}^6\mathrm{A}/{\mathrm{m}}^2 \end{gathered}$$

d)Gauge 10:

The values of diameter using equation (iv) and current from table:

$$\begin{gathered} D=102milsor2.5908\times {10}^{-3}m \\ i=25A \end{gathered}$$

The current density is then given by using the given data in equation (iii) as follows:

$$\begin{gathered} J=\frac{4\left(25\mathrm{A}\right)}{\left(\mathrm{\pi }\right){\left(2.5908\times {10}^{-3}\mathrm{m}\right)}^2} \\ =4.742227\times {10}^{6}A/{\mathrm{m}}^2 \\ \approx 4.7\times {10}^6\mathrm{A}/{\mathrm{m}}^2 \end{gathered}$$

e)Gauge 12:

The values of diameter using equation (iv) and current from table:

$$\begin{gathered} \mathrm{D}=81\mathrm{mils}\mathrm{or}2.0574\times {10}^{-3}\mathrm{m} \\ \mathrm{i}=20\mathrm{A} \end{gathered}$$

The current density is then given by using the given data in equation (iii) as follows:

role="math" localid="1661837595471" $$\begin{gathered} J=\frac{4\left(20\mathrm{A}\right)}{\left(\mathrm{\pi }\right){\left(2.0574\times {10}^{-3}\mathrm{m}\right)}^2} \\ =6.015928\times {10}^{6}A/{\mathrm{m}}^2 \\ \approx 6.0\times {10}^6\mathrm{A}/{\mathrm{m}}^2 \end{gathered}$$

f)Gauge 14:

The values of diameter using equation (iv) and current from table:

$$\begin{gathered} \mathrm{D}=64\mathrm{mils}\mathrm{or}1.6256\times {10}^{-3}\mathrm{m} \\ \mathrm{i}=15\mathrm{A} \end{gathered}$$

The current density is then given by using the given data in equation (iii) as follows:

$$\begin{gathered} J=\frac{4\left(15\mathrm{A}\right)}{\left(\mathrm{\pi }\right){\left(1.6256\times {10}^{-3}\mathrm{m}\right)}^2} \\ =7.227265\times {10}^{6}A/{\mathrm{m}}^2 \\ \approx 7.2\times {10}^6\mathrm{A}/{\mathrm{m}}^2 \end{gathered}$$

g)Gauge 16:

The values of diameter using equation (iv) and current from table:

$$\begin{gathered} D=51milsor1.2954\times {10}^{-3}m \\ i=6\mathrm{A} \end{gathered}$$

The current density is then given by using the given data in equation (iii) as follows:

$$\begin{gathered} J=\frac{4\left(6\mathrm{A}\right)}{\left(\mathrm{\pi }\right){\left(1.2954\times {10}^{-3}\mathrm{m}\right)}^2} \\ =4.552538\times {10}^{6}A/{\mathrm{m}}^2 \\ \approx 4.6\times {10}^6\mathrm{A}/{\mathrm{m}}^2 \end{gathered}$$

h)Gauge 18:

The values of diameter using equation (iv) and current from table:

$$\begin{gathered} \mathrm{D}=40\mathrm{mils}\mathrm{or}1.016\times {10}^{-3}\mathrm{m} \\ \mathrm{i}=3\mathrm{A} \end{gathered}$$

The current density is then given by using the given data in equation (iii) as follows:

$$\begin{gathered} J=\frac{4\left(3\mathrm{A}\right)}{\left(\mathrm{\pi }\right){\left(1.016\times {10}^{-3}\mathrm{m}\right)}^2} \\ =3.700359\times {10}^6\frac{A}{m^2} \\ \approx 3.7\times {10}^6\frac{\mathrm{A}}{{\mathrm{m}}^2} \end{gathered}$$

We can plot the safe current density as a function of diameter.