If the gauge number of a wire is increased by 6, the diameter is halved; if a gauge number is increased by 1, the diameter decreases by the factor 2^1/6 (see the table). Knowing this, and knowing thatof 10-gauge copper wire has a resistance of approximately, $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\Omega }$, estimate the resistance of 25 ftof 22-gauge copper wire.

1. If the gauge number is increased by 6, the diameter is halved.
2. $R=1.00\Omega$, for $L=1000\mathrm{ft}$of 10-gauge copper wire

Resistance is opposition to the flow of current. The resistance is calculated as the ratio of potential difference to the current in the conductor. The resistance of the material is directly proportional to the length of the conductor and inversely proportional to the cross-section area of the conductor.

First, we have to find the diameter of the copper-gauge wire using the formula for resistance. Using the conversion factor, we can find the diameter of -the gauge wire. Using this in the formula for resistance, we can find the required resistance of 25 ft of the copper-gauge wire.

Formulae:

The resistance of the wire material,$R=\frac{\rho L}{A}$ …(i)

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

The cross-sectional area of the wire in terms of diameter,$A=\frac{\pi d^2}{4}$ …(ii)

Here,$A$ is the area of the cross-section,$d$ is the diameter of the conductor.

The value for resistivity of copper, from the table$26-1,\mathrm{\rho }=1.69\times {10}^{-8}\mathrm{\Omega }·\mathrm{m}$

Also, the length through which the current flows:

$$\begin{gathered} L=1000\mathrm{ft}\times \frac{0.3048\mathrm{m}}{1\mathrm{ft}} \\ =304.8\mathrm{m} \end{gathered}$$

Substituting the value of area from equation (ii) in equation (i), we can get the value of the resistance of the 10 -gauge copper wire as follows:

$R=\frac{\rho L}{\left({\displaystyle \frac{\pi d^2}{4}}\right)}$ …(iii)

$$\begin{gathered} 1\Omega =\frac{1.69\times {10}^{-8}\Omega ·m\times 304.8m}{{\displaystyle \frac{\pi {d_{10}}^2}{4}}} \\ \frac{\pi {d_{10}}^2}{4}=1.69\times {10}^{-8}\mathrm{m}\times 304.8\mathrm{m} \\ {d_{10}}^2=\frac{4\times 1.69\times {10}^{-8}\mathrm{m}\times 304.8\mathrm{m}}{\mathrm{\pi }} \\ {d}_{10}=\sqrt{\frac{4\times 1.69\times {10}^{-8}\mathrm{m}\times 304.8\mathrm{m}}{\mathrm{\pi }}} \\ =2.56\times {10}^{-3}\mathrm{m} \end{gathered}$$

Now, as the copper wire is increased from-gauge to 22 -gauge,

Thus, the increase in gauge is,$22-10=12$

It is given that, if the gauge number of a wire is increased by 6, the diameter is halved.

Here, the gauge number is increased by 12 , so the diameter becomes ¼ of the original diameter.

Thus, the diameter for 22 -gauge copper wire will be,

$$\begin{gathered} d_{22}=\frac{1}{4}{d}_{10} \\ =\frac{1}{4}\times 2.56\times {10}^{-3}\mathrm{m} \\ =6.4\times {10}^{-4}\mathrm{m} \end{gathered}$$

The length of the 25ft 22 -gauge wire is given by,

$$\begin{gathered} L=25\mathrm{ft}\times \frac{0.3048\mathrm{m}}{1\mathrm{ft}} \\ =7.62\mathrm{m} \end{gathered}$$

Thus, the resistance of this wire can be calculated using the given data in equation (iii) as follows:

$$\begin{gathered} R=\frac{1.69\times {10}^{-8}\mathrm{\Omega }·\mathrm{m}\times 7.62\mathrm{m}}{\left({\displaystyle \frac{\mathrm{\pi }\times {\left(6.4\times {10}^{-4}\mathrm{m}\right)}^2}{4}}\right)} \\ =0.4\mathrm{\Omega } \end{gathered}$$

Hence, the value of the resistance is $0.4\mathrm{\Omega }$.