From our previous lesson on Resistance, we have used the $\rho$ as a proportionality constant. However, resistivity has a Resistivity is a measure of the natural property of the material of a given size to resist current. Since it is a natural property, it varies from material to material. The table below shows the resistivity of some common materials.
Table 1: Values of Resistivity, $\rho$ at Room Temperature $20\degg{C}$
| Material | $\textcolor{white}{\rho\,[\un{\ohm\cdot m}]}$ |
|---|---|
| Silver | $1.47\x10^{-8}$ |
| Copper | $1.72\x10^{-8}$ |
| Gold | $2.44\x10^{-8}$ |
| Aluminum | $2.75\x10^{-8}$ |
| Tungsten | $5.25\x10^{-8}$ |
| Iron | $9.7\x10^{-8}$ |
| Steel | $20\x10^{-8}$ |
| Lead | $22\x10^{-8}$ |
| Mercury | $95\x10^{-8}$ |
| Nichrome | $100\x10^{-8}$ |
| Pure carbon (graphite) | $3.5\x10^{-5}$ |
| Pure germanium | 0.60 |
| Pure silicon | 2300 |
| Amber | $5\x10^{14}$ |
| Glass | $10^{10} - 10^{14}$ |
| Lucite | $>10^{13}$ |
| Mica | $10^{11} - 10^{15}$ |
| Quartz (fused) | $75\x10^{16}$ |
| Sulfur | $10^{15}$ |
| Teflon | $>10^{13}$ |
| Wood | $10^{8} - 10^{11}$ |
$\example{1}$ The 18-gauge copper wire has a cross-sectional area of $8.20 \x 10^{-7} \un{m^2}$. It carries a current of 1.67 A. Find the resistance of a 50 m length of this wire.
$\given$
\(\begin{align*}
A &= 8.20 \x 10^{-7} \un{m^2} & \\
L &= 50\un{m} \\
I &= 1.67\un{A}
\end{align*}\)
$\solution$
In order to solve for the resistance, we must first determine the value of the resistivity $\rho$. Since the problem had stated that the material used for the wire is copper, we can use $\tref{1}$ to determine its resistivity. Thus, $\rho = 1.72\x10^{-8} \un{\ohm\cdot m}$. Then using the Equation of Resistance,
$\example{2}$ What is the length of the wire of the same size (area) in the previous example to maintain the same resistance if the wire is aluminum instead of copper.
$\given$
\(\begin{align*}
A &= 8.20 \x 10^{-7} \un{m^2} & \\
R &= 1.05\un{\ohm}
\end{align*}\)
$\solution$
From table $\tref{1}$, the resistivity of aluminum is $2.75\x10^{-8}\un{\ohm\cdot m}$. Hence,
$\example{3}$ If 750 m of 3.0 mm diameter wire have a resistance of 27.6 ohms, what length of wire made with similar material that has a diameter of 5.0 mm will have the same resistance?
$\given$
\(\begin{align*}
L_1 &= 750 \un{m} & \\
d_1 &= 3.0\un{mm} \\
d_2 &= 5.0\un{mm} \\
R_1 &= R_2 = 27.6\un{\ohm}
\end{align*}\)
$\solution$
In this example, the resistance of the two wires are the same. Hence,
Since the two wires are made up of the same material, their resistivity are the same, $\rho_1=\rho_2$. Thus, canceling the resistivity will not affect the equation.
\[\begin{align*} \frac{L_1}{A_1} &= \frac{L_2}{A_2} \\ L_2 &= L_1\,\frac{A_2}{A_1} \end{align*}\]We cannot directly plug the given parameters to the equation since the diameter is not in the formula. However, we can assume that the wire is circular in shape. With this, we can substitute the area in the equation with the formula of the area of a circle in terms of the diameter, $A=\frac{\pi}{4}d^2$.
\[\begin{align*} L_2 &= L_1\,\frac{\frac{\pi}{4}{d_2}^2}{\frac{\pi}{4}{d_1}^2} \\ &= L_1\,\frac{d_2^{\;2}}{d_1^{\;2}} \\ &= 750\un{m}\,\frac{(5.0\un{mm})^2}{(3.0\un{mm})^2} \\ &= 2,083.33\un{m} \tagans \end{align*}\]