Chapter 12.25, Problem 1QP

To determine

To compute:

(a) The electrical conductivity of a cylindrical silicon specimen.

(b) The resistance over the entire specimen length.

Answer to Problem 1QP

(a) The electrical conductivity of a cylindrical silicon specimen $12.2{\left(\Omega -\text{m}\right)}^{-1}$.

(b) The resistance over the entire specimen length is $R=121.4\Omega$.

Explanation of Solution

Given:

The diameter of cylindrical silicon specimen, $d\text{=7}\text{mm}\left(0.28\text{in}\text{.}\right)$.

The two probes separated by length, $l\text{=}45\text{mm}\left(1.75\text{in}\text{.}\right)$.

The current carried by silicon specimen in an axial direction, $I=25\text{A}$.

Minimum voltage across two probes, $V=24\text{V}$.

The entire length of the silicon specimen, $l\text{=}57\text{mm}$.

Explanation:

Write the equation for electrical conductivity $\sigma$.

$\begin{array}{l}\sigma =\frac{1}{\rho }\\ \rho =\frac{1}{\sigma }\end{array}$ (I).

Write the expression of electrical resistivity $\rho$.

$\rho =\frac{RA}{l}$ (II).

$R=\frac{l\rho }{A}$ (III).

Write the expression for voltage.

$\begin{array}{l}V=IR\\ R=\frac{V}{I}\end{array}$

Write the equation to calculate area $A$.

$A=\pi {\left(\frac{d}{2}\right)}^{\text{2}}$

Conclusion:

Substitute $\frac{V}{I}$ for $R$ in Equation (II).

$\rho =\frac{VA}{Il}$

Substitute $\frac{VA}{Il}$ for $\rho$ and $\pi {\left(\frac{d}{2}\right)}^{\text{2}}$ for $A$ in Equation (I).

$\begin{array}{c}\sigma =\frac{1}{\rho }\\ =\frac{Il}{V\pi {\left(\frac{d}{2}\right)}^{\text{2}}}\end{array}$ (IV).

Substitute $\left(0.25\text{A}\right)$ for $I$, $\left(45\text{mm}\right)$ for $l$, $\left(24\text{V}\right)$ for $V$, and $7\text{mm}$ for $d$ in Equation (IV).

$\begin{array}{c}\sigma =\frac{Il}{V\pi {\left(\frac{d}{2}\right)}^{\text{2}}}\\ =\frac{\left(0.25\text{A}\right)\left(45\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}{\left(24\text{V}\right)\pi {\left[\left(\frac{7\text{mm}}{2}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\right]}^{\text{2}}}\\ =12.2{\left(\Omega -\text{m}\right)}^{-1}\end{array}$

Thus, the electrical conductivity of a cylindrical silicon specimen $12.2{\left(\Omega -\text{m}\right)}^{-1}$.

Substitute $\frac{1}{\sigma }$ for $\rho$ and $\pi {\left(\frac{d}{2}\right)}^{\text{2}}$ for $A$ in Equation (III).

$\begin{array}{c}R=\frac{l\frac{1}{\sigma }}{\pi {\left(\frac{d}{2}\right)}^{\text{2}}}\\ =\frac{l}{\sigma \pi {\left(\frac{d}{2}\right)}^{\text{2}}}\end{array}$ (V).

Substitute $12.2{\left(\Omega -\text{m}\right)}^{-1}$ for $\sigma$, $\left(57\text{mm}\right)$ for $l$, and $7\text{mm}$ for $d$ in Equation (V).

$\begin{array}{l}R=\frac{l}{\sigma \pi {\left(\frac{d}{2}\right)}^{\text{2}}}\\ =\frac{\left(57\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}{12.2{\left(\Omega -\text{m}\right)}^{-1}\pi {\left[\left(\frac{7\text{mm}}{2}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\right]}^{\text{2}}}\end{array}$

Thus, the resistance over the entire specimen length is $121.4\Omega$.