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Chapter 18: Problem 1

(a) Compute the electrical conductivity of a 5.1-mm- (0.2-in.-) diameter cylindrical silicon specimen \(51 \mathrm{~mm}\) ( 2 in.) long in which a current of \(0.1\) A passes in an axial direction. A voltage of \(12.5 \mathrm{~V}\) is measured across two probes that are separated by \(38 \mathrm{~mm}\) (1.5 in.). (b) Compute the resistance over the entire \(51 \mathrm{~mm}\) ( \(2 \mathrm{in} .)\) of the specimen.

Short Answer

Expert verified
Answer: The electrical conductivity of the silicon specimen is 7.25 x 10^(-4) S/mm, and the resistance over the entire length of the specimen is 101.89 Ω.

Step by step solution

01

Problem data

We have the following data: - Diameter of the cylindrical silicon specimen: \(d = 5.1\) mm - Length of the specimen: \(L_{1} = 51\) mm (entire length), \(L_{2} = 38\) mm (between the voltage probes) - Current passing through the specimen: \(I = 0.1\) A - Voltage measured across the voltage probes: \(V = 12.5\) V
02

Find the cross-sectional area

First, let's find the cross-sectional area of the cylindrical silicon specimen: - Cross-sectional area \(A = \pi r^2\), where \(r\) is the radius Since the diameter \(d\) is given, we can find the radius as \(r = \frac{d}{2} = \frac{5.1 \, \text{mm}}{2} = 2.55 \, \text{mm}\). Now, we can find the cross-sectional area \(A\): $$A = \pi (2.55 \, \text{mm})^2 = 20.439 \, \text{mm}^2$$
03

Calculate the resistance R between probes

Next, we will use Ohm's Law to find the resistance between the voltage probes, which have a length of \(L_{2} = 38 \, \text{mm}\): - Ohm's Law: \(V = IR\) We can find the resistance \(R_{2}\) as: $$R_{2} = \frac{V}{I} = \frac{12.5 \, \text{V}}{0.1 \, \text{A}} = 125 \, \Omega$$
04

Calculate the electrical conductivity σ

Now we can find the electrical conductivity \(\sigma\) using the formula: - \(\sigma = \frac{L_{2}}{R_{2} A}\) Plugging in the values obtained in previous steps, we can calculate the electrical conductivity: $$\sigma = \frac{38 \, \text{mm}}{125 \, \Omega \cdot 20.439 \, \text{mm}^2} = 7.25 \times 10^{-4} \, \frac{\text{S}}{\text{mm}}$$ So, the electrical conductivity of the silicon specimen is \(7.25 \times 10^{-4} \, \frac{\text{S}}{\text{mm}}\).
05

Calculate the resistance R over the entire specimen

Finally, we will compute the resistance \(R_{1}\) over the entire length of the specimen \(L_{1} = 51 \, \text{mm}\). We will use the formula for the resistance of a cylinder: - \(R_{1} = \frac{L_{1}}{\sigma A}\) Using the electrical conductivity \(\sigma\) and the specimen dimensions obtained earlier, we can find \(R_{1}\): $$R_{1} = \frac{51 \, \text{mm}}{(7.25 \times 10^{-4} \, \frac{\text{S}}{\text{mm}})(20.439 \, \text{mm}^2)} = 101.89 \, \Omega$$ Thus, the resistance over the entire length of the specimen is \(101.89 \, \Omega\).

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Most popular questions from this chapter

At temperatures near room temperature, the temperature dependence of the conductivity for intrinsic germanium is found to equal $$ \sigma=C T^{-3 / 2} \exp \left(-\frac{E_{\mathrm{g}}}{2 k T}\right) $$ where \(C\) is a temperature-independent constant and \(T\) is in Kelvins. Using Equation 18.36, calculate the intrinsic electrical conductivity of germanium at \(150^{\circ} \mathrm{C}\).

Cite the differences in operation and application for junction transistors and MOSFETs.

(a) Using the data in Figure \(18.8\), determine the values of \(\rho_{0}\) and \(a\) from Equation \(18.10\) for pure copper. Take the temperature \(T\) to be in degrees Celsius. (b) Determine the value of \(A\) in Equation \(18.11\) for nickel as an impurity in copper, using the data in Figure \(18.8\). (c) Using the results of parts (a) and (b), estimate the electrical resistivity of copper containing \(1.75\) at \(\%\) Ni at \(100^{\circ} \mathrm{C}\).

(a) Compute the magnitude of the dipole moment associated with each unit cell of \(\mathrm{BaTiO}_{3}\), as illustrated in Figure \(18.35\). (b) Compute the maximum polarization that is possible for this material.

For \(\mathrm{NaCl}\), the ionic radii for \(\mathrm{Na}^{+}\)and \(\mathrm{Cl}^{-}\) ions are \(0.102\) and \(0.181 \mathrm{~nm}\), respectively. If an externally applied electric field produces a \(5 \%\) expansion of the lattice, compute the dipole moment for each \(\mathrm{Na}^{+}-\mathrm{Cl}^{-}\)pair. Assume that this material is completely unpolarized in the absence of an electric field.
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