162 lines
8.0 KiB
TeX
162 lines
8.0 KiB
TeX
\subsection{Resistance, Resistivity, and Ohm's Law}
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This subsection introduces resistance and resistivity, connects the macroscopic Ohm's law $V = IR$ to the microscopic relation $\vec{J} = \sigma \vec{E}$, and derives the geometric expression $R = \rho L / A$ for a uniform conductor.
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\dfn{Resistance}{Let a conducting element have a potential difference $V$ across its ends and carry a steady current $I$. The \emph{resistance} $R$ of the element is
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\[
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R = \frac{V}{I}.
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\]
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The SI unit of resistance is the \emph{ohm}, denoted $\Omega$, where $1\,\Omega = 1\,\mathrm{V/A}$.}
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\dfn{Resistivity, conductivity}{The \emph{resistivity} $\rho$ of a material is an intrinsic property that quantifies how strongly that material opposes the flow of electric current. The \emph{conductivity} $\sigma$ is the reciprocal of resistivity:
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\[
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\sigma = \frac{1}{\rho}.
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\]
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The SI unit of resistivity is the ohm-meter ($\Omega\!\cdot\!\mathrm{m}$). The SI unit of conductivity is $(\Omega\!\cdot\!\mathrm{m})^{-1}$, also called siemens per meter ($\mathrm{S/m}$).}
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\nt{A perfect conductor has $\rho = 0$ and $\sigma \to \infty$. A perfect insulator has $\rho \to \infty$ and $\sigma \to 0$. Metals have very low resistivity (typically $10^{-8}\,\Omega\!\cdot\!\mathrm{m}$); good insulators like glass have resistivity on the order of $10^{10}\,\Omega\!\cdot\!\mathrm{m}$ or higher.}
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\thm{Microscopic Ohm's law}{Let $\vec{E}$ denote the electric field inside a conducting material and let $\vec{J}$ denote the resulting current density at the same point. If the material is an \emph{ohmic conductor} -- one for which $\rho$ is independent of the magnitude of $\vec{E}$ -- then at every point inside the material,
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\[
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\vec{J} = \sigma \vec{E}
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\qquad \text{or equivalently} \qquad
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\vec{E} = \rho \vec{J},
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\]
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where $\sigma = 1/\rho$ is the conductivity.}
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\pf{Microscopic Ohm's law from the macroscopic form}{Consider a straight wire of uniform cross-sectional area $A$ and length $L$, made of a material with resistivity $\rho$. Suppose a potential difference $V$ is applied across the ends, producing a uniform field $\vec{E}$ along the wire and a uniform current density $\vec{J}$.
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From the macroscopic definition of resistance, $R = V/I$. For a uniform wire, the field and current density are related to the macroscopic quantities by $E = V/L$ and $J = I/A$. Substituting these into the resistance formula gives
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\[
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R = \frac{EL}{JA}.
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\]
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The resistance of a uniform wire is also known from experiment and geometry to be $R = \rho L/A$. Equating the two expressions for $R$ yields
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\[
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\frac{\rho L}{A} = \frac{EL}{JA},
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\]
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which simplifies to $\rho J = E$, or $\vec{E} = \rho \vec{J}$. This is the microscopic form of Ohm's law. The same relation holds pointwise even if the field and current density vary spatially, because resistivity is a local material property.}
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\cor{Ohm's law is empirical}{Not all materials obey Ohm's law. Semiconductors, diodes, and superconductors are non-ohmic: their $I$-$V$ characteristic is not linear. The microscopic relation $\vec{J} = \sigma \vec{E}$ holds only for ohmic conductors where $\sigma$ is independent of $\vec{E}$.}
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\mprop{Resistance of a uniform conductor}{A straight wire of length $L$, uniform cross-sectional area $A$, and resistivity $\rho$ has resistance
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\[
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R = \rho \,\frac{L}{A}.
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\]
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This relation follows from combining $V = IR$ with the microscopic Ohm's law $\vec{E} = \rho \vec{J}$ applied to a geometry where $\vec{E}$ and $\vec{J}$ are uniform and parallel to the wire axis.}
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\nt{The resistance of a uniform wire increases linearly with length and decreases inversely with cross-sectional area. This is the electrical analogue of fluid flow through a pipe: a longer pipe gives more resistance, and a wider pipe gives less.}
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\thm{Temperature dependence of resistivity}{For many materials, over a limited temperature range, the resistivity varies approximately as
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\[
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\rho = \rho_0 \,\bigl[1 + \alpha(T - T_0)\bigr],
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\]
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where $\rho_0$ is the resistivity at a reference temperature $T_0$, and $\alpha$ is the \emph{temperature coefficient of resistivity} for that material. The SI unit of $\alpha$ is $\mathrm{K}^{-1}$ (or equivalently $^\circ\mathrm{C}^{-1}$). For most metals, $\alpha > 0$, so resistivity increases with temperature.}
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\ex{Illustrative example}{A copper wire and an aluminum wire of the same length and cross-sectional area are connected to the same potential difference. Since copper has a lower resistivity than aluminum, the copper wire will carry more current and dissipate less power. The ratio of their resistivities at room temperature is approximately $\rho_{\mathrm{Cu}}/\rho_{\mathrm{Al}} \approx 1.7\times 10^{-8} / 2.8\times 10^{-8} \approx 0.61$.}
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\qs{Worked example}{A cylindrical copper wire of length
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\[
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L = 50\,\mathrm{m}
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\]
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and radius
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\[
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r = 1.0\,\mathrm{mm} = 1.0 \times 10^{-3}\,\mathrm{m}
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\]
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has a potential difference
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\[
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V = 10\,\mathrm{V}
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\]
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applied across its ends. The resistivity of copper at room temperature is
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\[
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\rho = 1.7 \times 10^{-8}\,\Omega\!\cdot\!\mathrm{m}.
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\]
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Assume the conduction-electron number density of copper is
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\[
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n = 8.5 \times 10^{28}\,\mathrm{m^{-3}},
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\]
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the elementary charge is $e = 1.60 \times 10^{-19}\,\mathrm{C}$, and the wire is uniform. Let the current flow from the high-potential end toward the low-potential end, and let $+\hat{\imath}$ point in the direction of the current.
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Find:
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\begin{enumerate}[label=(\alph*)]
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\item the resistance $R$ of the wire,
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\item the current $I$ through the wire,
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\item the current density magnitude $J$ and vector $\vec{J}$,
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\item the drift velocity magnitude $v_d$ of the electrons, and
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\item the power dissipated in the wire.
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\end{enumerate}}
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\sol \textbf{Part (a).} The cross-sectional area of the cylindrical wire is
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\[
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A = \pi r^2 = \pi (1.0 \times 10^{-3}\,\mathrm{m})^2 = \pi \times 1.0 \times 10^{-6}\,\mathrm{m^2} = 3.14 \times 10^{-6}\,\mathrm{m^2}.
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\]
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The resistance is
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\[
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R = \rho \,\frac{L}{A}.
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\]
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Substitute the values:
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\[
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R = \left(1.7 \times 10^{-8}\,\Omega\!\cdot\!\mathrm{m}\right)\,\frac{50\,\mathrm{m}}{3.14 \times 10^{-6}\,\mathrm{m^2}}.
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\]
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Compute the numerator:
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\[
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\left(1.7 \times 10^{-8}\right)(50) = 8.5 \times 10^{-7}\,\Omega\!\cdot\!\mathrm{m^2}.
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\]
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Then
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\[
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R = \frac{8.5 \times 10^{-7}}{3.14 \times 10^{-6}}\,\Omega = 0.271\,\Omega.
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\]
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\textbf{Part (b).} Ohm's law gives the current:
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\[
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I = \frac{V}{R} = \frac{10\,\mathrm{V}}{0.271\,\Omega} = 36.9\,\mathrm{A}.
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\]
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\textbf{Part (c).} The current density magnitude is
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\[
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J = \frac{I}{A} = \frac{36.9\,\mathrm{A}}{3.14 \times 10^{-6}\,\mathrm{m^2}} = 1.18 \times 10^7\,\mathrm{A/m^2}.
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\]
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The current flows in the $+\hat{\imath}$ direction (from high to low potential), so the current density vector is
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\[
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\vec{J} = \left(1.18 \times 10^7\,\mathrm{A/m^2}\right)\hat{\imath}.
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\]
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\textbf{Part (d).} The drift velocity of electrons relates to the current density by
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\[
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\vec{J} = nq_e\vec{v}_d,
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\]
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where $q_e = -e$ is the charge of an electron. Since the electrons are negatively charged, their drift velocity is opposite to the current direction. The magnitude is
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\[
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v_d = \frac{J}{ne} = \frac{1.18 \times 10^7\,\mathrm{A/m^2}}{\left(8.5 \times 10^{28}\,\mathrm{m^{-3}}\right)\left(1.60 \times 10^{-19}\,\mathrm{C}\right)}.
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\]
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The denominator is
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\[
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\left(8.5 \times 10^{28}\right)\left(1.60 \times 10^{-19}\right) = 1.36 \times 10^{10}\,\mathrm{C/m^3},
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\]
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so
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\[
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v_d = \frac{1.18 \times 10^7}{1.36 \times 10^{10}}\,\mathrm{m/s} = 8.69 \times 10^{-4}\,\mathrm{m/s}.
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\]
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Electrons drift opposite to $\vec{J}$, so $\vec{v}_d = -\left(8.69 \times 10^{-4}\,\mathrm{m/s}\right)\hat{\imath}$.
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\textbf{Part (e).} The power dissipated in the wire is
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\[
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P = IV = \left(36.9\,\mathrm{A}\right)\left(10\,\mathrm{V}\right) = 369\,\mathrm{W}.
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\]
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Equivalently, $P = I^2R = (36.9\,\mathrm{A})^2(0.271\,\Omega) = 369\,\mathrm{W}$, or $P = V^2/R = (10\,\mathrm{V})^2/(0.271\,\Omega) = 369\,\mathrm{W}$. All three give the same result.
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Therefore,
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\[
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R = 0.271\,\Omega,
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\qquad
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I = 36.9\,\mathrm{A},
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\qquad
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J = 1.18 \times 10^7\,\mathrm{A/m^2},
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\]
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\[
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\vec{J} = \left(1.18 \times 10^7\,\mathrm{A/m^2}\right)\hat{\imath},
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\qquad
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v_d = 8.69 \times 10^{-4}\,\mathrm{m/s},
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\qquad
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P = 369\,\mathrm{W}.
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\]
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