\subsection{Resistance, Resistivity, and Ohm's Law}

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.

\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
\[
R = \frac{V}{I}.
\]
The SI unit of resistance is the \emph{ohm}, denoted $\Omega$, where $1\,\Omega = 1\,\mathrm{V/A}$.}

\wc{Current is not ``used up'' by resistors}{Current $I$ is the same through every element in a series circuit. A resistor does not ``consume'' current --- it drops voltage (potential energy per charge). Think of the circuit as a closed loop: whatever current leaves the battery returns to it. What is ``used up'' is electrical potential energy (converted to heat), not charge.}

\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:
\[
\sigma = \frac{1}{\rho}.
\]
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 ($\mathrm{S/m}$).}

\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.}

\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,
\[
\vec{J} = \sigma \vec{E}
\qquad \text{or equivalently} \qquad
\vec{E} = \rho \vec{J},
\]
where $\sigma = 1/\rho$ is the conductivity.}

\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 $\vec{J}$.

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
\[
R = \frac{EL}{JA}.
\]
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
\[
\frac{\rho L}{A} = \frac{EL}{JA},
\]
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.}

\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 $\vec{E}$.}

\mprop{Resistance of a uniform conductor}{A straight wire of length $L$, uniform cross-sectional area $A$, and resistivity $\rho$ has resistance
\[
R = \rho \,\frac{L}{A}.
\]
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.}

\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.}

\thm{Temperature dependence of resistivity}{For many materials, over a limited temperature range, the resistivity varies approximately as
\[
\rho = \rho_0 \,\bigl[1 + \alpha(T - T_0)\bigr],
\]
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.}

\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 $\rho_{\mathrm{Cu}}/\rho_{\mathrm{Al}} \approx 1.7\times 10^{-8} / 2.8\times 10^{-8} \approx 0.61$.}

\qs{Worked example}{A cylindrical copper wire of length
\[
L = 50\,\mathrm{m}
\]
and radius
\[
r = 1.0\,\mathrm{mm} = 1.0 \times 10^{-3}\,\mathrm{m}
\]
has a potential difference
\[
V = 10\,\mathrm{V}
\]
applied across its ends. The resistivity of copper at room temperature is
\[
\rho = 1.7 \times 10^{-8}\,\Omega\!\cdot\!\mathrm{m}.
\]
Assume the conduction-electron number density of copper is
\[
n = 8.5 \times 10^{28}\,\mathrm{m^{-3}},
\]
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.

Find:
\begin{enumerate}[label=(\alph*)]
\item the resistance $R$ of the wire,
\item the current $I$ through the wire,
\item the current density magnitude $J$ and vector $\vec{J}$,
\item the drift velocity magnitude $v_d$ of the electrons, and
\item the power dissipated in the wire.
\end{enumerate}}

\sol \textbf{Part (a).} The cross-sectional area of the cylindrical wire is
\[
A = \pi r^2 = \pi (1.0 \times 10^{-3}\,\mathrm{m})^2 = \pi \times 1.0 \times 10^{-6}\,\mathrm{m^2} = 3.14 \times 10^{-6}\,\mathrm{m^2}.
\]
The resistance is
\[
R = \rho \,\frac{L}{A}.
\]
Substitute the values:
\[
R = \left(1.7 \times 10^{-8}\,\Omega\!\cdot\!\mathrm{m}\right)\,\frac{50\,\mathrm{m}}{3.14 \times 10^{-6}\,\mathrm{m^2}}.
\]
Compute the numerator:
\[
\left(1.7 \times 10^{-8}\right)(50) = 8.5 \times 10^{-7}\,\Omega\!\cdot\!\mathrm{m^2}.
\]
Then
\[
R = \frac{8.5 \times 10^{-7}}{3.14 \times 10^{-6}}\,\Omega = 0.271\,\Omega.
\]

\textbf{Part (b).} Ohm's law gives the current:
\[
I = \frac{V}{R} = \frac{10\,\mathrm{V}}{0.271\,\Omega} = 36.9\,\mathrm{A}.
\]

\textbf{Part (c).} The current density magnitude is
\[
J = \frac{I}{A} = \frac{36.9\,\mathrm{A}}{3.14 \times 10^{-6}\,\mathrm{m^2}} = 1.18 \times 10^7\,\mathrm{A/m^2}.
\]
The current flows in the $+\hat{\imath}$ direction (from high to low potential), so the current density vector is
\[
\vec{J} = \left(1.18 \times 10^7\,\mathrm{A/m^2}\right)\hat{\imath}.
\]

\textbf{Part (d).} The drift velocity of electrons relates to the current density by
\[
\vec{J} = nq_e\vec{v}_d,
\]
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
\[
v_d = \frac{J}{ne} = \frac{1.18 \times 10^7\,\mathrm{A/m^2}}{\left(8.5 \times 10^{28}\,\mathrm{m^{-3}}\right)\left(1.60 \times 10^{-19}\,\mathrm{C}\right)}.
\]
The denominator is
\[
\left(8.5 \times 10^{28}\right)\left(1.60 \times 10^{-19}\right) = 1.36 \times 10^{10}\,\mathrm{C/m^3},
\]
so
\[
v_d = \frac{1.18 \times 10^7}{1.36 \times 10^{10}}\,\mathrm{m/s} = 8.69 \times 10^{-4}\,\mathrm{m/s}.
\]
Electrons drift opposite to $\vec{J}$, so $\vec{v}_d = -\left(8.69 \times 10^{-4}\,\mathrm{m/s}\right)\hat{\imath}$.

\textbf{Part (e).} The power dissipated in the wire is
\[
P = IV = \left(36.9\,\mathrm{A}\right)\left(10\,\mathrm{V}\right) = 369\,\mathrm{W}.
\]
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.

Therefore,
\[
R = 0.271\,\Omega,
\qquad
I = 36.9\,\mathrm{A},
\qquad
J = 1.18 \times 10^7\,\mathrm{A/m^2},
\]
\[
\vec{J} = \left(1.18 \times 10^7\,\mathrm{A/m^2}\right)\hat{\imath},
\qquad
v_d = 8.69 \times 10^{-4}\,\mathrm{m/s},
\qquad
P = 369\,\mathrm{W}.
\]
