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\subsection{Current, Drift Velocity, and Current Density}
This subsection connects the macroscopic current $I$ in a wire to the microscopic motion of charge carriers through the current density $\vec{J}$ and the drift velocity $\vec{v}_d$.
\dfn{Current, drift velocity, and current density}{Let $\Delta q$ denote the net charge that crosses a chosen surface in a time interval $\Delta t$. The \emph{electric current} through that surface is
\[
I=\frac{\Delta q}{\Delta t}
\]
in the limit of very small time intervals.
Let $\vec{v}_d$ denote the average drift velocity of the charge carriers, let $n$ denote the number of carriers per unit volume, and let $q$ denote the charge of each carrier. The \emph{current density} $\vec{J}$ is the vector that describes current per unit area, with direction defined by the motion of positive charge. For an oriented surface element $d\vec{A}$,
\[
dI=\vec{J}\cdot d\vec{A}.
\]
Thus $\vec{J}$ links the local flow of charge to the total current through a cross section.}
\nt{Conventional current is defined to point in the direction that positive charges would move. Therefore $\vec{J}$ points in the conventional-current direction. In a metal wire, the mobile charge carriers are electrons, so $q=-e$ and the electron drift velocity $\vec{v}_d$ points opposite to $\vec{J}$. If the carriers were positive instead, then $\vec{v}_d$ and $\vec{J}$ would point in the same direction.}
\mprop{Microscopic-macroscopic current relations}{Let $n$ denote the carrier number density, let $q$ denote the charge of each carrier, let $\vec{v}_d$ denote the drift velocity, and let $S$ be a surface with oriented area element $d\vec{A}$. Then the current density is
\[
\vec{J}=nq\vec{v}_d.
\]
The total current through $S$ is
\[
I=\iint_S \vec{J}\cdot d\vec{A}.
\]
If $\vec{J}$ is uniform across a flat cross section of area $A$ and parallel to the area normal, then
\[
I=JA
\qquad \text{and} \qquad
J=\frac{I}{A}.
\]
For a straight wire with uniform carrier density, the corresponding magnitude relation is
\[
I=n|q|Av_d,
\]
where $v_d=|\vec{v}_d|$.}
\qs{Worked AP-style problem}{A long copper wire carries a steady current
\[
I=3.0\,\mathrm{A}
\]
to the right. The wire has radius
\[
r=0.80\,\mathrm{mm}=8.0\times 10^{-4}\,\mathrm{m},
\]
so its cross-sectional area is $A=\pi r^2$. Assume the conduction-electron number density is
\[
n=8.5\times 10^{28}\,\mathrm{m^{-3}},
\]
and the charge of each electron is
\[
q_e=-1.60\times 10^{-19}\,\mathrm{C}.
\]
Let $+\hat{\imath}$ point to the right.
Find:
\begin{enumerate}[label=(\alph*)]
\item the current density magnitude $J$,
\item the current density vector $\vec{J}$, and
\item the electron drift velocity vector $\vec{v}_d$.
\end{enumerate}}
\sol First compute the wire's cross-sectional area:
\[
A=\pi r^2=\pi(8.0\times 10^{-4}\,\mathrm{m})^2.
\]
Since
\[
(8.0\times 10^{-4})^2=6.4\times 10^{-7},
\]
we get
\[
A=\pi(6.4\times 10^{-7})\,\mathrm{m^2}=2.01\times 10^{-6}\,\mathrm{m^2}.
\]
For part (a), use
\[
J=\frac{I}{A}.
\]
Then
\[
J=\frac{3.0\,\mathrm{A}}{2.01\times 10^{-6}\,\mathrm{m^2}}=1.49\times 10^6\,\mathrm{A/m^2}.
\]
For part (b), the current is to the right, so conventional current and $\vec{J}$ point to the right:
\[
\vec{J}=(1.49\times 10^6\,\mathrm{A/m^2})\hat{\imath}.
\]
For part (c), use the microscopic relation
\[
\vec{J}=nq_e\vec{v}_d.
\]
Solve for the drift velocity:
\[
\vec{v}_d=\frac{\vec{J}}{nq_e}.
\]
Because $q_e$ is negative, $\vec{v}_d$ must point opposite to $\vec{J}$, so it points left. Its magnitude is
\[
v_d=\frac{J}{n|q_e|}.
\]
Substitute the values:
\[
v_d=\frac{1.49\times 10^6}{(8.5\times 10^{28})(1.60\times 10^{-19})}\,\mathrm{m/s}.
\]
The denominator is
\[
(8.5\times 10^{28})(1.60\times 10^{-19})=1.36\times 10^{10},
\]
so
\[
v_d=\frac{1.49\times 10^6}{1.36\times 10^{10}}\,\mathrm{m/s}=1.10\times 10^{-4}\,\mathrm{m/s}.
\]
Therefore the drift velocity vector is
\[
\vec{v}_d=-(1.10\times 10^{-4}\,\mathrm{m/s})\hat{\imath}.
\]
Therefore,
\[
J=1.49\times 10^6\,\mathrm{A/m^2},
\qquad
\vec{J}=(1.49\times 10^6\,\mathrm{A/m^2})\hat{\imath},
\]
\[
\vec{v}_d=-(1.10\times 10^{-4}\,\mathrm{m/s})\hat{\imath}.
\]

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\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 $1\,\Omega = 1\,\mathrm{V/A}$.}
\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,
\[
R = 0.271\,\Omega,
\qquad
I = 36.9\,\mathrm{A},
\qquad
J = 1.18 \times 10^7\,\mathrm{A/m^2},
\]
\[
\vec{J} = \left(1.18 \times 10^7\,\mathrm{A/m^2}\right)\hat{\imath},
\qquad
v_d = 8.69 \times 10^{-4}\,\mathrm{m/s},
\qquad
P = 369\,\mathrm{W}.
\]

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\subsection{Electric Power and Dissipation}
Electric power quantifies the rate at which electrical energy is transferred or dissipated in a circuit element. When a charge $q$ moves through a potential difference $\Delta V$, its electric potential energy changes by $\Delta U = q\,\Delta V$. The power delivered to (or dissipated by) a circuit element is the rate of this energy transfer.
\dfn{Electric power}{Let a circuit element have a potential difference $\Delta V$ across it and carry a current $I$ through it. The \emph{electric power} delivered to the element is
\[
P = I\,\Delta V.
\]
The SI unit of power is the watt ($1\,\mathrm{W}=1\,\mathrm{J/s}=1\,\mathrm{A\!\cdot\!V}$). When the element is a resistor, electrical energy is dissipated as thermal energy (Joule heating).}
\nt{Power is the rate of energy transfer. A battery \emph{supplies} power to the circuit, while resistors \emph{dissipate} it. In a steady circuit, the total power supplied equals the total power dissipated.}
\ex{Illustrative example}{A $60\,\mathrm{W}$ incandescent lightbulb is connected to a $120\,\mathrm{V}$ household outlet. Find (a) the current through the bulb and (b) the resistance of its filament.
\sol (a) From $P = I\,\Delta V$,
\[
I = \frac{P}{\Delta V} = \frac{60\,\mathrm{W}}{120\,\mathrm{V}} = 0.50\,\mathrm{A}.
\]
(b) From Ohm's law,
\[
R = \frac{\Delta V}{I} = \frac{120\,\mathrm{V}}{0.50\,\mathrm{A}} = 240\,\Omega. \qedhere
\]}
\nt{The $I^{2}R$ form shows why transmission lines use very high voltages: for a fixed power $P=IV$, raising the voltage lowers the current, and since resistive losses scale as $I^{2}R$, the dissipation drops dramatically.}
\thm{Microscopic power density}{In a continuous medium, the local rate of energy dissipation per unit volume is the dot product of the current density $\vec{J}$ and the electric field $\vec{E}$ at that point:
\[
p = \vec{J}\cdot\vec{E}.
\]
The total power delivered to a volume $\mathcal{V}$ is the integral
\[
P = \iiint_{\mathcal{V}} \vec{J}\cdot\vec{E}\; d\tau.
\]}
\pf{Microscopic to macroscopic}{Consider a straight wire of length $L$ and cross-sectional area $A$, with a uniform electric field $\vec{E}$ along its axis and a uniform current $I$. The field is related to the potential difference by $\Delta V = E\,L$. The current density is $J = I/A$, and by Ohm's law $E = \rho\,J$ (where $\rho$ is the resistivity). The total power dissipated is
\[
P = \iiint_{\mathcal{V}} \vec{J}\cdot\vec{E}\; d\tau = J\,E\,(A\,L)
\]
since $\vec{J}$ and $\vec{E}$ are parallel and uniform. Substituting $J = I/A$ and $E = \Delta V/L$,
\[
P = \left(\frac{I}{A}\right)\!\left(\frac{\Delta V}{L}\right)(A\,L) = I\,\Delta V,
\]
recovering the macroscopic expression. Alternatively, using $E = \rho\,J = \rho\,(I/A)$ and $R = \rho\,L/A$,
\[
P = J\,E\,(A\,L) = \frac{I^{2}}{A^{2}}\cdot\rho\frac{I}{A}\cdot A\,L
= I^{2}\,\frac{\rho\,L}{A} = I^{2}R. \qedhere
\]
\qs{Worked example}{A resistor is connected across a $12.0\,\mathrm{V}$ battery. The current through the resistor is measured to be $2.00\,\mathrm{A}$.
Find:
\begin{enumerate}[label=(\alph*)]
\item the power dissipated by the resistor,
\item the resistance of the resistor, and
\item the total energy dissipated as heat over a time interval of $5.00\,\mathrm{min}$.
\end{enumerate}}
\sol \textbf{Part (a).} The power dissipated by the resistor is given by
\[
P = I\,\Delta V.
\]
Substitute the given values:
\[
P = (2.00\,\mathrm{A})(12.0\,\mathrm{V}) = 24.0\,\mathrm{W}.
\]
\textbf{Part (b).} By Ohm's law,
\[
R = \frac{\Delta V}{I}.
\]
Substitute the values:
\[
R = \frac{12.0\,\mathrm{V}}{2.00\,\mathrm{A}} = 6.00\,\Omega.
\]
As a check, verify using $P = I^{2}R$:
\[
P = (2.00\,\mathrm{A})^{2}(6.00\,\Omega) = 4.00 \times 6.00 = 24.0\,\mathrm{W},
\]
which agrees with part (a).}
\textbf{Part (c).} The total energy dissipated is power multiplied by time:
\[
E = P\,t.
\]
Convert the time to seconds:
\[
t = 5.00\,\mathrm{min}\times 60.0\,\mathrm{s/min} = 300\,\mathrm{s}.
\]
Then
\[
E = (24.0\,\mathrm{W})(300\,\mathrm{s}) = 7200\,\mathrm{J}.
\]
Alternatively, using $E = I\,\Delta V\,t$ directly:
\[
E = (2.00\,\mathrm{A})(12.0\,\mathrm{V})(300\,\mathrm{s}) = 7200\,\mathrm{J}.
\]
\nt{Energy can also be written as $E = I^{2}Rt$ or $E = (\Delta V)^{2}t/R$. All three are equivalent and follow from $E = P\,t$ together with Ohm's law.}

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\subsection{Equivalent Resistance of Series and Parallel Circuits}
This subsection defines equivalent resistance for resistive networks, derives the series and parallel combination rules from Kirchhoff's laws and Ohm's law, and shows how to reduce compound circuits step by step to a single equivalent resistor.
\dfn{Equivalent resistance}{Consider a network of resistors connected to an ideal battery of emf $\mathcal{E}$. The \emph{equivalent resistance} $R_{\mathrm{eq}}$ of the network is defined by
\[
R_{\mathrm{eq}} = \frac{\mathcal{E}}{I_{\mathrm{total}}},
\]
where $I_{\mathrm{total}}$ is the total current delivered by the battery. The equivalent resistance is the resistance of a single resistor that would draw the same current from the same battery as the entire network.}
\nt{Equivalent resistance is a bookkeeping device: it replaces an entire resistive sub-network by a single resistor whose effect on the rest of the circuit is identical. The replacement is always done between two terminals, and it preserves the $I$--$V$ relationship at those terminals.}
\dfn{Series combination}{Two (or more) resistors are in \emph{series} when they share exactly one common node and no other element is connected to that node. Equivalently, the same current $I$ flows through each resistor in a series chain, and the total voltage is the sum of individual voltage drops:
\[
V = V_1 + V_2 + \cdots + V_n.
\]
The equivalent resistance of $n$ resistors in series is
\[
R_{\mathrm{eq}} = R_1 + R_2 + \cdots + R_n = \sum_{i=1}^{n} R_i.
\]}
\thm{Voltage divider rule}{Consider two resistors $R_1$ and $R_2$ in series connected to a potential difference $V_{\mathrm{in}}$. The voltage across $R_2$ alone is
\[
V_{\mathrm{out}} = V_{R_2} = V_{\mathrm{in}}\,\frac{R_2}{R_1 + R_2}.
\]
This relation follows from Ohm's law and the fact that the same current $I = V_{\mathrm{in}}/(R_1+R_2)$ flows through both resistors.}
\nt{The voltage divider distributes the input voltage proportionally to each resistance. A larger resistance drops more voltage. This rule is the electrical analogue of the weighted average: $V_{R_i}$ is the fraction of $V_{\mathrm{in}}$ carried by $R_i$.}
\dfn{Parallel combination}{Two (or more) resistors are in \emph{parallel} when they are connected between the same two nodes, so the potential difference $V$ across each is identical. The total current splits among the branches:
\[
I_{\mathrm{total}} = I_1 + I_2 + \cdots + I_n.
\]
The equivalent resistance of $n$ resistors in parallel is
\[
\frac{1}{R_{\mathrm{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n} = \sum_{i=1}^{n} \frac{1}{R_i}.
\]
For two resistors in parallel, this simplifies to
\[
R_{\mathrm{eq}} = \frac{R_1 R_2}{R_1 + R_2}.
\]}
\thm{Current divider rule}{Consider two resistors $R_1$ and $R_2$ in parallel connected to a total current $I_{\mathrm{total}}$. The current through $R_1$ is
\[
I_1 = I_{\mathrm{total}}\,\frac{R_2}{R_1 + R_2}.
\]
This follows from KCL and Ohm's law: the common voltage is $V = I_{\mathrm{total}}/G_{\mathrm{eq}}$ where $G_{\mathrm{eq}} = 1/R_1 + 1/R_2$, and $I_1 = V/R_1$.}
\nt{Current divides inversely to resistance: the smaller resistor carries more current. In the limit $R_1 \ll R_2$, essentially all current flows through $R_1$; we say $R_1$ \emph{shunts} $R_2$.}
\mprop{Series and parallel equivalent resistance}{For $n$ resistors, the equivalent resistance satisfies
\[
\text{Series:}\quad R_{\mathrm{eq}} = \sum_{i=1}^{n} R_i
\qquad\text{and}\qquad
\text{Parallel:}\quad \frac{1}{R_{\mathrm{eq}}} = \sum_{i=1}^{n} \frac{1}{R_i}.
\]
The current through and voltage across each individual resistor are obtained by applying Ohm's law $V_i = I_i R_i$ once $I_i$ or $V_i$ is known from the reduction analysis.}
\pf{Series combination}{Kirchhoff's voltage law (the loop rule) states that the sum of potential differences around any closed loop is zero. For two resistors in series with a battery of emf $\mathcal{E}$,
\[
\mathcal{E} - V_1 - V_2 = 0,
\]
so $V_1 + V_2 = \mathcal{E}$. Since the same current $I$ flows through both resistors, $V_1 = IR_1$ and $V_2 = IR_2$. Substituting gives $I(R_1 + R_2) = \mathcal{E}$, or $R_{\mathrm{eq}} = \mathcal{E}/I = R_1 + R_2$. The result extends immediately to $n$ resistors by induction.}
\pf{Parallel combination}{Kirchhoff's current law (the node rule) states that the sum of currents entering a node equals the sum of currents leaving. At the top node of the parallel connection, $I_{\mathrm{total}} = I_1 + I_2 + \cdots + I_n$. Since each branch has the same voltage $V$, we have $I_i = V/R_i$. Thus
\[
I_{\mathrm{total}} = V\sum_{i=1}^{n} \frac{1}{R_i}.
\]
But $I_{\mathrm{total}} = V/R_{\mathrm{eq}}$, so $1/R_{\mathrm{eq}} = \sum 1/R_i$.}
\cor{Bounds on equivalent resistance}{For any combination of series and parallel resistors, the equivalent resistance of a parallel block is always less than the smallest resistance in that block:
\[
R_{\mathrm{parallel}} < \min_i(R_i).
\]
Conversely, the equivalent resistance of a series chain is always greater than the largest individual resistance. These bounds follow from the positivity of all resistances and provide a useful sanity check on computed results.}
\ex{Illustrative example}{When resistors are identical, the formulas simplify. $n$ identical resistors of resistance $R$ in series give $R_{\mathrm{eq}} = nR$. In parallel, $R_{\mathrm{eq}} = R/n$. Thus three $12\,\Omega$ resistors in parallel give $R_{\mathrm{eq}} = 4\,\Omega$, while in series they give $36\,\Omega$.}
\nt{The strategy for analysing a mixed (compound) circuit is: (1) identify the innermost series or parallel groups. (2) Replace each group by its equivalent resistance. (3) Repeat until the entire network is reduced to a single resistor $R_{\mathrm{eq}}$. (4) Use $I_{\mathrm{total}} = \mathcal{E}/R_{\mathrm{eq}}$ to find the battery current, then work backwards through the reduction steps to find individual branch currents and voltages.}
\nt{On the AP Physics C E\&M exam, equivalent resistance problems test your ability to correctly identify series versus parallel connections and to apply Kirchhoff's laws systematically. The most common errors are misidentifying which elements share the same current (series) and which share the same voltage (parallel). Always trace the current paths and label the nodes to verify your classification.}
\qs{Worked example}{An ideal battery of emf
\[
\mathcal{E} = 24\,\mathrm{V}
\]
is connected to a resistive network consisting of four resistors arranged as follows. Resistors $R_1 = 2.0\,\Omega$ and $R_2 = 4.0\,\Omega$ are connected in series. This series combination is connected in parallel with resistor $R_3 = 6.0\,\Omega$. Finally, resistor $R_4 = 3.0\,\Omega$ is connected in series with this entire parallel block, and the combination is connected to the battery.
Find:
\begin{enumerate}[label=(\alph*)]
\item the total equivalent resistance $R_{\mathrm{eq}}$ of the network,
\item the total current $I_{\mathrm{total}}$ delivered by the battery,
\item the voltage drop across each resistor, and
\item the current through each resistor.
\end{enumerate}}
\sol \textbf{Part (a).} Begin by reducing the network from the inside out. Resistors $R_1$ and $R_2$ are in series, so their combined resistance is
\[
R_{12} = R_1 + R_2 = 2.0\,\Omega + 4.0\,\Omega = 6.0\,\Omega.
\]
This series combination is in parallel with $R_3 = 6.0\,\Omega$. The equivalent resistance of the parallel block is
\[
\frac{1}{R_p} = \frac{1}{R_{12}} + \frac{1}{R_3} = \frac{1}{6.0\,\Omega} + \frac{1}{6.0\,\Omega} = \frac{2}{6.0\,\Omega} = \frac{1}{3.0\,\Omega}.
\]
Hence $R_p = 3.0\,\Omega$.
The parallel block is in series with $R_4$, so the total equivalent resistance is
\[
R_{\mathrm{eq}} = R_p + R_4 = 3.0\,\Omega + 3.0\,\Omega = 6.0\,\Omega.
\]
\textbf{Part (b).} Ohm's law gives the total current from the battery:
\[
I_{\mathrm{total}} = \frac{\mathcal{E}}{R_{\mathrm{eq}}} = \frac{24\,\mathrm{V}}{6.0\,\Omega} = 4.0\,\mathrm{A}.
\]
\textbf{Parts (c) and (d).} Now work backwards through the reduction. The current through $R_4$ equals the total current, since $R_4$ is in series with the battery:
\[
I_4 = I_{\mathrm{total}} = 4.0\,\mathrm{A}.
\]
The voltage drop across $R_4$ is
\[
V_4 = I_4 R_4 = (4.0\,\mathrm{A})(3.0\,\Omega) = 12\,\mathrm{V}.
\]
The remaining voltage appears across the parallel block:
\[
V_p = \mathcal{E} - V_4 = 24\,\mathrm{V} - 12\,\mathrm{V} = 12\,\mathrm{V}.
\]
Since the resistors $R_1$, $R_2$, and $R_3$ are all connected to the parallel block, the voltage across each is $V_p = 12\,\mathrm{V}$ (for $R_3$ directly) and $V_p = 12\,\mathrm{V}$ (for the series pair $R_1$--$R_2$).
The current through $R_3$ is
\[
I_3 = \frac{V_p}{R_3} = \frac{12\,\mathrm{V}}{6.0\,\Omega} = 2.0\,\mathrm{A}.
\]
The current through the $R_1$--$R_2$ branch is
\[
I_{12} = \frac{V_p}{R_{12}} = \frac{12\,\mathrm{V}}{6.0\,\Omega} = 2.0\,\mathrm{A}.
\]
Since $R_1$ and $R_2$ are in series, the same current flows through both:
\[
I_1 = I_2 = I_{12} = 2.0\,\mathrm{A}.
\]
The individual voltage drops are
\[
V_1 = I_1 R_1 = (2.0\,\mathrm{A})(2.0\,\Omega) = 4.0\,\mathrm{V},
\]
\[
V_2 = I_2 R_2 = (2.0\,\mathrm{A})(4.0\,\Omega) = 8.0\,\mathrm{V}.
\]
\noindent\textbf{Checks:}
\begin{itemize}
\item KVL on the left loop: $V_1 + V_2 = 4.0\,\mathrm{V} + 8.0\,\mathrm{V} = 12\,\mathrm{V} = V_p$.
\item KCL at the junction: $I_{12} + I_3 = 2.0\,\mathrm{A} + 2.0\,\mathrm{A} = 4.0\,\mathrm{A} = I_{\mathrm{total}}$.
\item KVL on the full loop: $V_p + V_4 = 12\,\mathrm{V} + 12\,\mathrm{V} = 24\,\mathrm{V} = \mathcal{E}$.
\end{itemize}
Therefore,
\begin{align*}
R_{\mathrm{eq}} &= 6.0\,\Omega,\\
I_{\mathrm{total}} &= 4.0\,\mathrm{A},\\
(V_1, V_2, V_3, V_4) &= (4.0\,\mathrm{V}, 8.0\,\mathrm{V}, 12\,\mathrm{V}, 12\,\mathrm{V}),\\
(I_1, I_2, I_3, I_4) &= (2.0\,\mathrm{A}, 2.0\,\mathrm{A}, 2.0\,\mathrm{A}, 4.0\,\mathrm{A}).
\end{align*}

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\subsection{Kirchhoff's Junction and Loop Rules}
This subsection introduces Kirchhoff's two rules for analyzing electric circuits. The junction rule (Kirchhoff's current law) expresses charge conservation at circuit nodes. The loop rule (Kirchhoff's voltage law) expresses energy conservation around any closed loop. Together they provide a systematic method for finding unknown currents in multi-loop circuits.
\dfn{Electric circuit junction (node)}{A \emph{junction} (or \emph{node}) is a point in a circuit where three or more conductors meet. The current in each conductor leading to the junction is called a \emph{branch current}.}
\nt{In a steady-state dc circuit the charge at any junction is constant: no charge accumulates at a node. This is the physical reason the junction rule holds.}
\thm{Kirchhoff's junction rule (KCL)}{At any junction in a steady-state dc circuit, the sum of currents entering the junction equals the sum of currents leaving the junction:
\begin{equation}
\sum I_{\text{in}} = \sum I_{\text{out}}.
\end{equation}
Equivalently, the algebraic sum of all currents at a junction is zero:
\begin{equation}
\sum I = 0,
\end{equation}
where currents entering the junction are taken as positive and currents leaving are taken as negative.
\textbf{Calc 3 connection:} The junction rule is the circuit analogue of the steady-state continuity equation $\nabla \cdot \vec{J} = 0$ (charge conservation with no time-varying charge density). Integrating $\nabla \cdot \vec{J} = 0$ over a small volume enclosing the junction gives $\oiint \vec{J} \cdot d\vec{A} = 0$, which is exactly $\sum I = 0$.}
\dfn{Electric circuit loop}{A \emph{loop} is any closed path through a circuit, traversing a sequence of circuit elements and returning to the starting point.}
\dfn{Sign convention for traversing circuit elements}{When applying the loop rule, traverse the loop in a chosen direction and assign potential changes as follows:
\begin{itemize}
\item \emph{Resistor:} Traversing in the \emph{same} direction as the assumed current gives a potential \emph{drop} of $IR$ (contribute $-IR$). Traversing \emph{opposite} to the assumed current gives a potential \emph{rise} of $IR$ (contribute $+IR$).
\item \emph{Battery (emf):} Traversing from the \emph{negative} terminal to the \emph{positive} terminal gives a potential \emph{rise} of $\mathcal{E}$ (contribute $+\mathcal{E}$). Traversing from \emph{positive} to \emph{negative} gives a potential \emph{drop} of $\mathcal{E}$ (contribute $-\mathcal{E}$).
\end{itemize}
The assumed direction of each branch current must be declared before writing equations. If the solved value of a current is negative, the actual current flows opposite to the assumed direction.}
\thm{Kirchhoff's loop rule (KVL)}{For any closed loop in a circuit, the algebraic sum of the potential differences across all elements in the loop is zero:
\begin{equation}
\sum \Delta V = 0.
\end{equation}
\textbf{Calc 3 connection:} The loop rule follows from energy conservation: a test charge $q$ that moves around a closed path and returns to its starting point must have zero net change in potential energy, so $\oint \vec{E} \cdot d\vec{\ell} = 0$. In electrostatics this is a consequence of $\vec{E}$ being a conservative field. More generally, in quasi-static circuits with no changing magnetic flux through the loop, Faraday's law gives $\nabla \times \vec{E} = -\partial \vec{B}/\partial t = 0$, so the integral over any closed loop vanishes.}
\nt{The loop rule holds for \emph{any} closed loop, not just the obvious ``mesh'' loops of a circuit diagram. Any closed path through the elements counts. In practice one typically chooses the minimal loops (meshes) because they lead to the most economical system of equations.}
\nt{The sign of a solved current encodes direction. A negative result does \emph{not} mean the current is unphysical --- it simply means the actual current is opposite to the assumed direction. Always state the assumed direction and report the final direction clearly.}
\mprop{Algorithm for solving multi-loop circuits}{Given a multi-loop circuit with unknown branch currents:
\begin{enumerate}
\item Label every branch with a current variable and an assumed direction.
\item Identify all junctions. For $N$ junctions, write $N-1$ independent junction equations.
\item Choose enough independent loops (typically meshes) so that the total number of equations (junction + loop) equals the number of unknown currents.
\item Apply the loop rule to each chosen loop, using the sign conventions above.
\item Solve the resulting system of linear equations.
\item Check: any current with a negative value flows opposite to the assumed direction. Verify that all junction equations are satisfied.
\end{enumerate}
The number of independent equations needed equals the number of unknown branch currents.}
\ex{Illustrative example}{Consider a junction with three wires meeting. Current $I_1 = 3.0\,\mathrm{A}$ enters the junction and current $I_2 = 1.5\,\mathrm{A}$ leaves. The third branch carries current $I_3$. By the junction rule, $I_1 = I_2 + I_3$, so $I_3 = 1.5\,\mathrm{A}$ leaves the junction.}
\nt{For circuits with a single battery and resistors in simple series or parallel, the equivalent-resistance method is faster. Kirchhoff's rules are needed whenever the circuit cannot be reduced to simple series/parallel combinations --- for instance, when there are two or more batteries arranged in different branches, forming multiple loops.}
\qs{Worked example}{Consider the two-loop circuit shown in the diagram below. The circuit consists of a left loop and a right loop sharing a common middle branch.
\begin{itemize}
\item The \emph{left branch} contains a battery of emf $\mathcal{E}_1 = 12\,\mathrm{V}$ (positive terminal up) in series with a resistor $R_1 = 4.0\,\Omega$.
\item The \emph{middle branch} contains a resistor $R_3 = 3.0\,\Omega$.
\item The \emph{right branch} contains a battery of emf $\mathcal{E}_2 = 6.0\,\mathrm{V}$ (positive terminal up) in series with a resistor $R_2 = 6.0\,\Omega$.
\end{itemize}
\begin{center}
\textbf{Diagram description:} Two rectangular loops share a vertical middle branch. The left vertical branch has the $12\,\mathrm{V}$ battery (positive up) and $4.0\,\Omega$ resistor. The middle vertical branch has the $3.0\,\Omega$ resistor. The right vertical branch has the $6.0\,\mathrm{V}$ battery (positive up) and $6.0\,\Omega$ resistor. All three vertical branches connect at top and bottom horizontal wires (ideal conductors with zero resistance).
\end{center}
Let the top junction be $A$ and the bottom junction be $B$. Define three branch currents:
\begin{itemize}
\item $I_1$ flows \emph{upward} through the left branch (from $B$ to $A$).
\item $I_2$ flows \emph{upward} through the right branch (from $B$ to $A$).
\item $I_3$ flows \emph{downward} through the middle branch (from $A$ to $B$).
\end{itemize}
Assume these directions when applying Kirchhoff's rules.
Find:
\begin{enumerate}[label=(\alph*)]
\item the junction equation at node $A$,
\item the two independent loop equations (left loop and right loop),
\item the values of all three currents $I_1$, $I_2$, and $I_3$, and
\item the direction of each current (consistent with the assumed direction).
\end{enumerate}}
\sol \textbf{Part (a). Junction equation.} At junction $A$, the currents $I_1$ and $I_2$ both enter (they flow upward from $B$ to $A$ in their respective branches). The current $I_3$ leaves $A$ (it flows downward from $A$ to $B$). By the junction rule:
\[
I_1 + I_2 = I_3.
\]
This is our first equation.
\textbf{Part (b). Loop equations.}
\emph{Left loop} (traverse clockwise starting from junction $B$):
\begin{itemize}
\item Go up through the left branch, in the same direction as $I_1$: the battery $\mathcal{E}_1$ is traversed from $-$ to $+$, contributing $+\mathcal{E}_1 = +12\,\mathrm{V}$. The resistor $R_1$ is traversed in the same direction as $I_1$, contributing $-I_1 R_1 = -4.0\,I_1$.
\item Go across the top wire from the left branch to the middle branch (ideal wire, $\Delta V = 0$).
\item Go down through the middle branch, in the same direction as $I_3$: resistor $R_3$ is traversed in the direction of $I_3$, contributing $-I_3 R_3 = -3.0\,I_3$.
\item Go across the bottom wire back to $B$ (ideal wire, $\Delta V = 0$).
\end{itemize}
Summing around the loop:
\[
+12 - 4.0\,I_1 - 3.0\,I_3 = 0,
\]
\[
4.0\,I_1 + 3.0\,I_3 = 12.
\]
\label{eq:loop1}
\emph{Right loop} (traverse clockwise starting from junction $A$):
\begin{itemize}
\item Go down through the right branch, in the \emph{opposite} direction to $I_2$: the resistor $R_2$ is traversed opposite to $I_2$, contributing $+I_2 R_2 = +6.0\,I_2$. The battery $\mathcal{E}_2$ is traversed from $+$ to $-$, contributing $-\mathcal{E}_2 = -6\,\mathrm{V}$.
\item Go across the bottom wire from right to middle (ideal wire, $\Delta V = 0$).
\item Go up through the middle branch, in the \emph{opposite} direction to $I_3$: resistor $R_3$ is traversed opposite to $I_3$, contributing $+I_3 R_3 = +3.0\,I_3$.
\item Go across the top wire back to $A$ (ideal wire, $\Delta V = 0$).
\end{itemize}
Summing around the loop:
\[
+6.0\,I_2 - 6 + 3.0\,I_3 = 0,
\]
\[
6.0\,I_2 + 3.0\,I_3 = 6.
\]
\label{eq:loop2}
\textbf{Part (c). Solving for the currents.} We have three equations:
\begin{align}
I_3 &= I_1 + I_2, &\text{(junction)} \\
4.0\,I_1 + 3.0\,I_3 &= 12, &\text{(left loop)} \\
6.0\,I_2 + 3.0\,I_3 &= 6. &\text{(right loop)}
\end{align}
Substitute equation (1) into equation (2):
\[
4.0\,I_1 + 3.0\,(I_1 + I_2) = 12,
\]
\[
4.0\,I_1 + 3.0\,I_1 + 3.0\,I_2 = 12,
\]
\[
7.0\,I_1 + 3.0\,I_2 = 12.
\]
\label{eq:a}
Substitute equation (1) into equation (3):
\[
6.0\,I_2 + 3.0\,(I_1 + I_2) = 6,
\]
\[
6.0\,I_2 + 3.0\,I_1 + 3.0\,I_2 = 6,
\]
\[
3.0\,I_1 + 9.0\,I_2 = 6.
\]
Dividing by $3.0$:
\[
I_1 + 3.0\,I_2 = 2.0,
\]
so
\[
I_1 = 2.0 - 3.0\,I_2.
\]
\label{eq:b}
Substitute equation (\ref{eq:b}) into equation (\ref{eq:a}):
\[
7.0\,(2.0 - 3.0\,I_2) + 3.0\,I_2 = 12,
\]
\[
14.0 - 21.0\,I_2 + 3.0\,I_2 = 12,
\]
\[
14.0 - 18.0\,I_2 = 12,
\]
\[
-18.0\,I_2 = -2.0,
\]
\[
I_2 = \frac{2.0}{18.0} = \frac{1}{9}\,\mathrm{A}.
\]
From equation (\ref{eq:b}):
\[
I_1 = 2.0 - 3.0\left(\frac{1}{9}\right) = 2.0 - \frac{1}{3} = \frac{6}{3} - \frac{1}{3} = \frac{5}{3}\,\mathrm{A}.
\]
From the junction equation (1):
\[
I_3 = I_1 + I_2 = \frac{5}{3} + \frac{1}{9} = \frac{15}{9} + \frac{1}{9} = \frac{16}{9}\,\mathrm{A}.
\]
\textbf{Verification.} Check against the left-loop equation:
\[
4.0\left(\frac{5}{3}\right) + 3.0\left(\frac{16}{9}\right) = \frac{20}{3} + \frac{16}{3} = \frac{36}{3} = 12.
\]
Check against the right-loop equation:
\[
6.0\left(\frac{1}{9}\right) + 3.0\left(\frac{16}{9}\right) = \frac{6}{9} + \frac{48}{9} = \frac{54}{9} = 6.
\]
Both are satisfied.
\textbf{Part (d). Directions.} All three currents are positive, confirming they flow in the assumed directions:
\begin{itemize}
\item $I_1 = 5/3\,\mathrm{A}$ upward in the left branch,
\item $I_2 = 1/9\,\mathrm{A}$ upward in the right branch,
\item $I_3 = 16/9\,\mathrm{A}$ downward in the middle branch.
\end{itemize}
Therefore, the branch currents are:
\[
I_1 = \frac{5}{3}\,\mathrm{A},\qquad
I_2 = \frac{1}{9}\,\mathrm{A},\qquad
I_3 = \frac{16}{9}\,\mathrm{A},
\]
all flowing in the assumed directions.

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\subsection{RC Transients and the Time Constant}
This subsection introduces RC circuits -- circuits containing resistors and capacitors -- derives the first-order differential equation governing charge evolution during charging and discharging, solves it by separation of variables, and defines the time constant $\tau = RC$ as the characteristic timescale of the transient response.
\dfn{RC circuit and transient response}{A series \emph{RC circuit} consists of a resistor $R$, a capacitor $C$, and (optionally) a battery of emf $\mathcal{E}$, all connected in a single closed loop. When the circuit is first connected (or when the battery is disconnected), the capacitor neither holds its initial charge nor its final charge instantaneously; instead, its charge evolves over time. This time-dependent behavior is called a \emph{transient response}.}
\nt{At the instant a capacitor begins charging, it behaves like a short circuit: its voltage is zero and all of the battery voltage appears across the resistor. As the capacitor charges, its voltage increases and the current decreases. In the limit $t \to \infty$, the capacitor is fully charged to voltage $\mathcal{E}$ and the current drops to zero, so the capacitor acts like an open circuit.}
\thm{Charging a capacitor}{A capacitor of capacitance $C$ that is initially uncharged is connected in series at $t = 0$ with a resistor of resistance $R$ and a battery of emf $\mathcal{E}$, forming a single-loop circuit. The charge on the capacitor at time $t$ is
\[
q(t) = C\mathcal{E}\,\bigl(1 - e^{-t/RC}\bigr),
\]
and the current flowing through the resistor is
\[
I(t) = \frac{\mathcal{E}}{R}\,e^{-t/RC}.
\]
Here $q(0) = 0$ and $I(0) = \mathcal{E}/R$. As $t \to \infty$, $q \to C\mathcal{E}$ and $I \to 0$.}
\pf{Derivation of charging equations}{Apply Kirchhoff's voltage law around the loop. The potential drops across the resistor and capacitor sum to the battery emf:
\[
\mathcal{E} - IR - \frac{q}{C} = 0,
\]
where $I = \dfrac{dq}{dt}$ is the current (the rate at which charge accumulates on the capacitor). Substituting gives the first-order linear ODE
\[
R\,\frac{dq}{dt} + \frac{q}{C} = \mathcal{E}.
\]
Divide by $R$:
\[
\frac{dq}{dt} + \frac{1}{RC}\,q = \frac{\mathcal{E}}{R}.
\]
This is a first-order linear ODE. Use the integrating factor $\mu(t) = e^{t/RC}$:
\[
\frac{d}{dt}\!\left(q\,e^{t/RC}\right) = \frac{\mathcal{E}}{R}\,e^{t/RC}.
\]
Integrate both sides from $0$ to $t$, with $q(0) = 0$:
\[
q(t)\,e^{t/RC} - q(0) = \frac{\mathcal{E}}{R}\int_0^t e^{t'/RC}\,dt' = \mathcal{E}C\left(e^{t/RC} - 1\right).
\]
Solving for $q(t)$:
\[
q(t) = C\mathcal{E}\left(1 - e^{-t/RC}\right).
\]
The current is obtained from $I = dq/dt$:
\[
I(t) = \frac{d}{dt}\!\left[C\mathcal{E}\left(1 - e^{-t/RC}\right)\right]
= \frac{\mathcal{E}}{R}\,e^{-t/RC}.
\]
The initial current is $I(0) = \mathcal{E}/R$, and as $t \to \infty$ the exponential vanishes, so $q \to C\mathcal{E}$ and $I \to 0$.}
\thm{Discharging a capacitor}{A capacitor of capacitance $C$ that is initially charged to charge $q_0$ is connected at $t = 0$ with a resistor of resistance $R$, forming a single-loop circuit with no battery. The charge on the capacitor at time $t$ is
\[
q(t) = q_0\,e^{-t/RC},
\]
and the rate of change of charge is
\[
\frac{dq}{dt} = -\,\frac{q_0}{RC}\,e^{-t/RC}.
\]
The negative sign indicates that charge is \emph{decreasing}: the capacitor is discharging. The magnitude of the current through the resistor is
\[
|I(t)| = \left|\frac{dq}{dt}\right| = \frac{q_0}{RC}\,e^{-t/RC}.
\]
As $t \to \infty$, both $q \to 0$ and $I \to 0$.}
\pf{Derivation of discharging equations}{With no battery in the loop, Kirchhoff's voltage law gives
\[
-\,\frac{q}{C} - IR = 0,
\]
where $I$ is the current through the resistor and $q/C$ is the voltage across the capacitor. During discharge, charge flows off the capacitor, so $I = -\,dq/dt$ (current is positive while charge is decreasing). Substituting:
\[
-\,\frac{q}{C} - R\!\left(-\frac{dq}{dt}\right) = 0,
\]
which simplifies to
\[
\frac{dq}{dt} = -\,\frac{q}{RC}.
\]
Separate variables:
\[
\frac{dq}{q} = -\frac{dt}{RC}.
\]
Integrate from $0$ to $t$, with $q(0) = q_0$:
\[
\int_{q_0}^{q(t)}\frac{dq'}{q'} = -\frac{1}{RC}\int_0^t dt',
\qquad
\ln\!\left(\frac{q(t)}{q_0}\right) = -\frac{t}{RC}.
\]
Exponentiate:
\[
q(t) = q_0\,e^{-t/RC}.
\]
Differentiating gives the rate of change of charge:
\[
\frac{dq}{dt} = -\,\frac{q_0}{RC}\,e^{-t/RC}.
\]
The current flowing through the resistor (in the direction that discharges the capacitor) is $I = -\,dq/dt$, so
\[
I(t) = \frac{q_0}{RC}\,e^{-t/RC}.
\]
The voltage across the capacitor is $V_C = q/C = (q_0/C)e^{-t/RC}$, and the voltage across the resistor is $V_R = IR = (q_0/C)e^{-t/RC}$, so $V_R = V_C$ at every instant, consistent with the loop equation.}
\nt{During charging, the current is positive and flows onto the positively charged plate of the capacitor. During discharging, the current flows off the capacitor, and the instantaneous current through the resistor has the same magnitude as the rate at which charge leaves the capacitor: $|dq/dt| = I$.}
\dfn{Time constant of an RC circuit}{The \emph{time constant} of an RC circuit is
\[
\tau = RC,
\]
where $R$ is the resistance and $C$ is the capacitance. The SI unit of $\tau$ is the second (s). The time constant sets the characteristic timescale of the transient response: at $t = \tau$, the capacitor charge during charging reaches $1 - e^{-1} \approx 63.2\%$ of its final value $C\mathcal{E}$, and during discharging the charge falls to $e^{-1} \approx 36.8\%$ of its initial value $q_0$. After $5\tau$, the transient is essentially over: $1 - e^{-5} \approx 0.993$ of the final charge has been reached during charging, and $e^{-5} \approx 0.0067$ of the initial charge remains during discharging.}
\nt{The time constant is the product of two quantities with SI units $\Omega$ (ohms) and F (farads). Since $\Omega = \mathrm{V/A}$ and $\mathrm{F} = \mathrm{C/V}$, the product is $(\mathrm{V/A})(\mathrm{C/V}) = \mathrm{C/A} = \mathrm{C/(C/s)} = \mathrm{s}$, confirming that $\tau$ has units of time. A larger resistance slows the charge flow, giving a longer time constant. A larger capacitance stores more charge per volt, also requiring more time to charge or discharge, giving a longer time constant.}
\mprop{Charging and discharging at one time constant}{For a charging capacitor, at $t = \tau = RC$:
\[
q(\tau) = C\mathcal{E}\left(1 - \frac{1}{e}\right) \approx 0.632\,C\mathcal{E},
\qquad
I(\tau) = \frac{\mathcal{E}}{R}\cdot\frac{1}{e} \approx 0.368\,\frac{\mathcal{E}}{R}.
\]
For a discharging capacitor, at $t = \tau = RC$:
\[
q(\tau) = \frac{q_0}{e} \approx 0.368\,q_0,
\qquad
|I(\tau)| = \frac{q_0}{RC}\cdot\frac{1}{e} \approx 0.368\,\frac{q_0}{RC}.
\]
After $n$ time constants ($t = n\tau$):
\[
q_{\text{charge}} = C\mathcal{E}\left(1 - e^{-n}\right)
\qquad \text{and} \qquad
q_{\text{discharge}} = q_0\,e^{-n}.
\]
Thus, after $n = 5$, charging reaches $1 - e^{-5} \approx 99.3\%$ of full charge and discharging leaves only $e^{-5} \approx 0.67\%$ of the initial charge.}
\nt{The time constant is independent of the initial conditions and of the battery emf. It depends only on the circuit geometry (through $R$ and $C$). This is a hallmark of first-order linear systems: the timescale of the exponential decay is set by the coefficients of the differential equation, not by the particular solution's initial values.}
\thm{Energy during charging}{When an initially uncharged capacitor $C$ is charged through a resistor $R$ by a battery of emf $\mathcal{E}$, the total energy supplied by the battery is
\[
U_{\text{battery}} = C\mathcal{E}^2.
\]
The energy finally stored in the capacitor is
\[
U_C = \frac{1}{2}\,C\mathcal{E}^2.
\]
The remaining half,
\[
U_R = \frac{1}{2}\,C\mathcal{E}^2,
\]
is dissipated as Joule heat in the resistor during the charging process. The fraction dissipated in the resistor is exactly $50\%$, independent of the value of $R$.}
\pf{Energy during charging}{The battery supplies energy at rate $dU_{\text{battery}}/dt = \mathcal{E}I(t)$, so the total energy delivered during charging is
\[
U_{\text{battery}} = \int_0^\infty \mathcal{E}\,I(t)\,dt = \mathcal{E}\int_0^\infty \frac{\mathcal{E}}{R}\,e^{-t/RC}\,dt.
\]
The integral is
\[
\int_0^\infty e^{-t/RC}\,dt = RC,
\]
so
\[
U_{\text{battery}} = \frac{\mathcal{E}^2}{R}\cdot RC = C\mathcal{E}^2.
\]
The energy stored in the capacitor at full charge ($q = C\mathcal{E}$) is
\[
U_C = \frac{q^2}{2C} = \frac{(C\mathcal{E})^2}{2C} = \frac{1}{2}\,C\mathcal{E}^2.
\]
By energy conservation, the energy dissipated in the resistor is
\[
U_R = U_{\text{battery}} - U_C = C\mathcal{E}^2 - \frac{1}{2}\,C\mathcal{E}^2 = \frac{1}{2}\,C\mathcal{E}^2.
\]
One can verify this directly:
\[
U_R = \int_0^\infty I(t)^2R\,dt = \int_0^\infty \frac{\mathcal{E}^2}{R}\,e^{-2t/RC}\,dt
= \frac{\mathcal{E}^2}{R}\cdot\frac{RC}{2} = \frac{1}{2}\,C\mathcal{E}^2.
\]
The result is independent of $R$, because a larger $R$ gives less current but a proportionally longer charging time.}
\cor{Energy during discharging}{During complete discharge, the energy initially stored in the capacitor,
\[
U_{\text{initial}} = \frac{q_0^2}{2C},
\]
is entirely dissipated as Joule heat in the resistor:
\[
U_R = \frac{q_0^2}{2C}.
\]
This follows from
\[
U_R = \int_0^\infty I(t)^2R\,dt = \int_0^\infty \left(\frac{q_0}{RC}\,e^{-t/RC}\right)^{\!2}\!R\,dt
= \frac{q_0^2}{RC^2}\int_0^\infty e^{-2t/RC}\,dt = \frac{q_0^2}{RC^2}\cdot\frac{RC}{2} = \frac{q_0^2}{2C}.
\]}
\ex{Illustrative example}{If a capacitor of capacitance $C$ is charged through a resistor $R$ by a battery of emf $\mathcal{E}$, the time at which the current has dropped to half its initial value is found from $I(t) = (\mathcal{E}/R)e^{-t/RC} = (\mathcal{E}/2R)$, giving $e^{-t/RC} = 1/2$ and $t = RC\ln(2) \approx 0.693\,\tau$. At this instant, the charge on the capacitor is $q = C\mathcal{E}(1 - 1/2) = C\mathcal{E}/2$, exactly half of its final value. The energy stored in the capacitor is $\frac{1}{2}C(\mathcal{E}/2)^2 = \frac{1}{8}C\mathcal{E}^2 = 25\%$ of the energy that will ultimately be stored, while the battery has supplied $C\mathcal{E}\cdot(\mathcal{E}/2) = \frac{1}{2}C\mathcal{E}^2$, exactly half of the total energy it will supply.}
\qs{Worked example}{A series RC circuit consists of a battery of emf
\[
\mathcal{E} = 12.0\,\mathrm{V},
\]
a resistor of resistance
\[
R = 2.00\,\mathrm{k\Omega} = 2.00 \times 10^3\,\Omega,
\]
and an initially uncharged capacitor of capacitance
\[
C = 3.00\,\mathrm{\mu F} = 3.00 \times 10^{-6}\,\mathrm{F}.
\]
At $t = 0$, a switch is closed connecting all three elements in series. Assume ideal wires and components.
Find:
\begin{enumerate}[label=(\alph*)]
\item the time constant $\tau$ of the circuit,
\item the time $t_{1/2}$ at which the capacitor reaches $50.0\%$ of its final (maximum) charge, and
\item the current $I(t_{1/2})$ at that instant.
\end{enumerate}}
\sol \textbf{Part (a).} The time constant is
\[
\tau = RC = \left(2.00 \times 10^3\,\Omega\right)\!\left(3.00 \times 10^{-6}\,\mathrm{F}\right) = 6.00 \times 10^{-3}\,\mathrm{s}.
\]
In more convenient units,
\[
\tau = 6.00\,\mathrm{ms}.
\]
\textbf{Part (b).} The charge on the capacitor during charging is
\[
q(t) = C\mathcal{E}\,\bigl(1 - e^{-t/\tau}\bigr).
\]
The final (maximum) charge is
\[
q_{\text{max}} = C\mathcal{E} = \left(3.00 \times 10^{-6}\,\mathrm{F}\right)\!\left(12.0\,\mathrm{V}\right) = 36.0 \times 10^{-6}\,\mathrm{C} = 36.0\,\mathrm{\mu C}.
\]
We want the time $t_{1/2}$ at which $q(t_{1/2}) = q_{\text{max}}/2$. Setting the charge equation equal to $q_{\text{max}}/2$:
\[
C\mathcal{E}\,\bigl(1 - e^{-t_{1/2}/\tau}\bigr) = \frac{C\mathcal{E}}{2}.
\]
Cancel $C\mathcal{E}$ and solve:
\[
1 - e^{-t_{1/2}/\tau} = \frac{1}{2},
\qquad
e^{-t_{1/2}/\tau} = \frac{1}{2}.
\]
Taking the natural logarithm:
\[
-\frac{t_{1/2}}{\tau} = \ln\!\left(\frac{1}{2}\right) = -\ln(2),
\qquad
t_{1/2} = \tau\ln(2).
\]
Substitute $\tau = 6.00\,\mathrm{ms}$:
\[
t_{1/2} = (6.00\,\mathrm{ms})\ln(2) = (6.00 \times 10^{-3}\,\mathrm{s})\,(0.6931) = 4.16 \times 10^{-3}\,\mathrm{s}.
\]
Thus,
\[
t_{1/2} = 4.16\,\mathrm{ms}.
\]
\textbf{Part (c).} The current during charging is
\[
I(t) = \frac{\mathcal{E}}{R}\,e^{-t/\tau}.
\]
At $t = t_{1/2}$, we found $e^{-t_{1/2}/\tau} = 1/2$, so
\[
I(t_{1/2}) = \frac{\mathcal{E}}{R}\cdot\frac{1}{2} = \frac{1}{2}\cdot\frac{12.0\,\mathrm{V}}{2.00 \times 10^3\,\Omega}.
\]
Compute the initial current:
\[
\frac{\mathcal{E}}{R} = \frac{12.0}{2.00 \times 10^3}\,\mathrm{A} = 6.00 \times 10^{-3}\,\mathrm{A} = 6.00\,\mathrm{mA}.
\]
Therefore,
\[
I(t_{1/2}) = \frac{6.00\,\mathrm{mA}}{2} = 3.00\,\mathrm{mA}.
\]
\textbf{Check.} At $t_{1/2} = \tau\ln(2)$, the charge is $q = q_{\text{max}}(1 - e^{-\ln 2}) = q_{\text{max}}(1 - 1/2) = q_{\text{max}}/2 = 18.0\,\mathrm{\mu C}$, and the current is $I = (\mathcal{E}/R)e^{-\ln 2} = (\mathcal{E}/R)(1/2) = 3.00\,\mathrm{mA}$. Both are consistent with the expected behavior of a charging RC circuit.
Therefore,
\[
\tau = 6.00\,\mathrm{ms},
\qquad
t_{1/2} = 4.16\,\mathrm{ms},
\qquad
I(t_{1/2}) = 3.00\,\mathrm{mA}.
\]

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\subsection{Internal Resistance and Measurement Devices}
This subsection introduces the real-battery model (ideal emf in series with an internal resistance), derives the terminal-voltage relation under load, and discusses how practical measurement devices (ammeters and voltmeters) affect the circuits they measure.
\dfn{Real battery (emf and internal resistance)}{A \emph{real battery} is modelled as an ideal electromotive-force source $\mathcal{E}$ in series with an \emph{internal resistance} $r$. The emf $\mathcal{E}$ represents the work per unit charge the battery can deliver when no current flows. The internal resistance $r$ accounts for the finite conductivity of the electrolyte, electrode materials, and other dissipative processes inside the battery. The SI unit of emf is the volt (V), which is dimensionally equivalent to $\mathrm{J/C}$.}
\thm{Terminal voltage of a real battery}{Let a real battery with emf $\mathcal{E}$ and internal resistance $r$ deliver a current $I$ to an external load. The \emph{terminal voltage} $V$ across the battery's terminals is
\[
V = \mathcal{E} - Ir.
\]
When the battery is delivering current (discharging), the terminal voltage is \emph{less} than the emf by the voltage drop $Ir$ across the internal resistance. When $I = 0$ (open circuit), the terminal voltage equals the emf: $V = \mathcal{E}$. If the current is driven backwards through the battery (charging), then $I$ is negative and $V > \mathcal{E}$.}
\pf{Derivation of the terminal-voltage relation}{Consider a real battery connected to an external load resistor $R$. The equivalent circuit consists of the emf $\mathcal{E}$, the internal resistance $r$, and the load $R$ all in series. By Kirchhoff's loop rule, traversing the loop in the direction of current $I$:
\[
\mathcal{E} - Ir - IR = 0.
\]
Solving for the current gives
\[
I = \frac{\mathcal{E}}{r + R}.
\]
The terminal voltage $V$ is the potential drop across the load, which also equals the potential drop from the battery's positive to negative terminal:
\[
V = IR = \mathcal{E} - Ir.
\]
This proves the relation $V = \mathcal{E} - Ir$.}
\thm{Power delivered by a real battery}{The power $P$ delivered to an external load $R$ by a real battery of emf $\mathcal{E}$ and internal resistance $r$ is
\[
P = I^2 R = \mathcal{E}\,I - I^2 r,
\]
where $I = \mathcal{E}/(r + R)$. The first term $\mathcal{E}I$ is the total rate at which the battery converts chemical energy to electrical energy; the second term $I^2 r$ is the rate of internal dissipation as heat within the battery. The difference is the power delivered to the external circuit.}
\thm{Power transfer theorem (maximum power transfer)}{For a real battery with fixed $\mathcal{E}$ and $r$ connected to a variable load $R$, the power delivered to the load is
\[
P(R) = \frac{\mathcal{E}^2\,R}{(r + R)^2}.
\]
This power is maximized when $R = r$, giving $P_{\max} = \mathcal{E}^2 / (4r)$.}
\pf{Maximum power transfer}{From the preceding theorem, $P(R) = I^2 R$ with $I = \mathcal{E}/(r + R)$, so
\[
P(R) = \frac{\mathcal{E}^2 R}{(r + R)^2}.
\]
To find the maximum, differentiate with respect to $R$ and set the derivative to zero:
\[
\frac{dP}{dR} = \mathcal{E}^2 \,\frac{(r + R)^2 - R \cdot 2(r + R)}{(r + R)^4}
= \mathcal{E}^2 \,\frac{(r + R) - 2R}{(r + R)^3}
= \mathcal{E}^2 \,\frac{r - R}{(r + R)^3}.
\]
Setting $dP/dR = 0$ gives $R = r$. For $R < r$ the derivative is positive (power increases); for $R > r$ it is negative (power decreases). Hence $R = r$ is a maximum. Substituting $R = r$ into the power expression gives $P_{\max} = \mathcal{E}^2/(4r)$.}
\cor{Short circuit and open circuit limits}{When the load is zero ($R = 0$, \emph{short circuit}), the current is $I_{\text{sc}} = \mathcal{E}/r$ and the terminal voltage is $V = 0$. All power is dissipated internally: $P_{\text{load}} = 0$, and the battery heats up. When the load is infinite ($R \to \infty$, \emph{open circuit}), the current is $I = 0$ and $V = \mathcal{E}$. No power is delivered.}
\dfn{Ammeter}{An \emph{ammeter} measures the current through a branch of a circuit. It is inserted in \emph{series} with the branch. An ideal ammeter has zero resistance so it does not affect the circuit. A real ammeter has a small but finite resistance $R_A$ (the \emph{ammeter resistance}).}
\dfn{Voltmeter}{A \emph{voltmeter} measures the potential difference between two points in a circuit. It is connected in \emph{parallel} between those points. An ideal voltmeter has infinite resistance so no current flows through it. A real voltmeter has a large but finite resistance $R_V$ (the \emph{voltmeter resistance}).}
\thm{Loading effects of measurement devices}{When an ammeter of resistance $R_A$ is inserted in series with a circuit of total resistance $R_{\text{eq}}$ (not including $R_A$), the current measured is
\[
I_{\text{measured}} = \frac{I_{\text{true}}\,R_{\text{eq}}}{R_{\text{eq}} + R_A},
\]
where $I_{\text{true}}$ is the current that would flow without the ammeter. The reading is smaller than the true value by a factor of $R_{\text{eq}}/(R_{\text{eq}} + R_A)$.
When a voltmeter of resistance $R_V$ is connected across a component of resistance $R$ that has voltage $V$ across it (without the voltmeter), the measured voltage is
\[
V_{\text{measured}} = V \,\frac{R\,R_V/(R + R_V)}{R\,R_V/(R + R_V) + R_{\text{series}}}
= V \,\frac{R_V}{R_V + R_{\text{series}}\,(1 + R_V/R)}^{-1},
\]
where $R_{\text{series}}$ is the resistance in series with the component. In the common case where the component of interest is connected to a source with small internal resistance $r \ll R_V$, the correction is small: $V_{\text{measured}} \approx V\,(1 - r/R_V)$. The reading is smaller than the true voltage.}
\thm{Efficiency of power transfer}{The \emph{efficiency} $\eta$ of a real battery delivering power to a load $R$ is the ratio of power delivered to the load to the total power generated by the emf:
\[
\eta = \frac{P_{\text{load}}}{P_{\text{total}}}
= \frac{I^2 R}{\mathcal{E}I}
= \frac{IR}{\mathcal{E}}
= \frac{R}{R + r}.
\]
Efficiency increases as $R$ becomes large compared to $r$. When $R = r$, the efficiency is $50\%$: half the power is dissipated in the internal resistance.}
\nt{An ideal ammeter ($R_A = 0$) does not change the current in the branch it measures. A real ammeter always slightly \emph{reduces} the current. An ideal voltmeter ($R_V = \infty$) draws no current and does not disturb the circuit. A real voltmeter always slightly \emph{lowers} the voltage it is measuring because it provides an additional parallel current path. In well-designed circuits, $R_A \ll R_{\text{branch}}$ and $R_V \gg R_{\text{parallel}}$, so these perturbations are negligible.}
\ex{Illustrative example}{A $9\,\mathrm{V}$ battery with internal resistance $r = 1\,\Omega$ is connected to a load $R = 8\,\Omega$. The current is $I = 9\,\mathrm{V}/(1\,\Omega + 8\,\Omega) = 1.0\,\mathrm{A}$, and the terminal voltage is $V = 9\,\mathrm{V} - (1.0\,\mathrm{A})(1\,\Omega) = 8\,\mathrm{V}$.}
\qs{Worked example}{A battery has an emf
\[
\mathcal{E} = 12.0\,\mathrm{V}
\]
and an internal resistance
\[
r = 2.0\,\Omega.
\]
The battery is connected to an external load resistor $R = 10.0\,\Omega$, as shown in the circuit diagram below.
\begin{center}
\emph{(Diagram description: A single loop consisting of an ideal emf source $\mathcal{E}$, an internal resistor $r$ in series, and an external load resistor $R$ in series, forming a closed circuit.)}
\end{center}
Find:
\begin{enumerate}[label=(\alph*)]
\item the current $I$ in the circuit,
\item the terminal voltage $V$ across the battery,
\item the power $P_{\text{load}}$ delivered to the load resistor,
\item the power $P_{\text{int}}$ dissipated in the internal resistance,
\item the total power $P_{\text{total}}$ generated by the emf,
\item the efficiency $\eta$ of the battery, and
\item the value of $R$ that maximizes the power delivered to the load.
\end{enumerate}}
\sol \textbf{Part (a).} The circuit consists of $\mathcal{E}$, $r$, and $R$ in series. By Kirchhoff's loop rule,
\[
\mathcal{E} - Ir - IR = 0,
\]
so
\[
I = \frac{\mathcal{E}}{r + R}
= \frac{12.0\,\mathrm{V}}{2.0\,\Omega + 10.0\,\Omega}
= \frac{12.0\,\mathrm{V}}{12.0\,\Omega}
= 1.0\,\mathrm{A}.
\]
\textbf{Part (b).} The terminal voltage is
\[
V = \mathcal{E} - Ir
= 12.0\,\mathrm{V} - (1.0\,\mathrm{A})(2.0\,\Omega)
= 12.0\,\mathrm{V} - 2.0\,\mathrm{V}
= 10.0\,\mathrm{V}.
\]
Equivalently, $V = IR = (1.0\,\mathrm{A})(10.0\,\Omega) = 10.0\,\mathrm{V}$, which confirms the result.
\textbf{Part (c).} The power delivered to the load resistor is
\[
P_{\text{load}} = I^2 R
= (1.0\,\mathrm{A})^2(10.0\,\Omega)
= 10.0\,\mathrm{W}.
\]
Alternatively, $P_{\text{load}} = VI = (10.0\,\mathrm{V})(1.0\,\mathrm{A}) = 10.0\,\mathrm{W}$, giving the same result.
\textbf{Part (d).} The power dissipated in the internal resistance is
\[
P_{\text{int}} = I^2 r
= (1.0\,\mathrm{A})^2(2.0\,\Omega)
= 2.0\,\mathrm{W}.
\]
\textbf{Part (e).} The total power generated by the emf is
\[
P_{\text{total}} = \mathcal{E} I
= (12.0\,\mathrm{V})(1.0\,\mathrm{A})
= 12.0\,\mathrm{W}.
\]
Check: $P_{\text{total}} = P_{\text{load}} + P_{\text{int}} = 10.0\,\mathrm{W} + 2.0\,\mathrm{W} = 12.0\,\mathrm{W}$, so energy is conserved.
\textbf{Part (f).} The efficiency of the battery is
\[
\eta = \frac{P_{\text{load}}}{P_{\text{total}}}
= \frac{10.0\,\mathrm{W}}{12.0\,\mathrm{W}}
= 0.833
= 83.3\%.
\]
Equivalently, $\eta = R/(R + r) = 10.0\,\Omega/(10.0\,\Omega + 2.0\,\Omega) = 10/12 = 0.833 = 83.3\%$.
\textbf{Part (g).} By the maximum power transfer theorem, the power delivered to the load is maximized when the load resistance equals the internal resistance:
\[
R = r = 2.0\,\Omega.
\]
At this value, the maximum power delivered to the load is
\[
P_{\max} = \frac{\mathcal{E}^2}{4r}
= \frac{(12.0\,\mathrm{V})^2}{4(2.0\,\Omega)}
= \frac{144\,\mathrm{V}^2}{8.0\,\Omega}
= 18.0\,\mathrm{W}.
\]
Therefore,
\[
I = 1.0\,\mathrm{A},
\qquad
V = 10.0\,\mathrm{V},
\qquad
P_{\text{load}} = 10.0\,\mathrm{W},
\]
\[
P_{\text{int}} = 2.0\,\mathrm{W},
\qquad
P_{\text{total}} = 12.0\,\mathrm{W},
\qquad
\eta = 83.3\%,
\qquad
R_{\text{max power}} = 2.0\,\Omega.
\]