content(warnings): Add W17-W28, X8-X12, N7 — E&M misconceptions, cross-refs, notes

concepts/em/u11/e11-1-current-density.tex

@@ -14,6 +14,8 @@ dI=\vec{J}\cdot d\vec{A}.
 \]
 Thus $\vec{J}$ links the local flow of charge to the total current through a cross section.}
 
+\wc{Electrons flow opposite to conventional current}{The \emph{conventional current} direction is defined as the direction positive charges would move (from higher to lower potential). In metal wires, the actual charge carriers are electrons, which move \emph{opposite} to the conventional current direction. This historical convention does not affect circuit analysis results.}
+
 \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

concepts/em/u11/e11-2-resistance-ohm.tex

@@ -8,6 +8,8 @@ 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}.

concepts/em/u11/e11-5-kirchhoff.tex

@@ -49,6 +49,8 @@ The assumed direction of each branch current must be declared before writing equ
 \end{enumerate}
 The number of independent equations needed equals the number of unknown branch currents.}
 
+Kirchhoff's rules generalize the equivalent-resistance method from Section 11.4. For single-battery circuits with pure series/parallel arrangements, equivalent resistance is faster. Kirchhoff's rules become necessary when multiple batteries or loops make reduction impossible.
+
 \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.}

concepts/em/u11/e11-6-rc-circuits.tex

@@ -16,6 +16,8 @@ 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 $I \to 0$.}
 
+\nt{The RC time constant $\tau=RC$ has units of seconds: $[\Omega]\cdot[\mathrm{F}]=[\mathrm{V/A}]\cdot[\mathrm{C/V}]=[\mathrm{C/A}]=[\mathrm{C/(C/s)}]=[\mathrm{s}]$. After one time constant during charging, the capacitor reaches $1-e^{-1}\approx 63.2\%$ of its final voltage. After five time constants, it reaches $99.3\%$, which is the practical definition of ``fully charged'' in circuit analysis.}
+
 \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,