• Junction Rule (KCL): $\sum I_{\text{in}} = \sum I_{\text{out}}$. At any junction, label branch currents with arrows. If answer is negative, actual flow is opposite.
• Loop Rule (KVL): $\sum \Delta V = 0$ along any closed loop.
• Resistor $R$: Traversing WITH current arrow $\implies \mathbf{-IR}$ (drop). Traversing AGAINST arrow $\implies \mathbf{+IR}$ (rise).
• EMF Source $\mathcal{E}$: Traversing from $-$ (short) to $+$ (long) $\implies \mathbf{+\mathcal{E}}$. From $+$ to $-$ $\implies \mathbf{-\mathcal{E}}$ (independent of current direction!).
• Terminal Voltage $V_T$ with internal $r$:
- Discharging (supplying power): $V_T = \mathcal{E} - Ir \le \mathcal{E}$.
- Charging (driven backward into $+$): $\mathbf{V_T = \mathcal{E} + Ir \gt \mathcal{E}}$ (CQ 12!).
Setup: Left branch ($\mathcal{E}_1, R_1$), middle ($R_2$, $I_2\downarrow$), right ($\mathcal{E}_2$ opposing, $R_3$).
1. Junction: $I_1 = I_2 + I_3 \implies I_2 = I_1 - I_3$.
2. Loop 1 (Left, CW): $+\mathcal{E}_1 - I_1 R_1 - I_2 R_2 = 0 \implies \mathcal{E}_1 - I_1(R_1+R_2) + I_3 R_2 = 0$.
3. Loop 2 (Right, CCW): $+\mathcal{E}_2 - I_3 R_3 - I_2 R_2 = 0 \implies \mathcal{E}_2 + I_1 R_2 - I_3(R_2+R_3) = 0$.
4. Solve 2×2: Substitute $I_2$ into loops; solve for $I_1, I_3$.
• Opposing batteries in 1 loop: $I = \frac{\mathcal{E}_1 - \mathcal{E}_2}{R_{\text{tot}}}$. Larger EMF wins.
• Current Divider: $I_1 = I_{\text{tot}}\left(\frac{R_2}{R_1+R_2}\right)$. Smaller $R$ takes larger $I$.
• Bridge / Ladder: Work inward! Parallel pair $R_D\parallel R_E = \frac{R}{2}$; in series with $R_C \implies 1.5R$; in parallel with $R_B \implies \frac{R(1.5R)}{2.5R} = 0.6R$; in series with $R_A \implies R_{\text{tot}} = 1.6R$ (PE 12).
• Resistor Cube ($12$ equal $R$): Body diagonal: $\mathbf{R_{\text{eq}} = \frac{5}{6}R}$. Face diagonal: $\frac{3}{4}R$. Along one edge: $\frac{7}{12}R$.
• Charging ($Q(0)=0$): $\tau = RC$. $I_0 = \mathcal{E}/R$, $Q_{\text{max}} = C\mathcal{E}$.
$Q(t) = Q_{\text{max}}(1 - e^{-t/\tau}) \quad \bullet \quad I(t) = I_0 e^{-t/\tau}$
$V_C(t) = \mathcal{E}(1 - e^{-t/\tau}) \quad \bullet \quad V_R(t) = \mathcal{E} e^{-t/\tau}$
At $t = 0$: Capacitor acts as a short circuit ($V_C=0, I=\mathcal{E}/R$).
At $t \to \infty$: Capacitor acts as an open circuit ($I=0, V_C=\mathcal{E}$).
• Discharging ($Q(0)=Q_0$): $Q(t) = Q_0 e^{-t/\tau}$, $I(t) = \frac{Q_0}{RC}e^{-t/\tau}$, $V_C(t) = V_0 e^{-t/\tau}$.
• Key Timing Multipliers:
- $t = \tau \implies 63.2\%$ charged ($36.8\%$ remaining on discharge).
- Half-life: $\mathbf{t_{1/2} = \tau \ln(2) \approx 0.693 \tau}$.
- Reaching fraction $f$ of max: $\mathbf{t = -\tau \ln(1 - f)}$. ($90\% \implies 2.30\tau$; $99\% \implies 4.61\tau$).
- Discharge to fraction $p$: $\mathbf{t = \tau \ln(1/p)}$ (PE 11: $3\% \implies t = \tau\ln(33.3) = 3.51\tau$).
• Power Dissipation in $R$: $P(t) = I^2 R = P_0 e^{-2t/\tau}$ (effective time constant is $\mathbf{\tau/2}$; half-power at $\mathbf{t = \frac{1}{2}\tau\ln 2 \approx 0.347\tau}$).
• 50% Energy Rule: Battery outputs $W = C\mathcal{E}^2$. Capacitor stores $U = \frac{1}{2}C\mathcal{E}^2$. Resistor dissipates $E_R = \frac{1}{2}C\mathcal{E}^2$ (independent of $R$!).
• Max Power Transfer: Load $R_L$ gets max power when $\mathbf{R_L = r}$.
$P_{\text{max}} = \frac{\mathcal{E}^2}{4r}$. Circuit efficiency is exactly $\mathbf{\eta = 50\%}$.
• Extracting $\mathcal{E}$ and $r$ from $I$ vs $R$ Graph (GW 10.1):
$I = \frac{\mathcal{E}}{R+r} \implies \mathcal{E} = I(R+r)$. $R=0 \implies I_{\text{sc}} = \mathcal{E}/r$. $R=r \implies I = I_{\text{sc}}/2$.
• Voltmeter / Ammeter Loading: High $R$ ($R \gg \sqrt{R_V R_A}$) $\implies$ voltmeter across both ammeter and $R$ minimizes error.
• $m_p = \mathbf{1.67 \times 10^{-27}\text{ kg}}$ • $m_e = \mathbf{9.11 \times 10^{-31}\text{ kg}}$ • $e = \mathbf{1.60 \times 10^{-19}\text{ C}}$
• $k_e = \frac{1}{4\pi\epsilon_0} = \mathbf{8.99 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2}$ • $\epsilon_0 = \mathbf{8.85 \times 10^{-12}\text{ C}^2/\text{N}\cdot\text{m}^2}$
• $1\text{ eV} = \mathbf{1.60 \times 10^{-19}\text{ J}}$. Speed from energy: $\mathbf{v = \sqrt{2K/m}}$.
• Coulomb 2-Body Energy: $K_1 + K_2 + U = \text{const}$, $U = \frac{k q_1 q_2}{r}$.
Equal mass particles fired apart at $v_0$ (PE 14): $m v_0^2 - \frac{k e^2}{r_0} = -\frac{k e^2}{r_{\text{max}}}$.
Escape velocity: $K + U = 0 \implies \mathbf{v_{\text{esc}} = \sqrt{\frac{k e^2}{m r_0}}}$.
Electron + Proton fired apart: $m_e v_e = m_p v_p \implies v_p = \frac{m_e}{m_p}v_e$.
• Closest Approach ($v\to 0$): $\frac{1}{2}m v_0^2 = \frac{k Z e^2}{r_{\text{min}}} \implies d_{\text{surf}} = r_{\text{min}} - R_{\text{nucleus}}$.
• Turning Points in $V(x)$: $E_{\text{tot}} = qV(x_{\text{turn}}) \implies v_{\text{max}} = \sqrt{2E_{\text{tot}}/m}$ at $V=0$.
• Finite Uniform Rod on Axis (GW 7.5): Length $L$, charge $Q$ on $[-L/2, L/2]$:
$\mathbf{V(x) = \frac{k Q}{L}\ln\left(\frac{x + L/2}{x - L/2}\right)}$ for $x > L/2$.
• Ring of Radius $R$ on Axis $z$: $\mathbf{V(z) = \frac{k Q}{\sqrt{z^2+R^2}}}$; $E_z = \frac{k Q z}{(z^2+R^2)^{3/2}}$.
• Disk of Radius $R$ on Axis $z$: $\mathbf{V(z) = \frac{\sigma}{2\epsilon_0}\left(\sqrt{z^2+R^2} - z\right)}$.
• Conducting Sphere ($R$): Inside ($r \le R$): $E=0, \mathbf{V = \frac{kQ}{R} = \text{const}}$. Outside ($r \ge R$): $E = \frac{kQ}{r^2}, \mathbf{V = \frac{kQ}{r}}$. Surface potential is flat!
• Two Spheres ($R_1, R_2$ dist $D$): $V_a = k\left(\frac{q_1}{R_1} + \frac{q_2}{D-R_1}\right)$ (GW 7.4).
• Field to Potential: $\Delta V = -\int_{x_1}^{x_2} E(x)dx$. If $E(x) = ax^3 - b/x$, $\int = \frac{a}{4}(x_2^4-x_1^4) - b\ln(x_2/x_1)$ (PE 7).
• Dielectric Insertion: Always increases capacitance ($C = \kappa C_0$).
• Isolated ($Q = \text{const}$): $V = V_0/\kappa \downarrow$, $E = E_0/\kappa \downarrow$, $U = U_0/\kappa \downarrow$. Stored energy drops $\implies$ fringing fields do positive work pulling slab in.
• Connected to Battery ($V = \text{const}$): $Q = \kappa Q_0 \uparrow$, $E = E_0$, $U = \kappa U_0 \uparrow$. Battery delivers $\Delta W = 2\Delta U$. Force pulls slab in: $F = +\frac{1}{2}V^2 \frac{dC}{dx}$.
• Dielectric Slicing Topologies:
- Parallel boundary (series): $C_{\text{eff}} = \frac{2\kappa_1 \kappa_2}{\kappa_1+\kappa_2}C_0$ (Harmonic mean).
- Perpendicular boundary (parallel): $C_{\text{eff}} = \frac{\kappa_1+\kappa_2}{2}C_0$ (Arithmetic mean $\to$ always higher $C$!).
• Capacitor Charge Sharing / Switching (GW 8.4):
Initial charge $Q_0 = C_1 V_0$. Connect in parallel with $C_{\text{load}}$:
$\mathbf{V_{\text{final}} = \frac{Q_0}{C_1 + C_{\text{load}}}}$, $Q_1 = C_1 V_{\text{final}}$, $Q_{\text{load}} = C_{\text{load}} V_{\text{final}}$.
• Series Voltage Divider: $Q_1 = Q_2 \implies \mathbf{\frac{V_1}{V_2} = \frac{C_2}{C_1}}$ (smaller cap drops larger $V$!).
• Volume Conservation (Wire Drawn/Stretched): $\text{Vol} = AL = \text{const} \implies A = \text{Vol}/L$. Resistance: $\mathbf{R = \rho \frac{L^2}{\text{Vol}} \propto L^2 \propto \frac{1}{A^2} \propto \frac{1}{d^4}}$.
- Double length ($L \to 2L$) $\implies R \to 4R$.
- Half resistance ($R \to R/2$) $\implies \mathbf{L \to L/\sqrt{2} \approx 0.707 L}$ (PE 10).
• Non-Uniform Current Integration: $I = \int_0^R J(r)(2\pi r dr)$.
If $J(r) = J_0(r/R) \implies \mathbf{I = \frac{2}{3}\pi R^2 J_0} \implies \mathbf{J_0 = \frac{3I}{2\pi R^2}}$ (GW 9.4).
• Time-Varying Current: $I(t) = \frac{dQ}{dt}$. If $Q = Q_0(1-e^{-t/\tau}) \implies I = \frac{Q_0}{\tau}e^{-t/\tau}$.
• Drift vs Signal: $v_d = \frac{J}{e n_e} \sim 10^{-4}\text{ m/s}$ (snail). Power transported by Poynting EM field wave $v_{\text{sig}} \sim 10^8\text{ m/s}$.