Electric current is rate of charge flow through a cross-section.

$I = \frac{dq}{dt}$

Units: Amperes (A) = Coulombs/second (C/s)

Conventional current:

• Defined as flow of positive charge
• Direction opposite to actual electron flow

Current density: $J = \frac{I}{A}$

Where A = cross-sectional area (m2)

Relation to drift velocity: $I = nqAv_d$

Where:

• n = charge carrier density (carriers/m3)
• q = charge per carrier (C)
• A = cross-sectional area (m2)
• $v_d$ = drift velocity (m/s)

Current density (vector): $\vec{J} = nq\vec{v}_d$

Units: A/m2

Drift velocity: $v_d = \frac{I}{nqA}$

Typical values:

• Copper: current density ~106 A/m2, $v_d$ ~ 0.1 mm/s
• Drift velocity is small, but electric signal propagates near speed of light

Microscopic view:

• Electrons collide frequently with lattice
• Average velocity (drift) much smaller than thermal velocity
• Signal velocity = wave propagation speed, not electron drift speed

Resistivity ρ (material property): $R = \rho \frac{L}{A}$

Where:

• L = length (m)
• A = cross-sectional area (m2)

Units: Ω - m

Conductivity σ (reciprocal of resistivity): $\sigma = \frac{1}{\rho}$

Units: S/m (Siemens/m)

Temperature dependence: $\rho = \rho_0[1 + \alpha(T - T_0)]$

Where alpha = temperature coefficient

Microscopic Ohm's law: $\vec{J} = \sigma\vec{E}$

Where:

• $\vec{J}$ = current density
• $\vec{E}$ = electric field
• σ = conductivity

Relates current density to electric field at microscopic level.

Macroscopic Ohm's law: $V = IR$

Where:

• V = voltage across element (V)
• I = current through element (A)
• R = resistance (Ω)

Valid for ohmic materials (linear I-V relationship).

$\rho = \frac{RA}{L}$

Units: Ω - m

Material property determining resistance.

Temperature dependence: $\rho(T) = \rho_0[1 + \alpha(T - T_0)]$

Common values (Ω - m, 20 degrees C):

• Copper: 1.68 × 10-8
• Aluminum: 2.65 × 10-8
• Silicon: 2.3 × 103 (semiconductor)

$\sigma = \frac{1}{\rho}$

Units: S/m (Siemens per meter)

Relates to microscopic Ohm's law: $\vec{J} = \sigma\vec{E}$

Power (rate of energy dissipation):

$P = IV$

Using Ohm's law ($V = IR$):

$P = I^2R = \frac{V^2}{R}$

Units: Watts (W) = Joules/second

Physical meaning:

• Energy converted to heat in resistor
• P = rate of electrical energy -> thermal energy

Energy dissipated: $E = Pt = IVt = I^2Rt$

Electromotive force (EMF): $\mathcal{E} = \text{energy per unit charge provided by source}$

Terminal voltage with internal resistance r:

$V_{terminal} = \mathcal{E} - Ir$

Where:

• $\mathcal{E}$ = emf (ideal voltage)
• I = current
• r = internal resistance

Current in circuit with internal resistance: $I = \frac{\mathcal{E}}{R_{load} + r}$

Power dissipated internally: $P_{internal} = I^2r$

Components connected end-to-end.

Properties:

• Same current through all: $I_1 = I_2 = ... = I$
• Voltages add: $V = V_1 + V_2 + ...$
• Equivalent resistance: $R_{eq} = R_1 + R_2 + ...$

Voltage divider: $V_i = \frac{R_i}{R_{eq}}V_{total}$

Procedure:

1. Calculate equivalent resistance
2. Find total current
3. Use voltage division to find individual voltages

Components connected across same voltage.

Properties:

• Same voltage: $V_1 = V_2 = ... = V$
• Currents add: $I = I_1 + I_2 + ...$
• Equivalent resistance: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + ...$

Two resistors in parallel: $R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$

Current divider: $I_i = \frac{R_{eq}}{R_i}I_{total}$

At any junction (node), sum of currents entering equals sum leaving.

$\sum I_{in} = \sum I_{out}$

Or: $\sum I = 0$

Physical basis: charge conservation (charge cannot accumulate at node).

Sum of voltage changes around any closed loop equals zero.

$\sum V = 0$

Physical basis: electric field is conservative (work around closed loop = 0).

Sign convention:

• Voltage rise across battery: +$\mathcal{E}$
• Voltage drop across resistor: -IR
• Voltage drop across internal resistance: -Ir

Series capacitors: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + ...$

Same charge on each; voltages add.

Parallel capacitors: $C_{eq} = C_1 + C_2 + ...$

Same voltage across each; charges add.

Steady state (after long time, $t \gg RC$):

• Capacitor fully charged: $I_C = 0$
• Capacitor acts as open circuit
• Current through circuit determined by resistors only

Transient state (during charging/discharging):

• Capacitor acts like time-varying element
• Use differential equations
• Characterized by time constant $\tau = RC$

From Kirchhoff's loop rule:

$\mathcal{E} - IR - \frac{Q}{C} = 0$

Substituting $I = dQ/dt$:

$\mathcal{E} - R\frac{dQ}{dt} - \frac{Q}{C} = 0$

$R\frac{dQ}{dt} = \mathcal{E} - \frac{Q}{C}$

This is first-order linear differential equation.

Solution (with Q(0) = 0): $Q(t) = C\mathcal{E}\left(1 - e^{-t/RC}\right)$

Voltage across capacitor: $V_C(t) = \mathcal{E}\left(1 - e^{-t/RC}\right)$

Current: $I(t) = \frac{\mathcal{E}}{R}e^{-t/RC}$

From loop rule (no battery):

$-IR - \frac{Q}{C} = 0$

$R\frac{dQ}{dt} = -\frac{Q}{C}$

Solution (with Q(0) = Q0): $Q(t) = Q_0 e^{-t/RC}$

Voltage: $V_C(t) = V_0 e^{-t/RC}$

Current: $I(t) = \frac{V_0}{R}e^{-t/RC}$ (negative sign indicates discharge direction)

$\tau = RC$

Units: seconds

Physical meaning: characteristic charging/discharging time.

Key values:

• At t = τ: reaches 63.2% of final value (charging), drops to 36.8% (discharging)
• At t = 2τ: reaches 86.5% (charging), drops to 13.5% (discharging)
• At t = 5τ: essentially complete (99.3%)

Charging (initially):

• Capacitor uncharged: acts like short circuit (V_C = 0)
• Initial current: $I_0 = \mathcal{E}/R$ (maximum)
• Voltage increases exponentially: $V_C(t) = \mathcal{E}(1 - e^{-t/\tau})$
• Current decreases exponentially: $I(t) = I_0 e^{-t/\tau}$

Discharging (initially):

• Capacitor charged: $V_C = V_0$
• Initial current: $I_0 = V_0/R$ (maximum)
• Voltage decreases: $V_C(t) = V_0 e^{-t/\tau}$
• Current decreases: $I(t) = I_0 e^{-t/\tau}$

After long time ($t \gg \tau$):

Charging circuit:

• Capacitor fully charged: $V_C = \mathcal{E}$
• Current through capacitor: $I_C = 0$
• Capacitor acts as open circuit for DC

Discharging circuit:

• Capacitor fully discharged: $V_C = 0$
• No current flows: $I = 0$

Steady state analysis procedure:

1. Replace capacitors with open circuits
2. Analyze resulting resistive circuit
3. Find steady-state voltages and currents