where $n$ is charge carrier density, $q$ is charge per carrier, $\vec{v}_d$ is drift velocity.

$I = \int \vec{J} \cdot d\vec{A}$

Resistance and Ohm's Law

Ohm's law: $V = IR$

Resistance: $R = \frac{\rho L}{A}$

where:

• $\rho$ = resistivity (material property)
• $L$ = length
• $A$ = cross-sectional area

Conductivity: $\sigma = 1/\rho$

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

Temperature Dependence

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

where $\alpha$ is temperature coefficient of resistivity.

Power

Power dissipated: $P = IV = I^2R = \frac{V^2}{R}$

Energy: $E = Pt$

Resistors in Series

Same current through each:

$R_{eq} = R_1 + R_2 + \cdots$

$V_{total} = V_1 + V_2 + \cdots$

Resistors in Parallel

Same voltage across each:

$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$

Two resistors: $R_{eq} = \frac{R_1R_2}{R_1 + R_2}$

Kirchhoff's Rules

Junction rule (current): $\sum I_{in} = \sum I_{out}$

(Charge conservation)

Loop rule (voltage): $\sum \Delta V = 0$

(Energy conservation)

Sign Conventions

• Traversing resistor with current: $-IR$
• Traversing resistor against current: $+IR$
• Traversing battery from $-$ to $+$: $+\mathcal{E}$
• Traversing battery from $+$ to $-$: $-\mathcal{E}$

EMF and Internal Resistance

Real battery has internal resistance $r$:

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

Maximum power to load: When $R_{load} = r$ (matched impedance)

Multiloop Circuits

Strategy:

1. Assign current to each branch
2. Apply junction rule
3. Apply loop rule to independent loops
4. Solve system of equations

Ammeter and Voltmeter

Ammeter: Measures current, low resistance (ideally 0), in series

Voltmeter: Measures voltage, high resistance (ideally ∞), in parallel

Wheatstone Bridge

Balanced when: $\frac{R_1}{R_2} = \frac{R_3}{R_4}$

No current through galvanometer.

$R = \frac{\rho L}{A} = \frac{(1.7 \times 10^{-8})(2.0)}{7.07 \times 10^{-6}}$

$R = \boxed{4.81 \times 10^{-3} \text{ Ω} = 4.81 \text{ mΩ}}$

(b) Current density:

$J = \frac{I}{A} = \frac{5.0}{7.07 \times 10^{-6}}$

$J = \boxed{7.07 \times 10^5 \text{ A/m}^2}$

(c) Electric field:

Using Ohm's law in differential form: $\vec{J} = \sigma \vec{E}$ where $\sigma = 1/\rho$: $E = \rho J = (1.7 \times 10^{-8})(7.07 \times 10^5)$

$E = \boxed{0.012 \text{ V/m} = 12 \text{ mV/m}}$

Alternatively: $E = V/L$ where $V = IR = (5.0)(4.81 \times 10^{-3}) = 0.024$ V