Units: 1 ampere (A) = 1 coulomb/second

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

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.

$R_{12} = 2.4 \text{ Ω}$

Total resistance: $R_{eq} = R_{12} + R_3 = 2.4 + 12 = \boxed{14.4 \text{ Ω}}$

(b) Current through R₃:

Total current from battery: $I_{total} = \frac{V}{R_{eq}} = \frac{24}{14.4}$

$I_3 = I_{total} = \boxed{1.67 \text{ A}}$

(In series, all current flows through R₃)

(c) Power dissipated:

Voltage across parallel combination: $V_{12} = V - V_3 = V - I_3 R_3 = 24 - (1.67)(12) = 4.0 \text{ V}$

$P_1 = \frac{V_{12}^2}{R_1} = \frac{(4.0)^2}{6.0} = \boxed{2.67 \text{ W}}$

$P_2 = \frac{V_{12}^2}{R_2} = \frac{(4.0)^2}{4.0} = \boxed{4.0 \text{ W}}$

$P_3 = I_3^2 R_3 = (1.67)^2(12) = \boxed{33.3 \text{ W}}$

Total power: 2.67 + 4.0 + 33.3 = 40 W = VI ✓

(a) & (b) Finding currents:

Loop through battery 1 and R (clockwise): $\varepsilon_1 - I_1 r_1 - IR = 0$ $12 - 0.5I_1 - 3I = 0 \quad \text{...(2)}$

Loop through battery 2 and R (clockwise): $\varepsilon_2 - I_2 r_2 - IR = 0$ $6 - 0.3I_2 - 3I = 0 \quad \text{...(3)}$

From (2): $I = \frac{12 - 0.5I_1}{3}$

From (3): $I = \frac{6 - 0.3I_2}{3}$

Setting equal: $12 - 0.5I_1 = 6 - 0.3I_2$ $6 = 0.5I_1 - 0.3I_2 \quad \text{...(4)}$

From (1): $I_2 = I - I_1$

Substitute into (4): $6 = 0.5I_1 - 0.3(I - I_1) = 0.5I_1 - 0.3I + 0.3I_1 = 0.8I_1 - 0.3I$

From (2): $I = 4 - 0.167I_1$

Substitute: $6 = 0.8I_1 - 0.3(4 - 0.167I_1)$ $6 = 0.8I_1 - 1.2 + 0.05I_1$ $7.2 = 0.85I_1$

$I_1 = \boxed{8.47 \text{ A}}$

$I = 4 - 0.167(8.47) = \boxed{2.58 \text{ A}}$

$I_2 = 2.58 - 8.47 = \boxed{-5.89 \text{ A}}$

(Negative means battery 2 is being charged!)

(c) Terminal voltage:

$V = IR = (2.58)(3.0) = \boxed{7.74 \text{ V}}$