Current, Resistance, and DC Circuits

Ohm's law, Kirchhoff's rules, and power in circuits

Current, Resistance, and DC Circuits

Electric Current

Definition: $I = \frac{dQ}{dt}$

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:

Assign current to each branch
Apply junction rule
Apply loop rule to independent loops
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.

📚 Practice Problems

1Problem 1easy

❓ Question:

A cylindrical wire of radius r = 1.5 mm and length L = 2.0 m has resistivity ρ = 1.7 × 10⁻⁸ Ω·m. (a) Find the resistance of the wire. (b) If a current I = 5.0 A flows through it, find the current density J. (c) Find the electric field inside the wire.

💡 Show Solution

Given:

r = 1.5 mm = 1.5 × 10⁻³ m
L = 2.0 m
ρ = 1.7 × 10⁻⁸ Ω·m
I = 5.0 A

(a) Resistance:

Cross-sectional area: $A = \pi r^2 = \pi (1.5 \times 10^{-3})^2 = 7.07 \times 10^{-6} \text{ m}^2$

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

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

(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

2Problem 2medium

❓ Question:

In the circuit shown, a 24 V battery is connected to three resistors: R₁ = 6.0 Ω, R₂ = 4.0 Ω, and R₃ = 12 Ω. R₁ and R₂ are in parallel, and this combination is in series with R₃. Find: (a) the equivalent resistance, (b) the current through R₃, and (c) the power dissipated by each resistor.

💡 Show Solution

Given:

V = 24 V
R₁ = 6.0 Ω, R₂ = 4.0 Ω, R₃ = 12 Ω

(a) Equivalent resistance:

R₁ and R₂ in parallel: $\frac{1}{R_{12}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6.0} + \frac{1}{4.0} = \frac{5}{12}$

$R_{12} = 2.4 \text{ Ω}$

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 ✓

3Problem 3hard

❓ Question:

Two batteries (ε₁ = 12 V with internal resistance r₁ = 0.5 Ω, and ε₂ = 6.0 V with r₂ = 0.3 Ω) are connected in parallel to an external load R = 3.0 Ω. Use Kirchhoff's laws to find: (a) the current through each battery, (b) the current through R, and (c) the terminal voltage across R.

💡 Show Solution

Given:

ε₁ = 12 V, r₁ = 0.5 Ω
ε₂ = 6.0 V, r₂ = 0.3 Ω
R = 3.0 Ω

Setup:

Let I₁ = current from battery 1 (positive right) Let I₂ = current from battery 2 (positive right) Let I = current through R (positive down)

Kirchhoff's current law at top junction: $I_1 + I_2 = I \quad \text{...(1)}$

(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:

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

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