Current, Resistance, and Ohm's Law

Electric current, resistance, resistivity, and Ohm's law

🔌 Current, Resistance, and Ohm's Law

Electric Current

Electric current is the rate of charge flow:

$I = \frac{\Delta Q}{\Delta t}$

where:

$I$ = current (Ampere, A)
$\Delta Q$ = charge passing point (C)
$\Delta t$ = time interval (s)

1 Ampere = 1 Coulomb/second

Direction:

Conventional current: Direction positive charges flow (+ to -)
Electron flow: Opposite direction (- to +)

We use conventional current!

Drift Velocity

In conductors, electrons drift slowly:

$I = nAv_d e$

where:

$n$ = charge carrier density
$A$ = cross-sectional area
$v_d$ = drift velocity (~mm/s, very slow!)
$e$ = elementary charge

💡 Key: Current is fast (~speed of light), but individual electrons drift slowly!

Resistance

Resistance opposes current flow:

$R = \frac{V}{I}$

Units: Ohm (Ω) = V/A

Resistivity

Material property:

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

where:

$\rho$ = resistivity (Ω·m)
$L$ = length
$A$ = cross-sectional area

Good conductors: Low $\rho$ (copper: $1.7 \times 10^{-8}$ Ω·m) Insulators: High $\rho$

Temperature Dependence

$R = R_0(1 + \alpha \Delta T)$

where $\alpha$ is temperature coefficient.

Metals: $\alpha > 0$ (R increases with T)
Semiconductors: $\alpha < 0$ (R decreases with T)

Ohm's Law

$V = IR$

Ohmic materials: Constant R (linear V-I graph) Non-ohmic: R varies (curved V-I graph, like diodes)

Electric Power

Power dissipated in resistor:

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

Units: Watt (W) = J/s

Energy dissipated (heat): $E = Pt = I^2Rt = \frac{V^2}{R}t$

Electrical Energy Cost

Power companies charge by kilowatt-hour (kWh):

$1 \text{ kWh} = (1000 \text{ W})(3600 \text{ s}) = 3.6 \times 10^6 \text{ J}$

Cost = (Power in kW) × (time in hours) × (rate per kWh)

AC vs DC

DC (Direct Current): Constant direction (batteries) AC (Alternating Current): Oscillates (wall outlets, 60 Hz in US)

For AC: $V_{rms} = \frac{V_0}{\sqrt{2}}$ , $I_{rms} = \frac{I_0}{\sqrt{2}}$

Household: 120 V AC is $V_{rms}$

Problem-Solving Strategy

Identify knowns: V, I, or R
Choose formula:
- Ohm's Law: $V = IR$
- Power: $P = IV = I^2R = V^2/R$
- Resistance: $R = \rho L/A$
Watch units: A, V, Ω, W
Check reasonableness

Common Mistakes

❌ Confusing current direction (use conventional!) ❌ Using diameter instead of radius in area ( $A = \pi r^2$ ) ❌ Wrong power formula (choose based on what you know) ❌ Forgetting to convert units (mA → A, kΩ → Ω) ❌ Treating all materials as ohmic

📚 Practice Problems

1Problem 1easy

❓ Question:

A 12 V battery is connected to a 6.0 Ω resistor. (a) What is the current? (b) What power is dissipated?

💡 Show Solution

Given:

Voltage: $V = 12$ V
Resistance: $R = 6.0$ Ω

Part (a): Current

Use Ohm's Law: $I = \frac{V}{R} = \frac{12}{6.0} = 2.0 \text{ A}$

Part (b): Power dissipated

$P = IV = (2.0)(12) = 24 \text{ W}$

Or alternatively: $P = \frac{V^2}{R} = \frac{(12)^2}{6.0} = \frac{144}{6.0} = 24 \text{ W}$ ✓

Answer:

(a) I = 2.0 A
(b) P = 24 W

2Problem 2easy

❓ Question:

A 12 V battery is connected to a 6.0 Ω resistor. (a) What is the current? (b) What power is dissipated?

💡 Show Solution

Given:

Voltage: $V = 12$ V
Resistance: $R = 6.0$ Ω

Part (a): Current

Use Ohm's Law: $I = \frac{V}{R} = \frac{12}{6.0} = 2.0 \text{ A}$

Part (b): Power dissipated

$P = IV = (2.0)(12) = 24 \text{ W}$

Or alternatively: $P = \frac{V^2}{R} = \frac{(12)^2}{6.0} = \frac{144}{6.0} = 24 \text{ W}$ ✓

Answer:

(a) I = 2.0 A
(b) P = 24 W

3Problem 3medium

❓ Question:

A copper wire (ρ = 1.7 × 10⁻⁸ Ω·m) has length 2.0 m and diameter 1.0 mm. What is its resistance?

💡 Show Solution

Given:

Resistivity: $\rho = 1.7 \times 10^{-8}$ Ω·m
Length: $L = 2.0$ m
Diameter: $d = 1.0$ mm $= 1.0 \times 10^{-3}$ m

Solution:

Step 1: Find cross-sectional area. $r = \frac{d}{2} = 5.0 \times 10^{-4} \text{ m}$ $A = \pi r^2 = \pi (5.0 \times 10^{-4})^2 = 7.85 \times 10^{-7} \text{ m}^2$

Step 2: Calculate resistance. $R = \frac{\rho L}{A} = \frac{(1.7 \times 10^{-8})(2.0)}{7.85 \times 10^{-7}}$ $R = \frac{3.4 \times 10^{-8}}{7.85 \times 10^{-7}} = 0.043 \text{ Ω}$

Answer: R = 0.043 Ω (very low, good conductor!)

4Problem 4medium

❓ Question:

A copper wire (ρ = 1.7 × 10⁻⁸ Ω·m) has length 2.0 m and diameter 1.0 mm. What is its resistance?

💡 Show Solution

Given:

Resistivity: $\rho = 1.7 \times 10^{-8}$ Ω·m
Length: $L = 2.0$ m
Diameter: $d = 1.0$ mm $= 1.0 \times 10^{-3}$ m

Solution:

Step 1: Find cross-sectional area. $r = \frac{d}{2} = 5.0 \times 10^{-4} \text{ m}$ $A = \pi r^2 = \pi (5.0 \times 10^{-4})^2 = 7.85 \times 10^{-7} \text{ m}^2$

Step 2: Calculate resistance. $R = \frac{\rho L}{A} = \frac{(1.7 \times 10^{-8})(2.0)}{7.85 \times 10^{-7}}$ $R = \frac{3.4 \times 10^{-8}}{7.85 \times 10^{-7}} = 0.043 \text{ Ω}$

Answer: R = 0.043 Ω (very low, good conductor!)

5Problem 5easy

❓ Question:

A 12 V battery is connected to a 4.0 Ω resistor. (a) What is the current through the resistor? (b) What is the power dissipated? (c) How much energy is dissipated in 5.0 minutes?

💡 Show Solution

Solution:

Given: V = 12 V, R = 4.0 Ω, t = 5.0 min = 300 s

(a) Current (Ohm's Law): I = V/R = 12/4.0 = 3.0 A

(b) Power: P = VI = 12 × 3.0 = 36 W Or: P = V²/R = 144/4.0 = 36 W ✓ Or: P = I²R = (3.0)²(4.0) = 36 W ✓

6Problem 6easy

❓ Question:

A 12 V battery is connected to a 4.0 Ω resistor. (a) What is the current through the resistor? (b) What is the power dissipated? (c) How much energy is dissipated in 5.0 minutes?

💡 Show Solution

Solution:

Given: V = 12 V, R = 4.0 Ω, t = 5.0 min = 300 s

(a) Current (Ohm's Law): I = V/R = 12/4.0 = 3.0 A

(b) Power: P = VI = 12 × 3.0 = 36 W Or: P = V²/R = 144/4.0 = 36 W ✓ Or: P = I²R = (3.0)²(4.0) = 36 W ✓

7Problem 7medium

❓ Question:

A wire of length 2.0 m and cross-sectional area 1.0 × 10⁻⁶ m² has resistance 0.50 Ω. (a) What is the resistivity of the material? (b) If the wire is stretched to 3.0 m (volume constant), what is the new resistance?

💡 Show Solution

Solution:

Given: L = 2.0 m, A = 1.0 × 10⁻⁶ m², R = 0.50 Ω

(a) Resistivity: R = ρL/A ρ = RA/L = (0.50)(1.0 × 10⁻⁶)/2.0 ρ = 2.5 × 10⁻⁷ Ω·m

(b) New resistance after stretching: Volume constant: V = A₁L₁ = A₂L₂ A₂ = A₁L₁/L₂ = (1.0 × 10⁻⁶)(2.0)/3.0 = 6.67 × 10⁻⁷ m²

R₂ = ρL₂/A₂ = (2.5 × 10⁻⁷)(3.0)/(6.67 × 10⁻⁷) R₂ = 1.125 Ω or 1.1 Ω

Alternative: R ∝ L²/V, so R₂ = R₁(L₂/L₁)² = 0.50(3.0/2.0)² = 1.125 Ω ✓

8Problem 8medium

❓ Question:

💡 Show Solution

Solution:

Given: L = 2.0 m, A = 1.0 × 10⁻⁶ m², R = 0.50 Ω

(a) Resistivity: R = ρL/A ρ = RA/L = (0.50)(1.0 × 10⁻⁶)/2.0 ρ = 2.5 × 10⁻⁷ Ω·m

(b) New resistance after stretching: Volume constant: V = A₁L₁ = A₂L₂ A₂ = A₁L₁/L₂ = (1.0 × 10⁻⁶)(2.0)/3.0 = 6.67 × 10⁻⁷ m²

R₂ = ρL₂/A₂ = (2.5 × 10⁻⁷)(3.0)/(6.67 × 10⁻⁷) R₂ = 1.125 Ω or 1.1 Ω

Alternative: R ∝ L²/V, so R₂ = R₁(L₂/L₁)² = 0.50(3.0/2.0)² = 1.125 Ω ✓

9Problem 9hard

❓ Question:

A 1500 W electric heater operates on 120 V. (a) What is the current? (b) What is the resistance? (c) How much does it cost to run for 8 hours if electricity costs $0.12 per kWh?

💡 Show Solution

Given:

Power: $P = 1500$ W
Voltage: $V = 120$ V
Time: $t = 8$ hours
Rate: $0.12 per kWh

Part (a): Current

$P = IV$ $I = \frac{P}{V} = \frac{1500}{120} = 12.5 \text{ A}$

Part (b): Resistance

$P = \frac{V^2}{R}$ $R = \frac{V^2}{P} = \frac{(120)^2}{1500} = \frac{14,400}{1500} = 9.6 \text{ Ω}$

Or using Ohm's Law: $R = \frac{V}{I} = \frac{120}{12.5} = 9.6 \text{ Ω}$ ✓

Part (c): Cost

Energy used: $E = Pt = (1500 \text{ W})(8 \text{ h}) = 12,000 \text{ Wh} = 12 \text{ kWh}$

Cost: $\text{Cost} = (12 \text{ kWh})(\$0.12/\text{kWh}) = \$1.44$

Answer:

(a) I = 12.5 A
(b) R = 9.6 Ω
(c) Cost = $1.44

10Problem 10hard

❓ Question:

A 1500 W electric heater operates on 120 V. (a) What is the current? (b) What is the resistance? (c) How much does it cost to run for 8 hours if electricity costs $0.12 per kWh?

💡 Show Solution

Given:

Power: $P = 1500$ W
Voltage: $V = 120$ V
Time: $t = 8$ hours
Rate: $0.12 per kWh

Part (a): Current

$P = IV$ $I = \frac{P}{V} = \frac{1500}{120} = 12.5 \text{ A}$

Part (b): Resistance

$P = \frac{V^2}{R}$ $R = \frac{V^2}{P} = \frac{(120)^2}{1500} = \frac{14,400}{1500} = 9.6 \text{ Ω}$

Or using Ohm's Law: $R = \frac{V}{I} = \frac{120}{12.5} = 9.6 \text{ Ω}$ ✓

Part (c): Cost

Energy used: $E = Pt = (1500 \text{ W})(8 \text{ h}) = 12,000 \text{ Wh} = 12 \text{ kWh}$

Cost: $\text{Cost} = (12 \text{ kWh})(\$0.12/\text{kWh}) = \$1.44$

Answer:

(a) I = 12.5 A
(b) R = 9.6 Ω
(c) Cost = $1.44

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