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Current, Resistance, and Ohm's Law

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

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🔌 Current, Resistance, and Ohm's Law

Electric Current

Electric current is the rate of charge flow:

I=ΔQΔtI = \frac{\Delta Q}{\Delta t}

where:

  • II = current (Ampere, A)
  • ΔQ\Delta Q = charge passing point (C)
  • Δt\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=nAvdeI = nAv_d e

where:

  • nn = charge carrier density
  • AA = cross-sectional area
  • vdv_d = drift velocity (~mm/s, very slow!)
  • ee = elementary charge

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

Resistance

Resistance opposes current flow:

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

Units: Ohm (Ω) = V/A

Resistivity

Material property:

R=ρLAR = \frac{\rho L}{A}

where:

  • ρ\rho = resistivity (Ω·m)
  • LL = length
  • AA = cross-sectional area

Good conductors: Low ρ\rho (copper: 1.7×10−81.7 \times 10^{-8} Ω·m) Insulators: High ρ\rho

Temperature Dependence

R=R0(1+αΔT)R = R_0(1 + \alpha \Delta T)

where α\alpha is temperature coefficient.

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

Ohm's Law

V=IRV = 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=I2R=V2RP = IV = I^2R = \frac{V^2}{R}

Units: Watt (W) = J/s

Energy dissipated (heat): E=Pt=I2Rt=V2RtE = Pt = I^2Rt = \frac{V^2}{R}t

Electrical Energy Cost

Power companies charge by kilowatt-hour (kWh):

1 kWh=(1000 W)(3600 s)=3.6×106 J1 \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: Vrms=V02V_{rms} = \frac{V_0}{\sqrt{2}}, Irms=I02I_{rms} = \frac{I_0}{\sqrt{2}}

Household: 120 V AC is VrmsV_{rms}

Problem-Solving Strategy

  1. Identify knowns: V, I, or R
  2. Choose formula:
    • Ohm's Law: V=IRV = IR
    • Power: P=IV=I2R=V2/RP = IV = I^2R = V^2/R
    • Resistance: R=ρL/AR = \rho L/A
  3. Watch units: A, V, Ω, W
  4. Check reasonableness

Common Mistakes

❌ Confusing current direction (use conventional!) ❌ Using diameter instead of radius in area (A=πr2A = \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=12V = 12 V
  • Resistance: R=6.0R = 6.0 Ω

Part (a): Current

Use Ohm's Law: I=VR=126.0=2.0 AI = \frac{V}{R} = \frac{12}{6.0} = 2.0 \text{ A}

Part (b): Power dissipated

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

Or alternatively: P=V2R=(12)26.0=1446.0=24 WP = \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=12V = 12 V
  • Resistance: R=6.0R = 6.0 Ω

Part (a): Current

Use Ohm's Law: I=VR=126.0=2.0 AI = \frac{V}{R} = \frac{12}{6.0} = 2.0 \text{ A}

Part (b): Power dissipated

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

Or alternatively: P=V2R=(12)26.0=1446.0=24 WP = \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: ρ=1.7×10−8\rho = 1.7 \times 10^{-8} Ω·m
  • Length: L=2.0L = 2.0 m
  • Diameter: d=1.0d = 1.0 mm =1.0×10−3= 1.0 \times 10^{-3} m

Solution:

Step 1: Find cross-sectional area. r=d2=5.0×10−4 mr = \frac{d}{2} = 5.0 \times 10^{-4} \text{ m} A=πr2=π(5.0×10−4)2=7.85×10−7 m2A = \pi r^2 = \pi (5.0 \times 10^{-4})^2 = 7.85 \times 10^{-7} \text{ m}^2

Step 2: Calculate resistance. R=ρLA=(1.7×10−8)(2.0)7.85×10−7R = \frac{\rho L}{A} = \frac{(1.7 \times 10^{-8})(2.0)}{7.85 \times 10^{-7}} R=3.4×10−87.85×10−7=0.043 ΩR = \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: ρ=1.7×10−8\rho = 1.7 \times 10^{-8} Ω·m
  • Length: L=2.0L = 2.0 m
  • Diameter: d=1.0d = 1.0 mm =1.0×10−3= 1.0 \times 10^{-3} m

Solution:

Step 1: Find cross-sectional area. r=d2=5.0×10−4 mr = \frac{d}{2} = 5.0 \times 10^{-4} \text{ m} A=πr2=π(5.0×10−4)2=7.85×10−7 m2A = \pi r^2 = \pi (5.0 \times 10^{-4})^2 = 7.85 \times 10^{-7} \text{ m}^2

Step 2: Calculate resistance. R=ρLA=(1.7×10−8)(2.0)7.85×10−7R = \frac{\rho L}{A} = \frac{(1.7 \times 10^{-8})(2.0)}{7.85 \times 10^{-7}} R=3.4×10−87.85×10−7=0.043 ΩR = \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 ✓

(c) Energy: E = Pt = 36 × 300 = 10,800 J or 10.8 kJ

6Problem 6hard

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

Part (a): Current

P=IVP = IV I=PV=1500120=12.5 AI = \frac{P}{V} = \frac{1500}{120} = 12.5 \text{ A}

Part (b): Resistance

P=V2RP = \frac{V^2}{R} R=V2P=(120)21500=14,4001500=9.6 ΩR = \frac{V^2}{P} = \frac{(120)^2}{1500} = \frac{14,400}{1500} = 9.6 \text{ Ω}

Or using Ohm's Law: R=VI=12012.5=9.6 ΩR = \frac{V}{I} = \frac{120}{12.5} = 9.6 \text{ Ω} ✓

Part (c): Cost

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

Cost: Cost=(12 kWh)($0.12/kWh)=$1.44\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

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 8hard

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

Part (a): Current

P=IVP = IV I=PV=1500120=12.5 AI = \frac{P}{V} = \frac{1500}{120} = 12.5 \text{ A}

Part (b): Resistance

P=V2RP = \frac{V^2}{R} R=V2P=(120)21500=14,4001500=9.6 ΩR = \frac{V^2}{P} = \frac{(120)^2}{1500} = \frac{14,400}{1500} = 9.6 \text{ Ω}

Or using Ohm's Law: R=VI=12012.5=9.6 ΩR = \frac{V}{I} = \frac{120}{12.5} = 9.6 \text{ Ω} ✓

Part (c): Cost

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

Cost: Cost=(12 kWh)($0.12/kWh)=$1.44\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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