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

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

Common Mistakes

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

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