Lecture Examples — All Problems
Lec 7 · Aluminum wire — Find resistance
Given: Aluminum wire (σ=4×10⁷ S/m), length=100m, cross-section=1mm²=1×10⁻⁶ m².
R = L/(σS) = 100 / (4×10⁷ × 1×10⁻⁶)
R = 100/40 = 2.5 Ω
Lec 7 · AWG aluminum wire — Find current from E field
Given: Aluminum σ=4×10⁷ S/m, diameter=9.266mm, E=0.01 V/m inside wire. Find I.
J = σE = 4×10⁷ × 0.01 = 4×10⁵ A/m²
r = diameter/2 = 9.266/2 = 4.633mm = 4.633×10⁻³ m
S = πr² = π × (4.633×10⁻³)² = 6.748×10⁻⁵ m²
I = J×S = 4×10⁵ × 6.748×10⁻⁵
I = 27 A
Lec 7 · R=300Ω, P=480mW — Find Ig and Vg
Given: R=300Ω, Power P=480mW=480×10⁻³ W. Find Ig and Vg.
P = Vg×Ig = Vg²/R → Vg² = P×R = 480×10⁻³ × 300 = 144
Vg = √144 = 12 V
Ig = Vg/R = 12/300 = 0.04 A
Vg = 12 V | Ig = 40 mA
Lec 8 · Series-parallel R_eq — Complex circuit (from lecture)
Given: Circuit with 15Ω, 10Ω, 24Ω, 40Ω, 80Ω, 10Ω, 60Ω, 30Ω, 45Ω, 20Ω, short circuit. 80mV source. Find R_eq.
Step 1: 45Ω // 0Ω (short) = 0Ω (short circuit wins)
Step 2: R_eq1 = 15Ω // 30Ω = 15×30/(15+30) = 10Ω
Step 3: R_eq2 = 10+20 = 30Ω (wait — need to redo from circuit)
Simplified: 45//0=0, then 0+20=20, then (15//30=10) series with 20=30, then 30//60=20, then 20+24=44... continue per exact topology:
R_eq1 = 1/(1/60 + 1/80 + 1/30) = 16Ω
R_eq2 = 10+20=30 → 30//15=10 → 10+24+16=50Ω → 10//50=8.33... (topology-dependent)
From lecture solution: R_eq = depends on exact topology shown in figure
Method: Simplify from innermost parallel/series combination outward
Lec 9 · Battery circuit — Find I, power, reversed polarity
Given: ε₁=6V, ε₂=12V, R₁=8Ω, R₂=10Ω in single loop.
(A) Find current:
KVL: −ε₁ + IR₁ + ε₂ + IR₂ = 0
−6 + 8I + 12 + 10I = 0 → 18I = −6
I = −1/3 A (flows opposite to assumed direction)
(B) Powers:
P_R1 = I²R₁ = (1/3)²×8 = 8/9 W ≈ 0.889 W
P_R2 = I²R₂ = (1/3)²×10 = 10/9 W ≈ 1.11 W
P_ε2 = V×I = (−12)×(−1/3) = +4W → ε₂ absorbs power (being charged!)
P_ε1 = V×I = (6)×(−1/3) = −2W → ε₁ supplies power
Check: −2 + 4 = +2 = 8/9 + 10/9 ✓ Power supplied = power absorbed
(C) Reversed ε₂ polarity:
−6 + 8I − 12 + 10I = 0 → 18I = 18 → I = 1A
Now BOTH batteries aid each other (same direction)
(A) I = −1/3 A | (B) P_R1=0.89W, P_R2=1.11W, ε₁ supplies, ε₂ absorbs | (C) I = 1A (reversed → batteries add)
Lec 10 · AWG 0000 Aluminum Transmission Line
Given: Aluminum σ=4×10⁷ S/m, A=107mm²=107×10⁻⁶ m², total length=100km (two conductors). Delivers 10MW. Find R and power loss at 3 operating conditions.
R_TL = L/(Aσ) = (100×10³) / (107×10⁻⁶ × 4×10⁷) = 10⁵ / (107×40) = 10⁵/4280 = 23.36 Ω
| V (kV) | I (A) | P_loss = I²R | P_loss % |
|---|
| 200 | 50 | 50²×23.36 = 58,400 W | 0.58% |
| 100 | 100 | 100²×23.36 = 233,600 W | 2.28% |
| 50 | 200 | 200²×23.36 = 934,400 W | 8.55% |
✔ Key lessonSame power (10MW) but at higher voltage → lower current → dramatically less loss. This is WHY power grids use 220kV or 440kV for long-distance transmission!
Lec 9 · Internal resistance from v-i graph
Voltage source: Every horizontal division=3V, every vertical division=1A. Find r.
From the v-i characteristic of a real voltage source: VS = V + I×r
The slope of the line = −r (voltage drops as current increases)
Slope = ΔV/ΔI = −(3V)/(1A) = −3... but from the figure, horizontal div=3V, vertical=1A
r = |slope| = 3/5 = 0.6 Ω (reading two points from graph: ΔV=3V, ΔI=5A)
r_voltage_source = 0.6 Ω
Tutorial 5–6 · Problem 1 — Find Req and power delivered
Given: Circuit with multiple resistors. Find R_ab = 8Ω and power from source.
Simplify resistor combinations step by step (series/parallel) until single R_ab remains.
Result from tutorial: R_ab = 8Ω
Power = V²/R_ab (if voltage source) or I²×R_ab (if current source)
R_ab = 8 Ω