EEE U111 · Module 1 · Extras lesson
Module 1 extras: graphs, averages, ratings, wire
Four small skills that Tutorial 1 tests and the Bobrow sections in the handout don't teach. Forty minutes, once. Then the practice drill.
The handout lists Bobrow §1.1–1.8 for Module 1, and the Module 1 lesson covers those. But Tutorial 1 also draws on four "fundamentals" habits the instructors assume from 12th-grade physics: reading charge and energy off a graph, the average of a waveform, what a nameplate rating actually fixes, and the resistance of a piece of wire. None of these is hard, none is new to only you, and together they are 12 of the sheet's 15 questions. This page teaches exactly those four, with nothing extra.
1 · Reading a graph: area or slope?
Module 1 gave you three definitions that are really one idea in three costumes:
Each says "this quantity is the rate of change of that one." When the tutorial hands you a graph instead of a formula, the calculus turns into geometry:
- Want the rate (current from charge, voltage from inductor current)? Read the slope. On a straight piece the slope is rise ÷ run, constant along the piece.
- Want the accumulated total (charge from current, energy from power)? Take the area under the graph between the two times. Straight-line graphs give triangles and trapezoids, so no integration is ever needed.
The graph above (a Tutorial-1-style current) shows the two moves on one picture: the shaded area from 0 to 5 is the charge delivered in the first 5 μs; the slope of any piece is di/dt, which an inductor would turn into a voltage.
Two habits keep the marks: write the units of the answer before computing (A × μs = μC; mC ÷ s = mA), and cut the region at every corner of the graph — one trapezoid per straight piece, so a corner never hides inside a shape.
Current is 4 A for 0–2 s, then falls linearly to 0 at t = 6 s. Charge delivered in 6 s?
Rectangle 4·2 = 8 C plus triangle ½·4·4 = 8 C: 16 C. (If asked for the charge in the first 3 s: 8 + trapezoid (4+3)/2·1 = 11.5 C — the current at t = 3 is 3 A because the slope is −1 A/s.)
Energy from a current graph through R and L
A "coil of resistance 1 Ω and inductance 2 H" is two elements in series. They absorb energy differently:
- Resistor: p = i²R, so the energy is the area under i²R — and i² is not a straight line even when i is. For a ramp i = kt you do need ∫k²t² dt = k²t³/3.
- Inductor: it only stores: w = ½Li² at the end minus at the start. While the current is constant it absorbs nothing at all.
Current ramps 0 → 6 A over 2 s then holds for 2 s; R = 1 Ω, L = 2 H. Energy absorbed in 4 s?
Resistor: ∫₀² (3t)² dt = 9·8/3 = 24 J, then 36·2 = 72 J; 96 J. Inductor: ½·2·36 = 36 J. 132 J. (This is Tutorial 1 Q13; 144 J is the "6 A for all 4 s" mistake.)
2 · The average of a waveform
The average of anything over one period is
which is the same area idea again. Two shortcuts cover every tutorial case:
- A straight-line rise or fall averages to its midpoint. A triangle wave between 5 A and 10 A averages 7.5 A, whatever its period.
- A square wave of ±V has average voltage zero — but the resistor doesn't feel the average voltage, it feels v²/R, and v² = V² at every instant.
Average the power, not the voltage. Pavg = (v²)avg/R, never Vavg²/R. For ±10 V across 10 Ω that is 10 W, not 0 W. (When you meet RMS values in Module 4, this is exactly where they come from: Vrms² = (v²)avg.)
A current is 8 A for half a period and 2 A for the other half. Average current? Average power in 1 Ω?
Average current (8+2)/2 = 5 A. Average power (64+4)/2 = 34 W — not 5² = 25 W.
3 · What a nameplate rating fixes
"200 W, 220 V" on a lamp is a promise about one operating point: at 220 V it will draw 200 W. The physical thing that stays fixed when you move it to another voltage is its resistance:
Everything else follows from Ohm's law with that R. Three moves appear on the sheets:
- Different voltage: find R from the nameplate, then P = V²/R at the new voltage. Below rating → runs dim/cool, not damaged. Above rating → current exceeds Irated: damage.
- Several rated parts together: each becomes a resistor; series shares one current, parallel shares one voltage. Then the question is ordinary circuit reduction.
- Power ratings as limits: a "100 Ω, 1 W" resistor tolerates Imax = √(P/R) = 0.1 A. In series, the smallest Imax limits the chain; in parallel, the smallest Vmax = √(PR) limits it.
A 60 W, 120 V bulb is put on 240 V. Power drawn (for the instant before it fails)?
R = 120²/60 = 240 Ω. At 240 V: 240²/240 = 240 W, four times the rating (double the voltage, four times the power). Current 1 A vs rated 0.5 A.
Energy bills: the kWh
Energy = power × time. The unit on the bill is the kilowatt-hour: 1 kW running for 1 h, i.e. 1 kWh = 3.6 × 10⁶ J. Add up the watts, multiply by hours, divide by 1000. Motors are often rated in horsepower: 1 hp = 746 W (½ hp = 373 W).
Load of 2413 W runs at a quarter of that, on average, all day. kWh per day?
¼ · 2413 · 24 = 14 478 Wh = 14.5 kWh. (Tutorial 1 Q12.)
4 · Resistance of a wire
Longer → more resistance (in proportion); fatter → less (in proportion to area, so doubling the diameter quarters the resistance). ρ, the resistivity, is a property of the material alone (copper ≈ 1.7 × 10⁻⁸ Ω·m, constantan ≈ 49 × 10⁻⁸ Ω·m); tutorials give it when it's needed.
Two habits: convert area to m² before using ρ in Ω·m (1 mm² = 10⁻⁶ m²), and when two wires of the same material are compared, skip ρ with a ratio: R ∝ L/A, so A₂/A₁ = (L₂/L₁)(R₁/R₂).
A wire-wound resistor adds one step: the wire length that gives the required R is L = RA/ρ, and one turn on a cylinder of diameter D uses πD of it, so N = L/(πD).
100 m of 1 mm-diameter copper wire (ρ = 1.7 × 10⁻⁸ Ω·m). Resistance?
A = π(0.5 mm)² = 0.785 mm² = 0.785 × 10⁻⁶ m². R = 1.7 × 10⁻⁸ · 100 / 0.785 × 10⁻⁶ = 2.17 Ω.
5 · Minimal prerequisite kit
| Fact | Used for |
|---|---|
| Area of a triangle ½·base·height; trapezoid ½(a+b)·width | charge/energy from graphs (§1), averages (§2) |
| Slope = rise/run; a straight piece has one slope | current from a charge graph (§1) |
| ∫t² dt = t³/3 | resistor energy under a ramp (§1) |
| P = VI = I²R = V²/R, all three forms fluent | every ratings question (§3) |
| Prefixes: μ 10⁻⁶, m 10⁻³, k 10³; 1 mm² = 10⁻⁶ m² | everything |
| 1 hp = 746 W; 1 kWh = 3.6 MJ | household loads (§3) |
6 · What to practise
| Skill | Source | How much |
|---|---|---|
| Area/slope from a graph | T1 Q1, Q5, Q13 · drill families 1–2 | 6 generated + the 3 sheet questions |
| Averages | T1 Q9, Q11 · drill family 3 | 4 generated |
| Ratings | T1 Q2, Q3, Q6, Q7, Q12 · drill family 4 | 8 generated (all three variants) |
| Wire | T1 Q8, Q10 · drill family 5 | 4 generated |
1) "A current graph is piecewise linear with corners at … ; show me how to cut the area under it into triangles and trapezoids to get the charge up to t = …, with units at each step." 2) "A lamp rated P W at V V is connected to V′ V. Walk me through what stays fixed and what changes, and whether it is damaged." 3) "Why is the average power of a ±V square wave across R not zero, and how does that lead to the idea of an RMS value? Keep it to Bobrow §1–2 level."