EEE U111 · Electrical Sciences · Mid-sem prep · Module 3
Transient ladders: from a tiny circuit to the mid-sem question
The past-paper transient questions look advanced, but each one is a small number of simple steps stacked on top of each other. These two ladders take them apart. Each rung adds exactly one new idea to the rung below, and the top rung is the real past-paper question.
Why this page exists. The mid-sem transient questions (about 40% of past marks) are not harder ideas than the rest of Module 3. They stack four or five small ideas into one question, so if one idea is missing the whole thing looks impossible. On a ladder you can see exactly which idea is missing.
Nobody arrives with this. School physics and JEE touch RC charging lightly and do not do second-order RLC circuits with initial slopes at all, so the whole class learns this in Module 3.
How to use a rung: read the question, try it on paper (the "try it first" line says what to write), then open Answer. If yours differs, open Worked solution. Every rung is tagged with its one new idea.
Cold check (about 20 minutes)
Do these four things cold, before reading anything else on this page, with no notes. They tell you where to start climbing.
- Ladder 1, rung 1: a capacitor discharging through a resistor.
- Ladder 2, rung 1: two starting values of a circuit with an inductor and a capacitor.
- Oct 2024 Q3(b), first moves only (circuit at rung 5 of Ladder 1): write the capacitor voltage just before the switch, just after, and long after, and τ. Don't write the formula or solve further.
- Oct 2024 Q5(b), starting values only (circuit at the top of Ladder 2): write vC(0⁺) and iL(0⁺). Nothing else.
Answers to items 3 and 4
Item 3 (Oct 2024 Q3(b)): before the switch the capacitor is an open circuit, so 3 kΩ and 5 kΩ divide 24 V: v(0⁻) = 24 × 5/8 = 15 V. It can't jump: v(0⁺) = 15 V. Long after, only 30 V and 4 kΩ are connected: v(∞) = 30 V. τ = 4 kΩ × 0.5 mF = 2 s.
Item 4 (Oct 2024 Q5(b)): before the switch the capacitor sits across the 2 Ω of a 1 Ω / 2 Ω divider: vC(0⁺) = vC(0⁻) = 12 × 2/3 = 8 V. The inductor branch is not connected to anything before the switch moves, so iL(0⁺) = iL(0⁻) = 0.
Start one rung below the first one you couldn't do. A rung-1 item went wrong? Read the Module 3 lesson §2–3 first, then start that ladder at rung 1. Both rung-1 items right, but item 3 or 4 wrong? Start that ladder at rung 2. All four right? Start each ladder at rung 4 and climb to the past-paper rungs.
Ladder 1 · First order (one capacitor or one inductor)
Every first-order answer is the same sentence: start value, end value, and how fast. The rungs add those three pieces one at a time.
Rung 1A capacitor lets go
New on this rung a charged capacitor draining through a resistor, and τ = RC
Find v(0), the time constant τ, the formula for v(t), and the voltage at t = 1 s.
Try it first Four lines: v(0) = …, τ = … × … = …, v(t) = …, v(1) = ….
Answer
v(0) = 20 V, τ = 1 s, v(t) = 20e−t V, v(1) = 7.36 V.
Worked solution
1Start value
The capacitor is charged to 20 V before the switch closes, and a capacitor's voltage can't jump, so v(0) = 20 V.
2End value
Nothing refills it, so all the charge eventually drains through the resistor: v(∞) = 0.
3How fast
τ = RC = 10 000 Ω × 100 × 10−6 F = 1 s. (Units: ohms × farads = seconds.)
4The formula
v(t) = 0 + (20 − 0)e−t/1 = 20e−t V. At t = 1 s: 20 × e−1 = 20 × 0.3679 = 7.36 V. After one τ, 36.8% of the voltage is left; this is true for every first-order circuit.
Rung 2The three snapshots (no formula yet)
New on this rung a switch that changes the circuit: look before (C = open), just after (v can't jump), and long after
🎧 Watch first (3 min): the three snapshots, narrated.
Find only these: v(0⁻), v(0⁺), v(∞), and the current in the new 6 Ω at 0⁻ and at 0⁺.
Try it first Redraw the circuit twice: once for t < 0 with the capacitor as a gap, once for t → ∞ with the switch closed and the capacitor as a gap.
Answer
v(0⁻) = v(0⁺) = 6 V, v(∞) = 4.5 V. New 6 Ω: 0 A at 0⁻, jumps to 1 A at 0⁺.
Worked solution
1Before (t < 0): capacitor = open
It has sat for a long time, so no current flows into the capacitor and it acts as a gap. The switch is open, so the new 6 Ω carries nothing. What's left is a divider: 6 Ω on top, 3 Ω below. v(0⁻) = 18 × 3/(6 + 3) = 6 V.
2Just after (t = 0⁺): v can't jump
The capacitor voltage is still 6 V. The new 6 Ω now sits across it, so it suddenly carries 6/6 = 1 A. Resistor currents can jump; only the capacitor voltage (and inductor current) can't.
3Long after (t → ∞): capacitor = open again
The switch is closed, so 3 Ω and 6 Ω are in parallel below the node: 3 ∥ 6 = (3 × 6)/(3 + 6) = 2 Ω. New divider: v(∞) = 18 × 2/(6 + 2) = 4.5 V.
That is two of the three ingredients. Rung 3 adds the third, τ. (For the record: here the capacitor would see 6 ∥ 3 ∥ 6 = 1.5 Ω.)
Rung 3Charging through two resistors
New on this rung τ uses the resistance the capacitor sees (sources killed), not just the resistor in series with it
Find v(t) for t ≥ 0, and v at t = 60 ms.
Try it first The three snapshots from rung 2, then: kill the 12 V (replace it with a wire) and find the resistance between the capacitor's two terminals.
Answer
v(t) = 9(1 − e−t/0.06) V (τ = 60 ms); v(60 ms) = 5.69 V.
Worked solution
1Start and end values
Uncharged, so v(0⁺) = v(0⁻) = 0. Long after, the capacitor is a gap and 4 kΩ / 12 kΩ divide 12 V: v(∞) = 12 × 12/(4 + 12) = 9 V.
2Resistance the capacitor sees
Kill the source (a voltage source becomes a wire). Standing at the capacitor and looking out, there are two paths to the bottom rail: 12 kΩ straight down, and 4 kΩ through the dead source. They are in parallel: Rth = (4 × 12)/(4 + 12) = 3 kΩ.
3τ and the formula
τ = RthC = 3000 × 20 × 10−6 = 0.06 s = 60 ms. So v(t) = 9 + (0 − 9)e−t/0.06 = 9(1 − e−t/0.06) V.
4The value
At t = 0.06 s (exactly one τ): 9(1 − e−1) = 9 × 0.6321 = 5.69 V, which is 63% of the way from 0 to 9.
Using only the series 4 kΩ: τ = 4000 × 20 µF = 80 ms. Wrong: the 12 kΩ is also connected to the capacitor, so it counts.
Rung 4The inductor version, and "when does it reach…?"
New on this rung the inductor: L = short in steady state, current can't jump, τ = L/Rth; and solving for a time with ln
Find i(t), and the time at which the inductor current reaches 3 A.
Try it first Same three snapshots, but for i. Long after, the inductor is a wire, so ask what that wire does to the 12 Ω.
Answer
i(t) = 4(1 − e−2t) A (τ = 0.5 s); it reaches 3 A at t = ln 2 ≈ 0.693 s.
Worked solution
1Start value
No current before the switch, and an inductor's current can't jump: i(0⁺) = 0.
2End value
Long after, the inductor is a wire from the node to the bottom rail. That wire is in parallel with the 12 Ω and has zero resistance, so it takes all the current and the 12 Ω carries none. i(∞) = 24/6 = 4 A.
3τ
Kill the 24 V (wire). The inductor sees 6 Ω and 12 Ω in parallel: Rth = (6 × 12)/18 = 4 Ω, so τ = L/Rth = 2/4 = 0.5 s. (For an inductor it's L divided by R, not times: only L/R comes out in seconds.)
4Formula
i(t) = 4 + (0 − 4)e−t/0.5 = 4(1 − e−2t) A.
5When is i = 3 A?
Set 4(1 − e−2t) = 3. Divide by 4: 1 − e−2t = 0.75, so e−2t = 0.25. Take ln of both sides: −2t = ln 0.25 = −ln 4, so t = (ln 4)/2 = ln 2 ≈ 0.693 s.
Unpack this step
ln undoes e: ln(ek) = k, so ln(e−2t) = −2t. And ln 0.25 = ln(1/4) = −ln 4. Finally ln 4 = ln(2²) = 2 ln 2, so (ln 4)/2 = ln 2.
Rung 5 · past paperOct 2024 Q3(b): a two-position switch · 9 marks
New on this rung the switch hands the capacitor from one circuit to a completely different one: find v(0⁻) in the old circuit and v(∞), τ in the new one
Try it first Rung 2 for the start value (a divider), rung 3 for τ. Then the formula and the two values.
Answer
v = 15 V for t < 0; v(t) = 30 − 15e−t/2 V for t ≥ 0; v(1) = 20.902 V, v(4) = 27.970 V.
Worked solution
1Before (switch on A)
Capacitor = gap, so 3 kΩ and 5 kΩ divide 24 V: v(0⁻) = 24 × 5/(3 + 5) = 15 V = v(0⁺).
2After (switch on B)
The left half is disconnected. Long after, no current flows, so there is no drop across 4 kΩ: v(∞) = 30 V. The capacitor sees only 4 kΩ: τ = 4000 × 0.5 × 10−3 = 2 s.
3Formula and values
v(t) = 30 + (15 − 30)e−t/2 = 30 − 15e−t/2 V. v(1) = 30 − 15 × 0.60653 = 20.902 V; v(4) = 30 − 15 × 0.13534 = 27.970 V.
Full write-up: past papers, Oct 2024 Q3 · 🎧 narrated walkthrough (about 6.5 min)
Rung 6 · past paper, topMar 2025 Q2: a second switch at t = 4 s · 20 marks
New on this rung a second switching: restart the clock at t = 4 s, using i(4) as the new start value and a new end value and τ
Try it first Treat it as two rung-4 problems in a row. Interval 1 (0 to 4 s): start, end, τ. Then interval 2: its start value is wherever interval 1 left off at 4 s.
Answer
i = 2(1 − e−2t) A for 0 ≤ t ≤ 4 s; i = 1.364 + 0.636e−1.467(t − 4) A for t ≥ 4 s; i(2) = 1.963 A, i(5) = 1.510 A.
Worked solution
1Interval 1 (0 < t < 4): only S₁ closed
Before t = 0 nothing is connected, so i(0⁺) = 0. Long after, the inductor is a wire: one loop, i(∞) = 40/(8 + 12) = 2 A. The inductor sees 8 + 12 = 20 Ω: τ₁ = 10/20 = 0.5 s. So i = 2(1 − e−2t) A, and i(2) = 2(1 − e−4) = 1.963 A.
2Hand-over at t = 4 s
i(4) = 2(1 − e−8) = 1.9993 A, practically 2 A. The inductor current can't jump when S₂ closes, so this is the start value of interval 2.
3Interval 2: replace everything left of P by its Thevenin equivalent
Two branches meet at P: 40 V through 8 Ω, and 10 V through 4 Ω. Open-circuit voltage at P (KCL, no current leaving P): (40 − V)/8 = (V − 10)/4 gives Vth = 20 V; Rth = 8 ∥ 4 = 8/3 Ω.
4New end value and τ
Loop: i(∞) = 20/(12 + 8/3) = 20/(44/3) = 15/11 ≈ 1.364 A. Resistance seen by L: 44/3 Ω, so τ₂ = 10/(44/3) = 15/22 ≈ 0.682 s (1/τ₂ ≈ 1.467).
5Restart the clock
Write (t − 4) wherever t appeared: i = 1.364 + (1.9993 − 1.364)e−(t − 4)/0.682 = 1.364 + 0.636e−1.467(t − 4) A. At t = 5: 1.364 + 0.636 × e−1.467 = 1.364 + 0.636 × 0.2307 = 1.510 A.
Full write-up: past papers, Mar 2025 Q2 · 🎧 narrated walkthrough (about 6.5 min)
Ladder 2 · Second order (an inductor and a capacitor)
A second-order answer needs four things: two starting values (x(0⁺) and its slope), which case (from α and ω₀), and the form for that case. The rungs add them one at a time. The first two rungs are pure Module 1 thinking: KCL, KVL and "what can't jump".
Rung 1Two things can't jump
New on this rung with both an L and a C, there are two starting values: vC(0⁺) and iL(0⁺), both from the t < 0 circuit
Find vC(0⁺), iL(0⁺), and the current in the 4 Ω at 0⁺.
Try it first Redraw for t < 0 with C = gap and L = wire.
Answer
vC(0⁺) = 8 V, iL(0⁺) = 2 A, 4 Ω current at 0⁺ = 2 A.
Worked solution
1Before: C = gap, L = wire
The capacitor branch carries nothing and the inductor is a wire, so there is one loop: 12 V, 2 Ω, (wire), 4 Ω. iL(0⁻) = 12/(2 + 4) = 2 A. The capacitor sits across the 4 Ω, so vC(0⁻) = 2 × 4 = 8 V.
2Neither can jump
vC(0⁺) = 8 V and iL(0⁺) = 2 A, whatever the switch does. The 4 Ω is across the capacitor, so it still has 8 V: 8/4 = 2 A.
Rung 2The starting slopes
New on this rung dvC/dt(0⁺) = iC(0⁺)/C (KCL) and diL/dt(0⁺) = vL(0⁺)/L (KVL)
🎧 Watch first (3 min): where the starting slopes come from, narrated.
These come straight from the element laws: iC = C dvC/dt and vL = L diL/dt, read at t = 0⁺. So you never differentiate anything: find a current for the capacitor with KCL, and a voltage for the inductor with KVL, using the rung-1 values.
Same circuit as rung 1 (vC(0⁺) = 8 V, iL(0⁺) = 2 A). Find both starting slopes in two versions:
(a) the switch replaces the source with a wire (as drawn in rung 1);
(b) instead, the source stays and a switch disconnects the 4 Ω at t = 0 (drawn below).
Try it first For each version, draw the t = 0⁺ circuit with the capacitor as an 8 V battery and the inductor as a 2 A current source. Then KCL at node b and KVL around the loop.
Answer
(a) diL/dt(0⁺) = −12 A/s, dvC/dt(0⁺) = 0. (b) dvC/dt(0⁺) = 4 V/s, diL/dt(0⁺) = 0.
Worked solution
a1(a) Capacitor slope: KCL at b
In through the inductor: 2 A. Out through the 4 Ω: 8/4 = 2 A. What's left goes into the capacitor: iC = 2 − 2 = 0, so dvC/dt = 0/0.5 = 0.
a2(a) Inductor slope: KVL round the loop
The source is now a wire. Walk the loop: the 2 Ω drops 2 × 2 = 4 V and the capacitor holds 8 V, and together with vL they must add to zero (nothing drives the loop): vL = 0 − 4 − 8 = −12 V. So diL/dt = −12/1 = −12 A/s: the current starts falling.
b1(b) Capacitor slope
The 4 Ω is gone, so all 2 A goes into the capacitor: dvC/dt = 2/0.5 = 4 V/s.
b2(b) Inductor slope
The source is still there: vL = 12 − 4 − 8 = 0, so diL/dt = 0.
A slope is non-zero only if the switch changed something around that element. In (a) the loop lost its source, so the inductor feels it; the capacitor's node didn't change, so its slope is 0. In (b) it's the other way round. A zero slope is a legitimate answer (Oct 2024 Q5(b) has one), but you must show the KCL or KVL line that gives it.
Rung 3Classify only
New on this rung α and ω₀, which differ for series and parallel, and the three cases
| α (s⁻¹) | ω₀ (rad/s) | |
|---|---|---|
| Series RLC (one loop) | R/(2L) | 1/√(LC) |
| Parallel RLC (one pair of nodes) | 1/(2RC) | 1/√(LC) |
Then compare: α > ω₀ overdamped (two real roots), α = ω₀ critically damped (one repeated root), α < ω₀ underdamped (it oscillates, with ωd = √(ω₀² − α²)). The roots are s = −α ± √(α² − ω₀²).
For each: α, ω₀, the case, and the roots (or ωd).
Try it first Decide "series or parallel?" by looking: one loop, or everything across the same two nodes?
Answer
(i) α = 3, ω₀ = √5 ≈ 2.236 → overdamped, s = −1, −5. (ii) α = 1, ω₀ = 2 → underdamped, ωd = √3 ≈ 1.732. (iii) α = 1, ω₀ = 1 → critically damped, s = −1 (twice).
Worked solution
iSeries 6 Ω, 1 H, 0.2 F
α = 6/(2 × 1) = 3; ω₀ = 1/√(1 × 0.2) = 1/0.4472 = 2.236. 3 > 2.236: overdamped. s = −3 ± √(9 − 5) = −3 ± 2, so −1 and −5.
iiSeries 4 Ω, 2 H, 0.125 F
α = 4/(2 × 2) = 1; ω₀ = 1/√(2 × 0.125) = 1/√0.25 = 2. 1 < 2: underdamped. ωd = √(4 − 1) = √3 ≈ 1.732 rad/s.
iiiParallel 2 Ω, 4 H, 0.25 F
α = 1/(2 × 2 × 0.25) = 1; ω₀ = 1/√(4 × 0.25) = 1. Equal: critically damped, s = −1 twice.
Using the series α for the parallel circuit: R/(2L) = 2/8 = 0.25 < 1 would say "underdamped". Wrong case, wrong form, and most of the marks go with it. Look at the picture first.
Rung 4A full natural response (overdamped, easy roots)
New on this rung putting it together: the form A₁es₁t + A₂es₂t, with the two constants from rungs 1 and 2
Find vC(t) and i(t) for t ≥ 0, and vC(1 s).
Try it first Rung 3 (series): α, ω₀, roots. Rung 1: vC(0⁺). Rung 2: dvC/dt(0⁺) = iC/C. Then two equations for A₁, A₂.
Answer
vC(t) = 20e−t − 10e−2t V, i(t) = 10e−t − 10e−2t A, vC(1) = 6.00 V.
Worked solution
1Classify (rung 3)
Series: α = 3/(2 × 1) = 1.5, ω₀ = 1/√(1 × 0.5) = √2 ≈ 1.414. 1.5 > 1.414: overdamped. s = −1.5 ± √(2.25 − 2) = −1.5 ± 0.5, so s₁ = −1, s₂ = −2.
2The form
vC(t) = A₁e−t + A₂e−2t. There is no source, so there is no constant term: everything dies away.
3Starting values (rungs 1–2)
vC(0⁺) = 10 V. The loop current starts at 0, so the capacitor current is 0 and dvC/dt(0⁺) = 0.
4Two equations
At t = 0 every e… equals 1: A₁ + A₂ = 10. The slope of Aest is s · Aest, so at 0: −1·A₁ − 2·A₂ = 0.
Why the slope of est is sest: ex is its own slope, and the chain rule multiplies by the slope of the inside, st, which is s. At t = 0 that leaves just s × (the constant in front).Unpack this step
5Answer and the current
vC = 20e−t − 10e−2t V; check: at 0, 20 − 10 = 10 ✓. The current leaves the + plate, so i = −C dvC/dt = −0.5(−20e−t + 20e−2t) = 10e−t − 10e−2t A; check: i(0) = 0 ✓. vC(1) = 20 × 0.3679 − 10 × 0.1353 = 7.358 − 1.353 = 6.00 V.
Rung 5An underdamped one (clean numbers)
New on this rung the oscillating form e−αt(B₁ cos ωdt + B₂ sin ωdt), and radians mode
Find vC(t), i(t), and vC(0.5 s).
Try it first α, ω₀, case, ωd; then vC(0⁺) fixes B₁ and the starting slope fixes B₂.
Answer
vC(t) = 10e−t sin 3t V, i(t) = e−t(3 cos 3t − sin 3t) A, vC(0.5) = 6.05 V (radians!).
Worked solution
1Classify
Series: α = 2/(2 × 1) = 1, ω₀ = 1/√(1 × 0.1) = √10 ≈ 3.162. 1 < 3.162: underdamped. ωd = √(10 − 1) = 3 rad/s.
2Form
vC = e−t(B₁ cos 3t + B₂ sin 3t).
3B₁ from the value
At t = 0: e0 = 1, cos 0 = 1, sin 0 = 0, so vC(0) = B₁ = 0.
4B₂ from the slope
The 3 A flows into the + plate, so dvC/dt(0⁺) = i/C = 3/0.1 = 30 V/s. The slope of the form at t = 0 is −αB₁ + ωdB₂, so −1 × 0 + 3B₂ = 30 and B₂ = 10.
Where "slope at 0 = −αB₁ + ωdB₂" comes from: by the product rule the slope of e−αt × (bracket) is (−αe−αt) × bracket + e−αt × (slope of bracket). At t = 0 the bracket is B₁, and its slope is −ωdB₁ sin 0 + ωdB₂ cos 0 = ωdB₂.Unpack this step
5Answer and the current
vC = 10e−t sin 3t V. The current into the + plate is C dvC/dt = 0.1 × 10e−t(3 cos 3t − sin 3t) = e−t(3 cos 3t − sin 3t) A; check: i(0) = 3 ✓.
6The value, in radians
vC(0.5) = 10 × e−0.5 × sin(1.5) = 10 × 0.6065 × 0.9975 = 6.05 V. The 1.5 is in radians (ωd is in rad/s).
Calculator in degree mode: sin(1.5°) = 0.0262, giving 0.16 V. That's wrong by a factor of almost 40, and nothing on the page looks suspicious. Switch to RAD before the exam starts.
Rung 6 · past papers, topOct 2024 Q5(b) and Mar 2025 Q3
New on this rung nothing new: rung 1 on a switch that moves, rung 2 with KVL or KCL, rung 3, then rung 5's form
Oct 2024 Q5(b) · series RLC · 12 marks
Try it first Rung 1: vC(0⁺), iL(0⁺) (you did this in the cold check). Rung 2: diL/dt(0⁺) by KVL round the new loop. Rung 3: series α, ω₀. Rung 5: the form.
Answer
vC(t) = e−2t(8 cos 3.464t + 4.619 sin 3.464t) V, iL(t) = 0.9238e−2t sin 3.464t A.
Worked solution
1Starting values
vC(0⁺) = 12 × 2/3 = 8 V; iL(0⁺) = 0.
2Starting slopes
KVL round the new loop: the capacitor's 8 V drives the inductor and 10 Ω, and 10 Ω × 0 = 0, so vL(0⁺) = 8 V and diL/dt(0⁺) = 8/2.5 = 3.2 A/s. Capacitor current = iL = 0, so dvC/dt(0⁺) = 0.
3Classify
Series loop: α = 10/(2 × 2.5) = 2; ω₀ = 1/√(2.5/40) = 1/√0.0625 = 4. Underdamped; ωd = √(16 − 4) = √12 ≈ 3.464.
4Constants
iL = e−2t(B₁ cos + B₂ sin): B₁ = 0, and 3.464B₂ = 3.2 gives B₂ = 0.9238. vC = e−2t(8 cos + D sin): slope −2 × 8 + 3.464D = 0, so D = 4.619.
Full write-up: past papers, Oct 2024 Q5 · 🎧 narrated walkthrough (about 8 min)
Mar 2025 Q3 · parallel RLC · 20 marks
(a) v(0⁺), dv(0⁺)/dt
(b) v(t) for t ≥ 0.
(c) Plot the graph of v(t). [5+10+5=20]
Try it first Rung 1 (v(0⁺), iL(0⁺)); rung 2 with KCL at the top node; rung 3: it's parallel.
Answer
v(0⁺) = 4 V, dv/dt(0⁺) = −8 V/s; v(t) = e−t(4 cos 1.732t − 2.309 sin 1.732t) V; zero at 0.605 s, minimum −1.194 V at 1.209 s.
Worked solution
1Starting values
Before: capacitor = gap, so no current in 8 Ω and v(0⁻) = 4 V. The L ∥ 0.5 Ω pair has no source: iL(0) = 0.
2Starting slope: KCL at the top node
The capacitor feeds the 0.5 Ω (4/0.5 = 8 A) and the inductor (0 A): iC = −8 A, so dv/dt(0⁺) = −8/1 = −8 V/s.
3Classify (parallel!)
α = 1/(2 × 0.5 × 1) = 1; ω₀ = 1/√(0.25 × 1) = 2. Underdamped; ωd = √3 ≈ 1.732.
4Constants
B₁ = 4; slope: −1 × 4 + 1.732B₂ = −8, so B₂ = −4/1.732 = −2.309.
Full write-up: past papers, Mar 2025 Q3 · 🎧 narrated walkthrough (about 6 min)
After the ladders
- Went back down a rung? Redo the rung you fell off tomorrow, cold, before climbing again.
- Topped a ladder? Do the start drill (first-order and second-order families) for 15 minutes: first moves only.
- The ideas behind every rung, taught slowly: Module 3 lesson. Exam-level examples and the traps: Module 3 notes. All four papers: past papers, decoded.