Transients, taught from zero

1 · What a "transient" is — and why every question has the same shape

In Modules 1–2 every circuit was settled: batteries and resistors, nothing changing. Throw a switch in a purely resistive circuit and it simply jumps to its new settled state — instantly. Add a capacitor or an inductor and that's no longer true. These two elements store energy, and stored energy can't appear or vanish in zero time, so the circuit has to travel from its old settled state to its new one. The journey is called the transient, and the whole module is about describing it as a formula in time.

2 · The two elements with memory

A capacitor stores energy w = ½Cv² in its electric field; an inductor stores w = ½Li² in its magnetic field. Power is the rate of energy change. If a capacitor's voltage jumped from 0 V to 8 V in literally zero seconds, its energy would jump too — and changing energy by a finite amount in zero time takes infinite power. No real source supplies infinite power. So:

Same conclusion from the element laws you met in Module 1: i = C dv/dt — a jump in v would mean infinite slope, so infinite current. And v = L di/dt — a jump in inductor current would need infinite voltage. We write the rule with two time-stamps: t = 0⁻ is the instant just before the switch moves, t = 0⁺ the instant just after:

Only those two. Everything else — the current through a capacitor, the voltage across an inductor, every resistor's voltage and current — is free to jump. That is the single most important sentence of this module, and we'll see it happen in a moment.

The "settled for a long time" rule

"The switch has been closed for a long time" is the exam's way of saying nothing is changing any more (with DC sources). If nothing changes, then dv/dt = 0, so a capacitor carries i = C · 0 = 0 — it behaves like a gap (open circuit). And di/dt = 0, so an inductor has v = L · 0 = 0 across it — it behaves like a plain wire (short circuit). That turns every "before" and "long after" picture into a Module 2 resistor problem.

Let's watch all of this on one small circuit. Example A: a 12 V source feeds, through a switch and a 2 Ω resistor, a node where a 4 Ω resistor and a 0.5 F capacitor both hang down to the bottom rail. The switch has been closed for a long time; at t = 0 it opens.

Before (t = 0⁻). Settled, so the capacitor is a gap. What's left is 12 V across 2 Ω + 4 Ω in series: current 12/6 = 2 A, and the capacitor — sitting in parallel with the 4 Ω — sees the 4 Ω's voltage: v(0⁻) = 2 × 4 = 8 V.

Just after (t = 0⁺). The source is disconnected. The capacitor's voltage can't jump, so it's still 8 V — for this one instant it behaves like an 8 V battery. That battery drives the 4 Ω: 8/4 = 2 A down through the resistor, and that current has to come out of the capacitor.

Long after (t → ∞). With no source, the capacitor eventually drains completely through the 4 Ω: v(∞) = 0. So v starts at 8 V and ends at 0 V. What remains is how fast — the next section.

3 · Why the journey is an exponential

After the switch opens in Example A, the capacitor and the 4 Ω form one loop. The capacitor current is C dv/dt, the resistor current is v/R, and KCL at the top node says they add to zero:

Read that equation in words: the voltage falls at a rate proportional to how much voltage is left. Lots of charge → big current → fast drain; little charge → trickle. A quantity whose rate of change is proportional to itself is exactly what an exponential does:

The Greek letter τ (tau) is the time constant, measured in seconds (ohms × farads = seconds). For Example A: τ = 4 × 0.5 = 2 s, so v(t) = 8e^−t/2 V. Every τ seconds, what's left gets multiplied by e⁻¹ ≈ 0.368:

Two landmarks worth memorising: after one τ, 63% of the way; after five τ, effectively finished. MCQs love both.

With a source still connected: the general formula

If the circuit after the switch still contains a DC source, the voltage doesn't head to zero — it heads to its new settled value v(∞). The same argument (rate proportional to distance still to go) gives the one formula this whole module runs on:

Say it as: final + (start − final) × decay. At t = 0 the decay factor is 1 and you get the start; as t → ∞ it's 0 and you get the final. Here x can be any voltage or current in a first-order circuit — capacitor voltage, inductor current, even a resistor's current — as long as you use that quantity's own start and final values.

The inductor version

Replace the capacitor with an inductor L discharging through R. KVL around the loop: L di/dt + Ri = 0, so di/dt = −i/(L/R). Same exponential, different time constant:

Why is R on top for one and underneath for the other? A capacitor drains through R, so a bigger resistor chokes the drain: slower. An inductor's stored current is burned off by R (power i²R), so a bigger resistor burns it faster. Play with it:

Things to try: set the capacitor to 4 Ω, 0.5 F, start 8, final 0 — that's Example A. Then double R in capacitor mode and in inductor mode, and watch the curve stretch in one and squash in the other. Set start below final: the same shape, climbing instead of falling.

4 · The six-step routine

Now the whole method, in the order you'll write it in the answer script. This is the one thing on this page to know cold:

Steps 1, 3 and 4 are each a small Module 2 problem. That's why this module feels hard at first and then suddenly easy: the new idea is only steps 2 and 5.

Example B. A 12 V source drives a 2 Ω resistor and a 1 H inductor in series into node b; from b, a 4 Ω resistor goes to the bottom rail, and a second 4 Ω can be connected alongside it by a switch that has been open for a long time and closes at t = 0. Find the inductor current i(t).

1. 0⁻: switch open, inductor a wire. One loop: 12/(2 + 4) = 2 A. So i(0⁻) = 2 A.
2. Continuity: i(0⁺) = 2 A.
3. ∞: switch closed, inductor a wire. The two 4 Ω are in parallel: 4 ∥ 4 = 2 Ω. Current 12/(2 + 2) = 3 A. So i(∞) = 3 A.
4. τ: remove the inductor, short the 12 V source (killed), look in from the inductor's two ends: the 2 Ω on one side, and 4 ∥ 4 = 2 Ω on the other, in series through the dead source: R_th = 4 Ω. τ = L/R_th = 1/4 = 0.25 s.
5. Write it: i(t) = 3 + (2 − 3)e^−t/0.25 = 3 − e^−4t A, for t ≥ 0.
6. Anything else: the inductor voltage is L di/dt = 1 × 4e^−4t = 4e^−4t V. It was 0 before the switch and is 4 V just after — a jump, which is allowed. Sanity check by KVL at 0⁺: the 2 Ω drops 2 × 2 = 4 V, the parallel pair drops 2 × 2 = 4 V, leaving 12 − 4 − 4 = 4 V for the inductor ✓.

Step 4 when there's a dependent source

Module 2's warning applies here unchanged: never kill a dependent source. And when the network seen by the capacitor has no independent source left, you can't use V_oc/I_sc (both are zero). The fix is the test-source method: in place of the capacitor, push in a test current of 1 A, find the voltage v it creates, and R_th = v/1.

KCL at the top node: in = 1 + 0.5i_x, out = i_x. So i_x = 2 A, the node voltage is v = 4 × 2 = 8 V, and R_th = 8/1 = 8 Ω. With a 0.25 F capacitor, τ = 8 × 0.25 = 2 s. Had you killed the diamond, you'd have seen only the 4 Ω and written τ = 1 s — half the right answer, and every number after it wrong.

5 · Two twists the papers like

A second switch: restart the clock

Sometimes a second switch moves later, at t = t₀. Nothing new: the first formula tells you the value at t₀; that value is the start of chapter two (continuity again), and the circuit after the second switch has its own final value and its own τ′. The only change is that the decay starts counting at t₀, not at 0:

Take Example B and suppose the switch re-opens at t = 0.5 s. At that instant i(0.5) = 3 − e⁻² = 3 − 0.135 = 2.865 A. With the switch open again, the final value is back to 2 A and the resistance seen by the inductor is 2 + 4 = 6 Ω, so τ′ = 1/6 s:

"When does it reach …?" — solve with a logarithm

Set the formula equal to the target and isolate the exponential, then take ln of both sides. For Example A, when does the capacitor fall to 2 V?

The general shape, if you prefer one formula: t = τ ln[(start − final)/(target − final)]. The fraction inside is always positive when the target really lies between start and final — if it isn't, the curve never gets there, which is itself a correct answer.

6 · Second order: when a circuit has two memories

Put a capacitor and an inductor in the same circuit and something new can happen: energy sloshes between them — from the capacitor's electric field into the inductor's magnetic field and back — while the resistor bleeds a little off on every pass. The circuit can now overshoot and ring, which no single exponential can describe. Two storage elements give a second-order differential equation; its solutions come in three flavours.

Writing the equation

Series RLC (no source, natural response). KVL around the loop: L di/dt + Ri + v_C = 0, and the loop current is the capacitor current, i = C dv_C/dt. Substitute and divide by LC:

Parallel RLC. KCL at the top node: v/R + i_L + C dv/dt = 0, with v = L di_L/dt. Differentiate once to get rid of i_L and divide by C:

Both have the same skeleton, x″ + 2α x′ + ω₀² x = 0, with two named numbers:

Note the opposite roles of R again: in series, a big R means lots of damping; in parallel, a big R is nearly an open circuit and hardly damps at all.

From differential equation to quadratic

For first order the answer was e^−t/τ, so try an exponential again: x = e^st, with s unknown. Its derivatives are s e^st and s²e^st — the same function times a number — so every term shares the factor e^st, which is never zero and can be cancelled. The calculus problem collapses into algebra, the characteristic equation:

Everything now depends on the sign of what's under the square root — whether damping α beats the natural frequency ω₀:

where ω_d = √(ω₀² − α²) is the damped frequency — how fast it actually rings. (Some papers also quote the damping ratio ζ = α/ω₀: overdamped when ζ > 1, critical at 1, underdamped below 1. Same test, one number.)

Slide R slowly up from 0.2 Ω. The ringing dies faster and faster until, at exactly 4 Ω (where α = R/2L = 2 = ω₀), it stops overshooting altogether. Past that point, more resistance makes the response slower, not faster — which surprises everyone the first time.

7 · Pinning the two constants

Every form in that table has two unknown constants, so you need two facts at t = 0⁺: the value x(0⁺) and the slope x′(0⁺). The value comes from continuity, exactly as before. The slope is the new move, and it always comes from the same two element laws read backwards:

So: draw the 0⁺ circuit (capacitor as a battery at its old voltage, inductor as a current source at its old current), find the capacitor's current or the inductor's voltage with KCL/KVL, and divide.

Example C (natural, underdamped). Series RLC with R = 2 Ω, L = 1 H, C = 0.25 F. The capacitor holds 10 V and the loop current is zero when the loop is closed at t = 0. Find v_C(t).

1. Numbers: α = R/2L = 1, ω₀ = 1/√(0.25) = 2. Since 1 < 2: underdamped, ω_d = √(4 − 1) = √3 ≈ 1.732 rad/s.
2. Form: v_C = e^−t(B₁ cos √3t + B₂ sin √3t).
3. Value: v_C(0⁺) = 10. At t = 0, cos = 1 and sin = 0, so B₁ = 10.
4. Slope: the capacitor current is the loop current, i(0⁺) = 0 (the inductor protects it), so v_C′(0⁺) = 0/C = 0. Differentiating the form and setting t = 0 gives v_C′(0) = −αB₁ + ω_dB₂, so −10 + √3 B₂ = 0 → B₂ = 10/√3 ≈ 5.77.
5. Answer: v_C(t) = e^−t(10 cos √3t + 5.77 sin √3t) V, t ≥ 0.

Want the inductor's slope instead? At 0⁺, KVL gives v_L = −v_C − Ri = −10 − 0 = −10 V, so i′(0⁺) = −10/1 = −10 A/s: the current starts growing in the negative direction — the capacitor beginning to discharge. The exam's "find di/dt(0⁺)" parts are exactly this one line.

With a DC source: forced + natural

If a DC source is connected after the switch, the response settles at a final value instead of zero. The complete response is simply

where x(∞) (the forced response) comes from the settled circuit, C open and L short, exactly as in first order. Then fit the two constants to the complete x(0⁺) and x′(0⁺).

Example D. The overdamped circuit from the check (5 Ω, 1 H, 0.25 F), capacitor uncharged, no current, and a 12 V source switched into the loop at t = 0. Final value: the capacitor ends up with the full source voltage, v_C(∞) = 12 V. So v_C = 12 + A₁e^−t + A₂e^−4t. Value: 12 + A₁ + A₂ = 0. Slope: i(0⁺) = 0 so −A₁ − 4A₂ = 0. Solving: A₂ = 4, A₁ = −16:

8 · The derivation they ask for

One past paper asked, for 6 marks: derive the capacitor voltage and inductor current for an underdamped series RLC circuit, natural response. You've already done every step; here they are in the order to write them. Let the capacitor start at V₀ and the loop current at I₀.

1. KVL (no source): L di/dt + Ri + v_C = 0, with i = C dv_C/dt.
2. ODE: v_C″ + (R/L)v_C′ + v_C/(LC) = 0, i.e. v_C″ + 2αv_C′ + ω₀²v_C = 0 with α = R/2L, ω₀ = 1/√(LC).
3. Characteristic equation (try e^st): s² + 2αs + ω₀² = 0, so s = −α ± √(α² − ω₀²).
4. Underdamped means α < ω₀: the root of a negative number, s = −α ± jω_d, ω_d = √(ω₀² − α²).
5. Real form: v_C = e^−αt(K₁e^jω_dt + K₂e^−jω_dt) = e^−αt(B₁ cos ω_dt + B₂ sin ω_dt).
   Unpack this step

   Euler's formula: e^jθ = cos θ + j sin θ (and e^−jθ = cos θ − j sin θ). Substitute both and collect: the cos terms carry K₁ + K₂ (call it B₁), the sin terms j(K₁ − K₂) (call it B₂). For a real voltage both come out real. You'll meet j properly in Module 4 (AC).

6. Constants: v_C(0) = V₀ gives B₁ = V₀. v_C′(0) = I₀/C gives −αB₁ + ω_dB₂ = I₀/C, so B₂ = (I₀/C + αV₀)/ω_d.
7. Current (it's the inductor's current too, since it's one loop): i = C dv_C/dt. In the usual case I₀ = 0 this tidies to:

Sanity-check on Example C (V₀ = 10, α = 1, ω_d = √3, L = 1): α/ω_d = 1/√3, so B₂ = 10/√3 ✓; and i = −(10/√3)e^−t sin √3t, whose slope at 0 is −10 A/s — the v_L/L we found by KVL ✓. Two independent routes agreeing is the best check you have in an exam.

9 · You're ready — what to do next

That's lectures 9–12: one first-order routine, one quadratic, and a way to get two starting facts. Where to go from here:

• Work the notes page — exam-level worked examples in the exact shapes past papers use, the trap list, and what to practise.
• Drill the routine until step 1 is automatic. In every past paper the marks are front-loaded: the 0⁻ circuit and the continuity statement earn marks even if the algebra later goes wrong. Write them first, every time.
• Textbook: problems from §3.2–3.5 — aim for 6–8 first-order (at least two with a switch moving between positions, one with a dependent source) and 4–6 second-order (one of each damping case).