📘 Suchenow Academy | TGPSC AEE Electrical
Subject 1: Electric Circuits & Fields → Chapter 1.3: Transient Response of DC & AC Networks — with live animated curves 🎮
| Covers | Initial conditions, RC & RL first-order transients, time constant τ, RLC second-order, damping (ζ), natural & damped frequency |
| Expected questions | 3–5 every exam — τ numericals and damping conditions dominate |
| Interactive labs | 4 live animations below |
| Study time | 10 hours | Revision: 2 hours |
1. Why Transients Matter (and Why Examiners Love Them)
Every switching event in a power system — energising a transformer, clearing a fault, switching a capacitor bank — is a transient. TGPSC/GENCO/TRANSCO exams test this chapter through three reliable question types: (1) inductor/capacitor behaviour at t=0⁺ and t=∞, (2) time-constant numericals, (3) RLC damping classification. Master these three and the chapter is yours.
2. The Golden Rules — Element Behaviour at Switching
2.1 Continuity Conditions (Cannot Change Instantly)
INDUCTOR: iL(0⁺) = iL(0⁻) — current through L cannot jump (needs infinite voltage: v = L di/dt)
CAPACITOR: vC(0⁺) = vC(0⁻) — voltage across C cannot jump (needs infinite current: i = C dv/dt)
Everything else (iC, vL, resistor quantities) CAN change instantly.
2.2 Equivalent Behaviour Table (★ most repeated question)
| Element | At t = 0⁺ (just after switching, zero initial energy) | At t = ∞ (DC steady state) |
|---|---|---|
| Inductor L | Open circuit (opposes sudden current) | Short circuit (di/dt = 0 → vL = 0) |
| Capacitor C | Short circuit (opposes sudden voltage) | Open circuit (dv/dt = 0 → iC = 0) |
| L with initial current I₀ | Current source I₀ | Short circuit |
| C with initial voltage V₀ | Voltage source V₀ | Open circuit |
3. First-Order Circuits — The Universal Solution
3.1 ✍️ Derivation — General First-Order Response
Any first-order circuit reduces to: τ·dx/dt + x = x(∞). Solving with integrating factor:
x(t) = x(∞) + [x(0⁺) − x(∞)]·e^(−t/τ)
This ONE formula solves every RC and RL transient — DC source or source-free. Identify three numbers: initial value x(0⁺), final value x(∞), time constant τ. Done. ∎
x(t) = x(∞) + [x(0⁺) − x(∞)]·e^(−t/τ)
This ONE formula solves every RC and RL transient — DC source or source-free. Identify three numbers: initial value x(0⁺), final value x(∞), time constant τ. Done. ∎
UNIVERSAL FORMULA: x(t) = x(∞) + [x(0⁺) − x(∞)] e^(−t/τ)
τ (RC circuit) = R_th · C (R seen by the capacitor with sources killed)
τ (RL circuit) = L / R_th (R seen by the inductor with sources killed)
3.2 RC Charging (from zero) — Classical Results
vC(t) = V(1 − e^(−t/RC)) iC(t) = (V/R) e^(−t/RC)
At t = τ: vC = 63.2% of V | i drops to 36.8% of initial
At t = 5τ: vC ≈ 99.3% → practically fully charged
Initial slope: if maintained, would reach V in exactly τ seconds (tangent construction)
3.3 RC Discharging & RL Growth/Decay
RC discharge: vC(t) = V₀ e^(−t/RC)
RL growth: iL(t) = (V/R)(1 − e^(−Rt/L)) | vL(t) = V e^(−Rt/L)
RL decay: iL(t) = I₀ e^(−Rt/L)
At t = τ: growth reaches 63.2% of final | decay falls to 36.8% of initial
🎮 Interactive Lab 1 — RC Charging/Discharging Curve (Live)
V = 10V. Drag R and C — watch τ and the curve respond. Then hit DISCHARGE:
🎮 Interactive Lab 2 — RL Current Growth & Inductor Voltage
V = 10V. Watch iL grow while vL collapses — they cross at t = τ·ln2 = 0.693τ:
4. Series RLC — Second-Order Transients
4.1 ✍️ Derivation — Characteristic Equation
KVL for source-free series RLC: L di/dt + Ri + (1/C)∫i dt = 0. Differentiate:
L d²i/dt² + R di/dt + i/C = 0 → s² + (R/L)s + 1/LC = 0
Roots: s₁,₂ = −α ± √(α² − ω₀²), where α = R/2L (damping factor, Np/s) and ω₀ = 1/√(LC) (undamped natural frequency). ∎
L d²i/dt² + R di/dt + i/C = 0 → s² + (R/L)s + 1/LC = 0
Roots: s₁,₂ = −α ± √(α² − ω₀²), where α = R/2L (damping factor, Np/s) and ω₀ = 1/√(LC) (undamped natural frequency). ∎
4.2 The Three Damping Cases (★ direct MCQ every exam)
| Condition | ζ = α/ω₀ | Case | Response shape | Roots |
|---|---|---|---|---|
| R > 2√(L/C) | ζ > 1 | Overdamped | Slow, no oscillation | Real, distinct, negative |
| R = 2√(L/C) | ζ = 1 | Critically damped | Fastest without overshoot | Real, equal: s = −α |
| R < 2√(L/C) | ζ < 1 | Underdamped | Decaying oscillation at ω_d | Complex conjugate |
| R = 0 | ζ = 0 | Undamped | Sustained oscillation at ω₀ | Pure imaginary ±jω₀ |
ζ (series RLC) = (R/2)·√(C/L) Critical resistance R_cr = 2√(L/C)
ω_d = ω₀√(1 − ζ²) (damped ringing frequency, underdamped only)
PARALLEL RLC: α = 1/(2RC) → ζ = (1/2R)·√(L/C) — R appears INVERTED vs series!
⚠️ Series vs Parallel trap: In series RLC, INCREASING R increases damping. In parallel RLC, increasing R DECREASES damping (α = 1/2RC). Examiners flip between the two to catch rote learners.
🎮 Interactive Lab 3 — RLC Damping Explorer ⭐
Series RLC: L = 1H, C = 0.25F → R_critical = 2√(L/C) = 4Ω. Drag R across the boundary and watch the response transform:
🎮 Interactive Lab 4 — The Universal τ Milestones
Every first-order rise passes the same checkpoints. Click a milestone to highlight it — memorise these five numbers:
Click any milestone button. Rule of thumb: 5τ = steady state for all practical purposes.
5. AC Transients — The Switching-Angle Result
When a sinusoidal source v = Vmsin(ωt + θ) is switched onto an RL circuit, the current has a steady-state term plus a decaying DC offset:
i(t) = (Vm/Z)·sin(ωt + θ − φ) − (Vm/Z)·sin(θ − φ)·e^(−Rt/L), φ = tan⁻¹(ωL/R)
NO transient (symmetrical switching): θ = φ → switch when source phase equals impedance angle
MAXIMUM transient (full DC offset): θ = φ ± 90° → basis of asymmetrical fault current &
transformer inrush in power systems
💡 This single result explains why circuit-breaker fault duty is rated for the asymmetrical (offset) current — a guaranteed interview and mains-descriptive topic for GENCO/TRANSCO.
6. Solved Problem Bank — 12 Exam-Grade Problems
P1 (Behaviour at t=0⁺ — classic). A series R-L circuit with switch closed at t=0, source 20V, R=4Ω, L=2H. Find i(0⁺), vL(0⁺), di/dt(0⁺).
i(0⁺)=i(0⁻)=0 (L continuity). vL(0⁺) = 20 − 0×4 = 20 V. di/dt(0⁺) = vL/L = 10 A/s.
i(0⁺)=i(0⁻)=0 (L continuity). vL(0⁺) = 20 − 0×4 = 20 V. di/dt(0⁺) = vL/L = 10 A/s.
P2 (τ identification with complex R network). A capacitor C=2μF sees, after killing the source: 3kΩ in series with (6kΩ ∥ 6kΩ). τ = ?
R_th = 3k + 3k = 6kΩ. τ = 6k × 2μ = 12 ms.
R_th = 3k + 3k = 6kΩ. τ = 6k × 2μ = 12 ms.
P3 (Universal formula). RC circuit: vC(0⁺)=2V, vC(∞)=10V, τ=4ms. Find vC(6ms).
vC = 10 + (2−10)e^(−6/4) = 10 − 8(0.2231) = 8.215 V.
vC = 10 + (2−10)e^(−6/4) = 10 − 8(0.2231) = 8.215 V.
P4 (Time to reach a value). In P3, when does vC reach 9V?
9 = 10 − 8e^(−t/4ms) → e^(−t/τ) = 1/8 → t = 4·ln8 = 8.32 ms.
9 = 10 − 8e^(−t/4ms) → e^(−t/τ) = 1/8 → t = 4·ln8 = 8.32 ms.
P5 (RL decay). Coil L=0.5H carrying 4A is suddenly shorted through total R=10Ω. Current after 0.1s?
τ = 0.05 s. i = 4e^(−0.1/0.05) = 4e^(−2) = 0.541 A.
τ = 0.05 s. i = 4e^(−0.1/0.05) = 4e^(−2) = 0.541 A.
P6 (Energy in transient). In P5, total energy dissipated in R from t=0 to ∞?
All stored energy dissipates: W = ½LI² = ½(0.5)(16) = 4 J — independent of R!
All stored energy dissipates: W = ½LI² = ½(0.5)(16) = 4 J — independent of R!
P7 (Capacitor with initial voltage — opposing). C charged to 5V is connected at t=0 to a 15V source through 1kΩ, C=100μF, polarity aiding source... vC(t)?
τ = 0.1 s, vC(∞) = 15, vC(0⁺) = 5. vC(t) = 15 − 10e^(−10t) V. At t = 0.1s: 15 − 10(0.368) = 11.32 V.
τ = 0.1 s, vC(∞) = 15, vC(0⁺) = 5. vC(t) = 15 − 10e^(−10t) V. At t = 0.1s: 15 − 10(0.368) = 11.32 V.
P8 (Damping classification — direct). Series RLC: R=6Ω, L=1H, C=0.04F. Nature of response?
R_cr = 2√(1/0.04) = 2×5 = 10Ω. R=6 < 10 → underdamped. ζ = 6/10 = 0.6; ω₀ = 5 rad/s; ω_d = 5√(1−0.36) = 4 rad/s.
R_cr = 2√(1/0.04) = 2×5 = 10Ω. R=6 < 10 → underdamped. ζ = 6/10 = 0.6; ω₀ = 5 rad/s; ω_d = 5√(1−0.36) = 4 rad/s.
P9 (Critical damping design). Series RLC with L=2H, C=8μF. Find R for critical damping.
R = 2√(L/C) = 2√(2/8×10⁻⁶) = 2×500 = 1000 Ω.
R = 2√(L/C) = 2√(2/8×10⁻⁶) = 2×500 = 1000 Ω.
P10 (Parallel RLC trap). Parallel RLC: R=25Ω, L=1H, C=100μF. Damping?
ζ = (1/2R)√(L/C) = (1/50)·√(10⁴) = 100/50 = 2 → ζ = 2 > 1 → overdamped. Note the inverted role of R!
ζ = (1/2R)√(L/C) = (1/50)·√(10⁴) = 100/50 = 2 → ζ = 2 > 1 → overdamped. Note the inverted role of R!
P11 (AC switching for zero transient). RL load: R=30Ω, ωL=40Ω, source v=311sin(ωt+θ). θ for transient-free switching?
φ = tan⁻¹(40/30) = 53.13°. Switch at θ = 53.13° → zero DC offset.
φ = tan⁻¹(40/30) = 53.13°. Switch at θ = 53.13° → zero DC offset.
P12 (Second switch — two-stage transient, TGPSC-style). RC circuit charges toward 12V with τ₁=2s. At t=2s a second resistor halves R (τ₂=1s). vC at t=3s?
Stage 1: vC(2) = 12(1−e^(−1)) = 7.585 V. Stage 2 (new initial value, same final 12V, τ=1s): vC(3) = 12 + (7.585−12)e^(−1) = 12 − 4.415(0.368) = 10.38 V. The universal formula handles multi-stage switching effortlessly.
Stage 1: vC(2) = 12(1−e^(−1)) = 7.585 V. Stage 2 (new initial value, same final 12V, τ=1s): vC(3) = 12 + (7.585−12)e^(−1) = 12 − 4.415(0.368) = 10.38 V. The universal formula handles multi-stage switching effortlessly.
7. PYQ Bank — Pattern Questions
- [TSGENCO 2015] At t=0⁺ an uncharged capacitor behaves as — short circuit; at t=∞ — open circuit.
- [TSSPDCL 2018] Time constant of RL circuit = L/R; of RC = RC.
- [TSTRANSCO 2018] At t = τ, a charging capacitor reaches — 63.2% of final voltage.
- [APPSC AEE 2016] Condition for critically damped series RLC — R = 2√(L/C).
- [GATE-style → TGPSC] Inductor current is a continuous function of time because — energy (½Li²) cannot change instantaneously.
- [ESE pattern] A series RLC with ζ = 0 gives — sustained oscillation at ω₀ = 1/√(LC).
- [TGPSC 2022 pattern] Practical full charge time of an RC circuit ≈ 5τ.
8. Examiner Traps ⚠️
| # | Trap | Correct |
|---|---|---|
| 1 | "iC(0⁺) = iC(0⁻) always" | Only vC is continuous; capacitor CURRENT can jump |
| 2 | "vL(0⁺) = vL(0⁻) always" | Only iL is continuous; inductor VOLTAGE can jump |
| 3 | τ = RC using total circuit R | τ uses R_th seen by C/L with sources killed |
| 4 | Increasing R always increases damping | True for series RLC; OPPOSITE for parallel RLC |
| 5 | Energy lost in R depends on R (RL decay) | Total dissipated = initial stored ½LI², independent of R |
| 6 | 63% confused with 37% | Rising quantity → 63.2% at τ; decaying quantity → 36.8% at τ |
| 7 | ω_d = ω₀ used for underdamped ringing | ω_d = ω₀√(1−ζ²) < ω₀ |
9. Memory Hooks 🧠
- "CIVIL" — in a Capacitor, I leads V; in an inductor (L), V leads I. Also encodes which quantity is "stubborn": C holds V, L holds I.
- "L = Lazy current, C = Calm voltage" — the continuous quantities.
- 63-37 rule: "63 up at τ, 37 down at τ" — rising hits 63.2%, falling hits 36.8%.
- "5τ = done" — 99.3%, treated as steady state.
- Damping ladder: ζ>1 slow crawl, ζ=1 perfect sprint, ζ<1 bouncy, ζ=0 forever swing.
10. One-Page Cheat Sheet 📄
CONTINUITY: iL(0⁺)=iL(0⁻) | vC(0⁺)=vC(0⁻) | everything else may jump
SWITCH TABLE (zero init): L: t=0⁺ OPEN → t=∞ SHORT | C: t=0⁺ SHORT → t=∞ OPEN
With initial energy: L → current source I₀ | C → voltage source V₀ (at t=0⁺)
UNIVERSAL 1st ORDER: x(t) = x(∞) + [x(0⁺) − x(∞)] e^(−t/τ)
τ_RC = R_th·C | τ_RL = L/R_th (kill sources, find R seen by the element)
Milestones: 1τ→63.2% | 2τ→86.5% | 3τ→95% | 4τ→98.2% | 5τ→99.3% ≈ done
Time to any level: t = τ·ln[(x0−x∞)/(x−x∞)]
RC charge: vC = V(1−e^(−t/RC)), i = (V/R)e^(−t/RC) | discharge: vC = V₀e^(−t/RC)
RL growth: iL = (V/R)(1−e^(−Rt/L)), vL = Ve^(−Rt/L) | decay: iL = I₀e^(−Rt/L)
RL decay energy in R = ½LI₀² (independent of R!)
SERIES RLC: s² + (R/L)s + 1/LC = 0 | α = R/2L | ω₀ = 1/√(LC) | ζ = (R/2)√(C/L)
R_cr = 2√(L/C): R>R_cr over | R=R_cr critical (fastest, no overshoot) | R<R_cr under
Underdamped ringing: ω_d = ω₀√(1−ζ²)
PARALLEL RLC: α = 1/(2RC) | ζ = (1/2R)√(L/C) ← R role INVERTED vs series
AC SWITCHING (RL): zero transient at θ = φ = tan⁻¹(ωL/R) | max DC offset at θ = φ±90°
→ asymmetrical fault current, breaker duty, transformer inrush
11. FAQ
Why exactly 63.2% at one time constant?
1 − e⁻¹ = 1 − 0.3679 = 0.632. The number is pure mathematics of the exponential — identical for every first-order system in nature, from circuits to thermal heating to radioactive decay.
Why is critical damping "fastest without overshoot"?
Overdamped responses contain a slow exponential (small |root|) that drags settling. Underdamped responses overshoot and ring. ζ=1 places both roots at −α — the quickest approach that never crosses the final value. This is why measuring instruments (PMMC) and relay dashpots are designed near critical damping.
Where do RLC transients appear in the power system job?
Capacitor-bank switching (restrike), transmission line energisation, TRV (transient recovery voltage) across breaker contacts, and ferroresonance — every one is an RLC transient. AE interview boards love asking "what happens when you switch a capacitor bank?"
📗 Next: Chapter 1.4 — Sinusoidal Steady-State Analysis & Resonance (phasors, series/parallel resonance, Q-factor, bandwidth — with an interactive resonance-curve lab).
🎯 Chapter 1.3 Quiz — 10 Questions
⏱ Exam Timer Drill — 5Q · 6 Min
6:00
D1. Time constant of RL circuit τ =
a) RCb) L/Rc) R/Ld) LCRL: τ=L/R. RC: τ=RC. At t=τ, response reaches 63.2% of final value.
D2. At t=5τ, transient is considered:
a) 50% completeb) 63% completec) Practically complete (99.3%)d) Just startedAt 5τ: e^(-5)=0.007 → 99.3% complete. Transient is considered over at 5τ.
D3. Inductor current at t=0⁺ (just after switching):
a) Jumps to maximumb) Equals value at t=0⁻c) Becomes zerod) UndefinedInductor: current cannot change instantaneously. iL(0⁺)=iL(0⁻). Initial condition.
D4. Overdamped RLC: roots of characteristic equation are:
a) Real and unequalb) Complex conjugatesc) Equal reald) Imaginary onlyOverdamped (ζ>1): two real unequal roots. Critically damped (ζ=1): equal real. Underdamped (ζ<1): complex conjugates.
D5. R=2Ω, L=4H DC circuit. τ=? At t=τ current is what % of final?
a) τ=8s, 50%b) τ=2s, 63.2%c) τ=0.5s, 36.8%d) τ=2s, 36.8%τ=L/R=4/2=2s. At t=τ: i=I_final×(1-e⁻¹)=63.2% of final.
📊 My Progress — Subject 1
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