📘 Suchenow Academy | TGPSC AEE Electrical
Subject 1 → Chapter 1.4: Sinusoidal Steady-State Analysis & Resonance — now with self-test quiz 🎯 & Telugu hooks 🌾
| Covers | Phasors, impedance, series/parallel resonance, Q-factor, bandwidth, selectivity, basic filters |
| Expected questions | 4–6 every exam — resonance is a TGPSC certainty |
| New in this chapter | 2 animations + 10-question interactive quiz + hidden "Try first" solutions + Telugu concept lines |
1. Phasors — The Language of AC
A sinusoid v(t) = Vmsin(ωt + φ) is fully described by two numbers: amplitude and phase. A phasor is that pair drawn as a rotating vector — freeze the rotation and AC circuit analysis becomes complex-number algebra instead of differential equations.
🌾 తెలుగులో: ఫేజర్ అంటే తిరుగుతున్న బాణం (rotating arrow). దాని పొడవు = వోల్టేజ్ పరిమాణం, దాని కోణం = ఫేజ్. సైన్ వేవ్ అంటే ఈ బాణం నీడ (projection) మాత్రమే!
Element impedances: Z_R = R∠0° | Z_L = jωL = ωL∠90° | Z_C = 1/jωC = (1/ωC)∠−90°
CIVIL: in C, I leads V by 90° | in L (L=coil), V leads I by 90°
RMS: V_rms = Vm/√2 (sine) | Average (full sine over half cycle) = 2Vm/π
Form factor (sine) = 1.11 | Peak factor = √2 = 1.414
🎮 Interactive Lab 1 — The Rotating Phasor → Sine Wave
Watch the phasor rotate at ω and trace the sine wave. Change phase φ and see the wave shift:
v(t) = Vm·sin(ωt + 0°) — the projection of the rotating arrow IS the sine wave.
2. Series RLC Impedance & the Road to Resonance
Z = R + j(X_L − X_C) = R + j(ωL − 1/ωC)
|Z| = √(R² + (X_L−X_C)²) | φ = tan⁻¹((X_L−X_C)/R)
X_L > X_C → inductive (I lags) | X_L < X_C → capacitive (I leads) | X_L = X_C → RESONANCE
3. Series Resonance ⭐ (The Heart of the Chapter)
3.1 ✍️ Derivation — Resonant Frequency
At resonance, Im(Z) = 0: ωL = 1/ωC → ω₀² = 1/LC →
ω₀ = 1/√(LC) | f₀ = 1/(2π√(LC)) ∎
Consequences at ω₀: Z = R (minimum) → I = V/R (maximum) → circuit is purely resistive → unity power factor.
ω₀ = 1/√(LC) | f₀ = 1/(2π√(LC)) ∎
Consequences at ω₀: Z = R (minimum) → I = V/R (maximum) → circuit is purely resistive → unity power factor.
🌾 తెలుగులో: రెసొనెన్స్ అంటే — L మరియు C ఒకదాన్ని ఒకటి రద్దు (cancel) చేసుకుంటాయి. మిగిలేది R మాత్రమే. అందుకే కరెంట్ గరిష్టం (maximum), పవర్ ఫ్యాక్టర్ = 1.
3.2 Q-Factor, Bandwidth & Voltage Magnification
Q = ω₀L/R = 1/(ω₀CR) = (1/R)√(L/C) ("quality" = energy stored / energy lost per cycle × 2π)
Bandwidth BW = f₀/Q = R/(2πL) (between half-power points f₁, f₂ where I = Imax/√2)
f₀ = √(f₁·f₂) (geometric mean!) | Selectivity ∝ Q
VOLTAGE MAGNIFICATION at resonance: V_L = V_C = Q × V_source ← can be dangerously high!
⚠️ Trap: At series resonance the source sees only R, yet the capacitor and inductor each carry Q times the supply voltage (equal and opposite). With Q = 50 and 230V supply, V_C = 11.5 kV — this is why series resonance is called voltage resonance and is a real insulation hazard in power systems.
🎮 Interactive Lab 2 — Resonance Curve Explorer ⭐
Series RLC: L = 100 mH, C = 10 μF → f₀ = 159.2 Hz fixed. Drag R — watch Q sharpen the peak and squeeze the bandwidth:
4. Parallel Resonance (Anti-Resonance)
| Property | Series Resonance | Parallel Resonance |
|---|---|---|
| Impedance at f₀ | Minimum (= R) | Maximum (= L/CR, "dynamic resistance") |
| Current at f₀ | Maximum | Minimum (line current) |
| Known as | Acceptor circuit / voltage resonance | Rejector circuit / current resonance |
| Magnification | Voltage: V_L = QV | Current: I_L = Q·I_line |
| Resonant frequency (coil R) | f₀ = 1/2π√(LC) | f₀ = (1/2π)√(1/LC − R²/L²) |
Parallel (tank) circuit dynamic resistance: Z_max = L/(CR)
Parallel f₀ with lossy coil: f₀ = (1/2π)·√(1/LC − R²/L²) — resonance impossible if R > √(L/C)!
🌾 తెలుగులో: సిరీస్ రెసొనెన్స్ = కరెంట్కి తలుపు తెరుస్తుంది (acceptor). పేరలల్ రెసొనెన్స్ = కరెంట్ని అడ్డుకుంటుంది (rejector). ఇది గుర్తుంచుకుంటే సగం మార్కులు మీవే!
5. Basic Filter Concepts (Syllabus Line: "basic filter concepts")
| Filter | Passes | Simplest RC/RL realisation | Cut-off |
|---|---|---|---|
| Low-pass | f < f_c | Series R, output across C | f_c = 1/(2πRC) |
| High-pass | f > f_c | Series C, output across R | f_c = 1/(2πRC) |
| Band-pass | f₁ < f < f₂ | Series RLC, output across R | centred at f₀, width f₀/Q |
| Band-stop (notch) | all except f₁–f₂ | Output across series LC arm | notch at f₀ |
At cut-off: |gain| = 1/√2 = 0.707 = −3 dB, output power = half of maximum. A series resonant circuit IS a band-pass filter — the resonance curve in Lab 2 is literally its frequency response.
6. Solved Problem Bank — 12 Problems (Try First! 🎯)
P1 (f₀ direct). L = 100 mH, C = 10 μF. Find f₀.
f₀ = 1/(2π√(0.1×10⁻⁵)) = 1/(2π×10⁻³) = 159.15 Hz.
P2 (Q and BW). In P1 with R = 10 Ω, find Q and bandwidth.
Q = (1/R)√(L/C) = (1/10)√(0.1/10⁻⁵) = (1/10)(100) = 10. BW = f₀/Q = 159.15/10 = 15.9 Hz.
P3 (Voltage magnification — hazard calc). A 230 V series RLC has Q = 40 at resonance. Find V_C.
V_C = Q×V = 40×230 = 9200 V. The capacitor insulation must withstand 9.2 kV from a 230 V supply — the classic danger of series resonance.
P4 (Half-power frequencies). f₀ = 500 kHz, BW = 20 kHz. Find f₁ and f₂ (Q large).
For high Q: f₁ ≈ f₀ − BW/2 = 490 kHz, f₂ ≈ f₀ + BW/2 = 510 kHz. Exact relation: f₀ = √(f₁f₂) = √(490×510) = 499.9 kHz ✓ (geometric mean).
P5 (Design for given BW). Design R for a series circuit with L = 50 mH to have BW = 400 Hz.
BW = R/(2πL) → R = 2πL×BW = 2π(0.05)(400) = 125.7 Ω. Note: BW is independent of C!
P6 (Impedance phase). R = 30 Ω, X_L = 70 Ω, X_C = 30 Ω at some f. Find |Z| and φ; does I lead or lag?
X = 70−30 = 40 Ω (net inductive). |Z| = √(900+1600) = 50 Ω, φ = tan⁻¹(40/30) = 53.13°. Inductive → current LAGS by 53.13°.
P7 (Frequency below/above f₀ nature). A series RLC operates at f < f₀. Circuit nature?
Below f₀: X_C = 1/ωC dominates (grows as ω falls) → capacitive, current leads. Above f₀ → inductive, current lags. At f₀ → resistive, unity pf.
P8 (Parallel dynamic resistance). Tank: L = 0.2 H, C = 20 μF, coil R = 10 Ω. Find Z at resonance.
Z_dyn = L/(CR) = 0.2/(20×10⁻⁶ × 10) = 1000 Ω — purely resistive maximum.
P9 (Parallel f₀ with lossy coil). Same tank as P8 — find f₀.
f₀ = (1/2π)√(1/LC − R²/L²) = (1/2π)√(1/(4×10⁻⁶) − 100/0.04) = (1/2π)√(250000−2500) = (1/2π)(497.5) = 79.2 Hz. (Ideal-coil value would be 79.6 Hz — coil loss lowers f₀ slightly.)
P10 (Resonance impossible condition). Tank with L = 1 mH, C = 1 μF. Max coil R for resonance to exist?
Condition: R < √(L/C) = √(10⁻³/10⁻⁶) = √1000 = 31.6 Ω. If R ≥ 31.6 Ω, the tank never resonates (f₀ becomes imaginary).
P11 (RC filter cut-off). Low-pass RC: R = 1.59 kΩ, C = 0.1 μF. Find f_c and gain at f = f_c.
f_c = 1/(2πRC) = 1/(2π×1.59k×0.1μ) ≈ 1 kHz. Gain at f_c = 1/√2 = 0.707 (−3 dB), output power halves.
P12 (Combined — TGPSC full-length pattern). A series RLC across 100 V has R = 5 Ω, and at resonance the capacitor voltage is 400 V. If f₀ = 50 Hz, find L and C.
Q = V_C/V = 400/100 = 4. Q = ω₀L/R → L = QR/ω₀ = (4×5)/(2π×50) = 63.7 mH. ω₀² = 1/LC → C = 1/(ω₀²L) = 1/((314.16)²×0.0637) = 159 μF. Verify: Q = (1/R)√(L/C) = (1/5)√(400.6) ≈ 4 ✓
7. PYQ Bank — Pattern Questions
- [TSGENCO 2015] At series resonance, power factor = unity; impedance = R (minimum).
- [TSSPDCL 2018] Q of series RLC = (1/R)√(L/C) = ω₀L/R.
- [TSTRANSCO 2018] Resonant frequency f₀ = √(f₁f₂) — geometric mean of half-power frequencies.
- [APPSC AEE 2016] Parallel resonant circuit is called — rejector circuit; impedance at resonance = L/CR (maximum).
- [GATE-style] Below resonance a series RLC behaves as — capacitive circuit.
- [ESE pattern] Increasing R in series RLC — f₀ unchanged, Q decreases, BW increases.
- [TGPSC 2022 pattern] At half-power frequencies, current = 0.707 I_max and power = half of maximum.
8. Examiner Traps ⚠️
| # | Trap | Correct |
|---|---|---|
| 1 | R changes the resonant frequency | In SERIES RLC, f₀ depends only on L, C. R changes only Q and BW |
| 2 | f₀ = arithmetic mean of f₁, f₂ | Geometric mean: f₀ = √(f₁f₂) |
| 3 | V_L or V_C can't exceed supply | Each equals Q×V at resonance — often far above supply! |
| 4 | Parallel resonance current is maximum | LINE current is MINIMUM; circulating tank current is Q×I_line |
| 5 | Lossy-coil tank always resonates | Only if R < √(L/C) |
| 6 | Bandwidth depends on C | BW = R/2πL — independent of C (series) |
| 7 | −3dB means output = 1/3 | −3 dB = 0.707 voltage = HALF power |
9. Memory Hooks 🧠
- "Series Sips minimum Z, Parallel Piles maximum Z" at resonance.
- "Acceptor accepts current (series), Rejector rejects it (parallel)".
- "Q kicks the voltage up" — V_C = QV (series), "Q circulates the current" — I_tank = QI (parallel).
- "R rules the width, LC rule the spot" — R sets BW; L, C set f₀.
- √(f₁f₂): "resonance sits at the geometric middle."
🌾 గుర్తుపెట్టుకోండి: "R పెరిగితే Q తగ్గుతుంది, BW పెరుగుతుంది — కానీ f₀ మారదు!" — ఇది ప్రతి పరీక్షలో వచ్చే ప్రశ్న.
10. One-Page Cheat Sheet 📄
PHASORS: Z_R = R∠0 | Z_L = ωL∠90° | Z_C = (1/ωC)∠−90° | CIVIL
Sine: RMS = Vm/√2 | Avg = 2Vm/π | FF = 1.11 | PF(peak factor) = 1.414
SERIES RLC: Z = R + j(ωL − 1/ωC)
f₀ = 1/(2π√LC) (independent of R!)
At f₀: Z = R min | I = V/R max | pf = 1 | V_L = V_C = Q·V (magnification!)
Q = ω₀L/R = 1/ω₀CR = (1/R)√(L/C) | BW = f₀/Q = R/2πL (independent of C)
f₀ = √(f₁·f₂) | half-power: I = 0.707 Imax, P = Pmax/2, −3 dB
f < f₀ capacitive (I leads) | f > f₀ inductive (I lags)
PARALLEL (tank, coil R): Z_dyn = L/CR (MAX) | line I min | I_tank = Q·I_line
f₀ = (1/2π)√(1/LC − R²/L²) | resonance exists only if R < √(L/C)
Acceptor = series | Rejector = parallel
FILTERS: LP/HP cut-off f_c = 1/2πRC | at f_c: 0.707 gain, −3 dB, half power
Series RLC = band-pass centred f₀, width f₀/Q | notch = band-stop
🎯 Chapter Quiz — 10 Questions (Instant Feedback)
Attempt honestly — explanations appear after each answer. Score shows at the end.
11. FAQ
Why is series resonance dangerous in power systems?
Voltage magnification: capacitor and inductor voltages reach Q times the supply. Cable capacitance resonating with transformer inductance can create kV-level overvoltages from a normal supply — a real cause of insulation failure and the reason detuning reactors are used with capacitor banks.
Why does R not affect the series resonant frequency?
Resonance is the condition Im(Z) = 0, i.e., ωL = 1/ωC. R sits in the real part only. It controls how sharp the peak is (Q, BW) but never where it sits.
Where is parallel resonance used deliberately?
Tank circuits in oscillators, induction-furnace power supplies, and harmonic filters at HV substations — a shunt filter tuned to the 5th/7th harmonic presents low series impedance to that harmonic and traps it. Directly relevant to TSTRANSCO substation practice.
📗 Next: Chapter 1.5 — Two-Port Networks (Z, Y, h, ABCD parameters with interconnection rules and an interactive parameter converter).
⏱ Exam Timer Drill — 5Q · 6 Min
6:00
D1. At resonance, series RLC impedance is:
a) R (purely resistive)b) Zeroc) Maximumd) InductiveAt resonance: XL=XC → Z=R (minimum, purely resistive). Current is maximum.
D2. Q-factor of series RLC = ω₀L/R. If R doubles, Q:
a) Doublesb) Halvesc) Unchangedd) QuadruplesQ=ω₀L/R. If R doubles → Q halves. Higher Q = sharper resonance = better selectivity.
D3. Bandwidth BW = f₂−f₁ =
a) f₀×Qb) f₀/Qc) Q/f₀d) f₀²/QBW=f₀/Q=R/(2πL). Narrower BW = higher Q = more selective filter.
D4. Parallel resonance: impedance at resonance is:
a) Minimumb) Maximum (=L/CR)c) Zerod) Equal to RParallel resonance: Z=MAXIMUM=L/CR (tank circuit). Current from source is minimum.
D5. L=10mH, C=10μF. ω₀=?
a) 1000 rad/sb) 3162 rad/sc) 100 rad/sd) 316 rad/sω₀=1/√(LC)=1/√(10×10⁻³×10×10⁻⁶)=1/√(10⁻⁷)=3162 rad/s.
📊 My Progress — Subject 1
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