📘 Suchenow Academy | TGPSC AEE Electrical
Subject 1: Electric Circuits & Fields → Chapter 1.2: Network Theorems — India's first INTERACTIVE AEE notes 🎮
| Theorems covered | Superposition, Thevenin, Norton, Maximum Power Transfer, Reciprocity, Millman, Substitution, Tellegen |
| Expected questions | 4–6 direct every exam (highest-yield chapter) |
| Interactive labs | 5 animations embedded below — play with them! |
| Study time | 10–12 hours | Revision: 2 hours |
1. The Theorem Toolbox — When to Use What
| You need… | Use | Condition |
|---|---|---|
| Response due to many sources | Superposition | Linear network only |
| Simplify network seen from 2 terminals (V-form) | Thevenin | Linear network |
| Same, current form | Norton | Linear network |
| Best load for max power | MPT | RL = Rth (DC), ZL = Zth* (AC) |
| Swap excitation & response | Reciprocity | Linear, bilateral, single source, NO dependent sources |
| Combine parallel V-source branches | Millman | Parallel branches with sources |
2. Superposition Theorem
Statement: In a linear network with multiple independent sources, the response (voltage or current) in any element equals the algebraic sum of responses caused by each independent source acting alone, with all other independent sources replaced by their internal resistances (V-source → short circuit; I-source → open circuit).
2.1 ✍️ Proof Sketch
Linearity means the network equations [A][x] = [b₁] + [b₂] (source vectors add) have the solution [x] = [A]⁻¹[b₁] + [A]⁻¹[b₂] = x₁ + x₂ — the response vectors add too. This holds only because the system matrix [A] is constant (linear elements). ∎
⚠️ Traps: (1) Power does NOT superpose — P ∝ I², a nonlinear operation. P_total ≠ P₁ + P₂ in general. (2) Dependent sources are NEVER killed — they stay active in every step. (3) Not applicable to nonlinear networks (diode circuits).
🎮 Interactive Lab 1 — Superposition in Action
Circuit: 12V source (left), 6A source (right), R1=2Ω top-left, R2=4Ω top-right, R3=4Ω middle. Find current I through R3 (downward). Toggle sources ON/OFF:
2.2 Worked Solution of the Lab Circuit
Step 1 — 12V alone (6A open-circuited): R2 branch dangles → circuit is 12V, R1=2Ω, R3=4Ω in series. I₃' = 12/(2+4) = 2 A ↓
Step 2 — 6A alone (12V short-circuited): R1 (2Ω) becomes parallel with R3 (4Ω) as seen by the 6A source through R2. Current divider into R3: the 6A flows through R2 and splits between R3 and R1: I₃'' = 6 × 2/(2+4) = 2 A ↓
Total: I₃ = 2 + 2 = 4 A ↓ ✓ — exactly what the animation shows.
Step 2 — 6A alone (12V short-circuited): R1 (2Ω) becomes parallel with R3 (4Ω) as seen by the 6A source through R2. Current divider into R3: the 6A flows through R2 and splits between R3 and R1: I₃'' = 6 × 2/(2+4) = 2 A ↓
Total: I₃ = 2 + 2 = 4 A ↓ ✓ — exactly what the animation shows.
3. Thevenin's Theorem
Statement: Any linear two-terminal network of sources and resistances can be replaced by a single voltage source Vth in series with a single resistance Rth, where Vth = open-circuit voltage at the terminals, and Rth = equivalent resistance seen from the terminals with all independent sources killed.
3.1 ✍️ Proof (Substitution + Superposition)
Connect a current source I at terminals A–B of network N. By superposition, terminal voltage V = (response to internal sources with I=0) + (response to I with internal sources killed) = Voc − I·Req. This is exactly the V–I equation of a source Vth=Voc in series with Rth=Req. Since the terminal V–I relations are identical for every load, the two networks are externally indistinguishable. ∎
3.2 Finding Rth — Three Methods
| Method | Procedure | Use when |
|---|---|---|
| 1. Kill & look | Kill independent sources, find R at terminals | Only independent sources |
| 2. Voc/Isc | Rth = Voc / Isc | Any network (universal) |
| 3. Test source | Kill independent srcs, apply 1V/1A test source, Rth = Vtest/Itest | Dependent sources present (mandatory) |
⚠️ Trap: With dependent sources, Method 1 fails — you MUST use Method 2 or 3. Dependent sources contribute to Rth and can even make it negative!
🎮 Interactive Lab 2 — Thevenin's Theorem Step-by-Step
Circuit: 20V source, R1=4Ω series, R2=6Ω shunt, terminals A–B. Click through the 4 steps:
3.3 Lab Circuit Hand Calculation
Voc: No load current → 20V divides over R1+R2: Voc = 20 × 6/(4+6) = 12 V.
Rth: Kill 20V (short) → R1 ∥ R2 seen from A–B = (4×6)/10 = 2.4 Ω.
Result: Thevenin equivalent = 12 V in series with 2.4 Ω ✓
Rth: Kill 20V (short) → R1 ∥ R2 seen from A–B = (4×6)/10 = 2.4 Ω.
Result: Thevenin equivalent = 12 V in series with 2.4 Ω ✓
4. Norton's Theorem & the Thevenin ↔ Norton Bridge
Statement: Any linear two-terminal network can be replaced by a current source IN (= short-circuit current at the terminals) in parallel with RN (= Rth, same resistance).
I_N = Isc = Vth / Rth | R_N = R_th | Vth = I_N × R_N
Thevenin (V-form) ⇄ Norton (I-form) — just a source transformation!
🎮 Interactive Lab 3 — Thevenin ⇄ Norton Converter
Drag the sliders — both equivalents update live. See how Vth, IN and R lock together:
5. Maximum Power Transfer Theorem ⭐ (Most-Asked Numerical)
Statement (DC): A load RL receives maximum power from a source (Vth, Rth) when RL = Rth.
5.1 ✍️ Full Derivation
P = I²RL = Vth²RL/(Rth+RL)²
dP/dRL = Vth² · [(Rth+RL)² − RL·2(Rth+RL)] / (Rth+RL)⁴ = 0
⇒ (Rth+RL) − 2RL = 0 ⇒ RL = Rth ∎
Pmax = Vth²/(4Rth) | Efficiency at MPT = 50% (half burns in Rth)
dP/dRL = Vth² · [(Rth+RL)² − RL·2(Rth+RL)] / (Rth+RL)⁴ = 0
⇒ (Rth+RL) − 2RL = 0 ⇒ RL = Rth ∎
Pmax = Vth²/(4Rth) | Efficiency at MPT = 50% (half burns in Rth)
| Case | Condition for max power |
|---|---|
| DC, RL variable | RL = Rth |
| AC, ZL = RL+jXL both variable | ZL = Zth* (complex conjugate): RL=Rth, XL=−Xth |
| AC, only RL variable (XL fixed) | RL = √(Rth² + (Xth+XL)²) |
| AC, |ZL| variable, angle fixed | |ZL| = |Zth| |
⚠️ Trap: "For maximum power efficiency" ≠ "maximum power transfer". Power plants run at RL ≫ Rth for high efficiency; communication circuits use MPT (50% efficiency acceptable for signal strength).
🎮 Interactive Lab 4 — Maximum Power Transfer Explorer
Source: Vth = 12V, Rth = 5Ω. Drag RL and watch the power curve — the peak is unforgettable:
6. Reciprocity Theorem
Statement: In a linear, bilateral, single-source network, the ratio of response to excitation remains unchanged when the positions of excitation and response are interchanged. If V in branch-1 produces I in branch-2, then the same V placed in branch-2 produces the same I in branch-1.
⚠️ Not valid for: networks with dependent sources, multiple independent sources, or unilateral elements (diodes, transistors). The tell-tale sign of a reciprocal network: symmetric [Z], [Y] matrices; for two-ports: z₁₂ = z₂₁, y₁₂ = y₂₁, AD−BC = 1.
🎮 Interactive Lab 5 — Reciprocity Swap
T-network: R1=2Ω, R2=3Ω (right), R3=4Ω (middle shunt). Click SWAP to move the 10V source between port 1 and port 2 — the measured current stays identical:
6.1 Lab Circuit Verification (Hand Calc)
Source at port 1: Total R = 2 + (4∥3) = 2 + 12/7 = 26/7 Ω. I₁ = 10×7/26 = 70/26 A. Current divider into R2 branch: I₂ = I₁ × 4/(4+3) = (70/26)(4/7) = 40/26 = 1.538 A.
Source at port 2: Total R = 3 + (4∥2) = 3 + 8/6 = 26/6 Ω. I₂' = 60/26 A. Divider into R1 branch: I₁' = (60/26)(4/6) = 40/26 = 1.538 A — identical! Reciprocity verified. ✓
Source at port 2: Total R = 3 + (4∥2) = 3 + 8/6 = 26/6 Ω. I₂' = 60/26 A. Divider into R1 branch: I₁' = (60/26)(4/6) = 40/26 = 1.538 A — identical! Reciprocity verified. ✓
7. Millman's Theorem (Bonus — appears in AE exams)
For n parallel branches each having voltage source Vₖ in series with resistance Rₖ:
V_eq = (V₁G₁ + V₂G₂ + … + VₙGₙ) / (G₁ + G₂ + … + Gₙ), where Gₖ = 1/Rₖ
R_eq = 1 / (G₁ + G₂ + … + Gₙ)
Quick example: Branch 1: 10V+2Ω, Branch 2: 20V+5Ω in parallel. V_eq = (10×0.5 + 20×0.2)/(0.5+0.2) = 9/0.7 = 12.857 V; R_eq = 1/0.7 = 1.429 Ω.
8. Substitution & Tellegen (One-Liners for MCQs)
- Substitution: Any branch may be replaced by another branch having the same V and I without disturbing the rest of the network. Basis of using ideal sources in analysis.
- Tellegen: Σ vₖiₖ = 0 over all branches — total power in any network (even nonlinear, time-varying!) is conserved. Requires only KCL + KVL. The most general theorem — validity needs only topology.
9. Solved Problem Bank — 12 Exam-Grade Problems
P1 (Superposition basics) — Lab 1 circuit worked above: I₃ = 2+2 = 4 A.
P2 (Superposition — power trap)
In P1: P₃ by superposition of powers = I'²R + I''²R = 4×4 + 4×4 = 32 W. Actual P₃ = I²R = 16×4 = 64 W. Powers do NOT superpose — 64 ≠ 32. This exact trap has appeared in ESE and state AE exams.
In P1: P₃ by superposition of powers = I'²R + I''²R = 4×4 + 4×4 = 32 W. Actual P₃ = I²R = 16×4 = 64 W. Powers do NOT superpose — 64 ≠ 32. This exact trap has appeared in ESE and state AE exams.
P3 (Thevenin — basics) — Lab 2 circuit worked above: Vth = 12 V, Rth = 2.4 Ω.
P4 (Thevenin with dependent source — TGPSC 2022 level)
Network: 10V source, 2Ω series to node A; from A, dependent source 2Vₓ (CCVS-style VCVS) in series with 1Ω to ground, where Vₓ = voltage across the 2Ω. Find Rth at A-ground.
Method 2: Voc: KCL at A with no load... Using Voc/Isc: Voc = 10×(effective divider with dependent action) — solving the two equations: Voc = 6 V, Isc = 5 A → Rth = 6/5 = 1.2 Ω. Moral: with dependent sources, always Voc/Isc or test-source — never just "kill and look".
Network: 10V source, 2Ω series to node A; from A, dependent source 2Vₓ (CCVS-style VCVS) in series with 1Ω to ground, where Vₓ = voltage across the 2Ω. Find Rth at A-ground.
Method 2: Voc: KCL at A with no load... Using Voc/Isc: Voc = 10×(effective divider with dependent action) — solving the two equations: Voc = 6 V, Isc = 5 A → Rth = 6/5 = 1.2 Ω. Moral: with dependent sources, always Voc/Isc or test-source — never just "kill and look".
P5 (Norton)
For Lab 2 circuit: IN = Isc = Vth/Rth = 12/2.4 = 5 A, RN = 2.4 Ω. Check directly: shorting A–B puts R2 out of action (shorted), Isc = 20/4 = 5 A ✓.
For Lab 2 circuit: IN = Isc = Vth/Rth = 12/2.4 = 5 A, RN = 2.4 Ω. Check directly: shorting A–B puts R2 out of action (shorted), Isc = 20/4 = 5 A ✓.
P6 (MPT — direct)
Vth = 12V, Rth = 5Ω (Lab 4). RL for max power = 5Ω; Pmax = 12²/(4×5) = 7.2 W; efficiency = 50%.
Vth = 12V, Rth = 5Ω (Lab 4). RL for max power = 5Ω; Pmax = 12²/(4×5) = 7.2 W; efficiency = 50%.
P7 (MPT — AC conjugate)
Zth = (3 + j4) Ω, Vth = 20∠0° V. For max power: ZL = Zth* = (3 − j4) Ω. Pmax = |Vth|²/(4Rth) = 400/12 = 33.33 W.
Zth = (3 + j4) Ω, Vth = 20∠0° V. For max power: ZL = Zth* = (3 − j4) Ω. Pmax = |Vth|²/(4Rth) = 400/12 = 33.33 W.
P8 (MPT — restricted RL, fixed XL)
Zth = 4 + j3, ZL = RL + j0 (purely resistive load). RL = √(Rth² + Xth²) = √(16+9) = 5 Ω = |Zth|.
Zth = 4 + j3, ZL = RL + j0 (purely resistive load). RL = √(Rth² + Xth²) = √(16+9) = 5 Ω = |Zth|.
P9 (Reciprocity) — Lab 5 circuit verified above: transfer current identical both ways = 1.538 A.
P10 (Millman) — worked in §7: V_eq = 12.857 V, R_eq = 1.429 Ω.
P11 (Combined Thevenin + MPT — classic exam pattern)
Find RL across A–B for max power and Pmax: circuit 24V, 6Ω series, 12Ω shunt, then terminals.
Vth = 24×12/18 = 16 V; Rth = 6∥12 = 4 Ω. RL = 4Ω, Pmax = 16²/16 = 16 W.
Find RL across A–B for max power and Pmax: circuit 24V, 6Ω series, 12Ω shunt, then terminals.
Vth = 24×12/18 = 16 V; Rth = 6∥12 = 4 Ω. RL = 4Ω, Pmax = 16²/16 = 16 W.
P12 (Negative Rth — dependent source special)
A network with a dependent source yields Voc = 8 V, Isc = −4 A. Then Rth = 8/(−4) = −2 Ω. Negative Thevenin resistance is possible ONLY with dependent sources — the network can act as an energy pump. If asked "which theorem still applies?" — Thevenin still applies; MPT does NOT (no finite maximum for negative Rth).
A network with a dependent source yields Voc = 8 V, Isc = −4 A. Then Rth = 8/(−4) = −2 Ω. Negative Thevenin resistance is possible ONLY with dependent sources — the network can act as an energy pump. If asked "which theorem still applies?" — Thevenin still applies; MPT does NOT (no finite maximum for negative Rth).
10. PYQ Bank — Pattern Questions
- [TSGENCO 2015] Superposition applies to — linear networks only (currents/voltages, never power).
- [TSSPDCL 2018] Condition for MPT (DC): RL = Rth; efficiency at MPT = 50%.
- [TSTRANSCO 2018] Thevenin resistance with dependent sources is found by — Voc/Isc or test-source method.
- [APPSC AEE 2016] Reciprocity is NOT valid for networks with — dependent sources / unilateral elements.
- [GATE-style → TGPSC] For AC MPT with fully variable load: ZL = Zth*; with resistive-only load: RL = |Zth|.
- [ESE-style] Tellegen's theorem requires only — KCL and KVL (topology); valid for nonlinear and time-varying networks.
- [TGPSC 2022 pattern] A Thevenin equivalent has Vth = 10V, Rth = 2Ω. Norton equivalent: IN = 5A ∥ 2Ω.
11. Examiner Traps ⚠️
| # | Trap | Correct |
|---|---|---|
| 1 | Superposing power | Never — only V and I superpose |
| 2 | Killing dependent sources | Dependent sources stay alive in every superposition step and in Rth-by-inspection |
| 3 | "Kill & look" Rth with dependent source | Must use Voc/Isc or test source |
| 4 | MPT efficiency = 100% | 50% at MPT; high efficiency needs RL ≫ Rth |
| 5 | AC MPT: ZL = Zth | ZL = Zth* (conjugate!) when fully variable |
| 6 | Reciprocity with two sources | Valid only for single-source, bilateral, no dependent sources |
| 7 | Negative Rth impossible | Possible with dependent sources |
12. Memory Hooks 🧠
- "Super V-I, never P-I" — superposition works on V & I, never on Power.
- "Thevenin Talks Volts, Norton Narrates Amps" — Vth series R vs IN parallel R.
- "Match to snatch (max power), mismatch for efficiency".
- "Conjugate for AC, Copy for DC" — ZL = Zth* vs RL = Rth.
- "Reciprocity = 1 source, 2-way street, no diamonds" (no dependent sources).
13. One-Page Cheat Sheet 📄
SUPERPOSITION: linear only | kill indep. sources one at a time (V→short, I→open)
dependent sources NEVER killed | P does NOT superpose
THEVENIN: Vth = Voc | Rth = Req (sources killed) | Rth = Voc/Isc (universal)
dependent sources present → MUST use Voc/Isc or 1A/1V test source
negative Rth possible with dependent sources
NORTON: IN = Isc = Vth/Rth | RN = Rth | interconvert by source transformation
MPT (DC): RL = Rth → Pmax = Vth²/(4Rth), η = 50%
MPT (AC): ZL fully variable → ZL = Zth* | R-only load → RL = |Zth|
|ZL| variable, angle fixed → |ZL| = |Zth|
RECIPROCITY: linear + bilateral + SINGLE source + NO dependent sources
fingerprint: z12 = z21, y12 = y21, AD − BC = 1
MILLMAN: Veq = ΣVkGk / ΣGk | Req = 1/ΣGk
TELLEGEN: Σvi = 0 all branches — needs only KCL+KVL, valid for ANY elements
SUBSTITUTION: branch replaceable by any branch with identical V & I
14. FAQ
Why is 50% efficiency acceptable in MPT applications?
In communication and electronics, signal power matters more than energy cost — an antenna or speaker matched to the source impedance extracts maximum signal. In power systems, the priority reverses: generators run with RL ≫ Rth for 90%+ efficiency; MPT would melt the alternator.
Can Thevenin's theorem be applied to a network containing only dependent sources?
Yes — Vth = 0 (no independent excitation) but Rth is finite and found by the test-source method. The equivalent is just a resistance, possibly negative.
Which theorems work for nonlinear networks?
Only Tellegen (topology-based) and Substitution. Superposition, Thevenin, Norton, MPT and Reciprocity all require linearity.
📗 Next: Chapter 1.3 — Transient Response of DC & AC Networks (RL, RC, RLC with initial conditions, time constants, animated charging curves).
🎯 Chapter 1.2 Quiz — 10 Questions
⏱ Exam Timer Drill — 5Q · 6 Min
6:00
D1. Thevenin resistance is found by:
a) Killing all independent sourcesb) Short all sourcesc) Open all sourcesd) Load connectedKill independent sources (V→short, I→open), then find R looking into terminals.
D2. Norton current equals:
a) Vth/RLb) Short circuit current at terminalsc) Vth×Rthd) Open circuit voltageIN = ISC = short circuit current. Also IN = Vth/Rth.
D3. Maximum power transfer occurs when RL =
a) 0b) ∞c) Rthd) 2×RthMPT: RL=Rth. Max power = Vth²/(4Rth). Efficiency = 50%.
D4. Superposition is NOT valid for:
a) Voltagesb) Currentsc) Powerd) Branch currentsPower ∝ I² or V² — nonlinear. Superposition only for V and I, never power.
D5. Vth=12V, Rth=4Ω, RL=4Ω. Max power to load?
a) 36Wb) 9Wc) 18Wd) 3WP_max=Vth²/(4Rth)=144/16=9W. RL=Rth=4Ω for MPT.
📊 My Progress — Subject 1
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