📘 Suchenow Academy | TGPSC AEE Electrical
Subject 1: Electric Circuits & Fields → Chapter 1.1: Network Fundamentals (Complete Theory with Diagrams & Derivations)
| Expected questions | 3–5 direct + foundation for 15+ more |
| Study time | 8–10 hours | Revision: 90 min |
| Target exams | TGPSC AEE, TSGENCO AE, TSTRANSCO AE, TSSPDCL/TGNPDCL AE |
| Syllabus line | Network graph, KCL, KVL, Node and mesh analysis |
1. Why This Chapter Decides Your Rank
Network Fundamentals is the load-bearing wall of Paper-II. KCL, KVL, mesh and nodal analysis are used inside Electrical Machines (equivalent circuits), Power Systems (load flow = giant nodal analysis), Control Systems (network functions) and Measurements (bridge balance). If your fundamentals are automatic, you save 30–45 seconds per question across the entire paper — the real difference between merit list and waiting list.
2. Circuit Elements — Classification with Tests
2.1 Linear vs Nonlinear
An element is linear if it satisfies both homogeneity (input k·x → output k·y) and additivity (superposition). The V–I characteristic must be a straight line through the origin.
⚠️ Examiner Trap: V = 2I + 3 is NOT linear — it doesn't pass through origin (fails homogeneity). V = 2I is linear. This exact trap has appeared in state AE exams.
2.2 Full Element Classification Table
| Basis | Type 1 | Type 2 | Test / Example |
|---|---|---|---|
| Energy | Active — delivers net energy | Passive — absorbs/stores | Sources, transistors vs R, L, C |
| Direction | Bilateral — same V–I both ways | Unilateral | R, L, C vs diode |
| Size vs λ | Lumped — size ≪ wavelength | Distributed | Circuit R vs transmission line |
| Time | Time-invariant | Time-variant | Fixed R vs carbon-mic resistance |
| Memory | Memoryless (R) | With memory (L, C) | L, C output depends on history |
💡 Passive ≠ dissipative. L and C are passive (can't deliver more than stored) but store energy. Only R dissipates. A charged capacitor can temporarily deliver energy yet it remains passive — a classic conceptual question.
3. Sources — Ideal, Practical, Dependent
3.1 Four Types of Sources with Diagrams
3.2 Dependent (Controlled) Sources — All 4 Types
| Type | Relation | Gain & Unit | Real Device |
|---|---|---|---|
| VCVS | v = μ·vx | μ — dimensionless | Op-amp |
| VCCS | i = gm·vx | gm — Siemens (S) | FET / MOSFET |
| CCVS | v = r·ix | r — Ohms (Ω) | Trans-resistance amp |
| CCCS | i = β·ix | β — dimensionless | BJT (hFE) |
🧠 Memory trick: Diamond symbol = dependent. First letter = what controls, third letter = what is produced. VCVS = voltage-controlled voltage source. The diamond shape looks like a "D" for "Dependent".
4. Network Graph Theory — Full Treatment with Derivations
4.1 From Circuit to Graph
Replace every element by a line segment (branch) between its two terminals (nodes). The graph captures only the topology (how things connect), not the element values.
4.2 Tree, Twigs, Links, Co-tree — Visual
4.3 ✍️ Derivation 1 — Why Twigs = n − 1
Build the tree node by node: Start with 1 node and 0 branches. To add each new node, attach it with exactly one new branch. A second branch to an existing pair would create a loop — which a tree forbids by definition. After connecting all n nodes: we added (n−1) branches. Therefore Twigs = n − 1. ∎
4.4 ✍️ Derivation 2 — Why Links = b − n + 1 = Independent KVL Equations
Step 1: Total branches b = twigs + links → links = b − (n−1) = b − n + 1.
Step 2 (independence): The tree provides exactly one path between any two nodes. Adding one link creates exactly one new loop — the fundamental loop (f-loop / tie-set). Each f-loop contains one link that no other f-loop contains. Therefore no f-loop equation can be derived from the others — they are mutually independent.
Conclusion: Number of independent KVL equations = number of links = b − n + 1. ∎
4.5 ✍️ Derivation 3 — Why Independent KCL Equations = n − 1
Write KCL at all n nodes, then add all n equations. Every branch current appears exactly twice — once leaving one node (+) and once entering another (−). The total sum is identically zero: the n equations are linearly dependent with exactly one redundancy. Removing any one node equation (grounding the reference node) leaves n − 1 independent equations. ∎
4.6 The Three Matrices of Graph Theory
| Matrix | Symbol | Order | Rank | Built on | Gives |
|---|---|---|---|---|---|
| Reduced Incidence | [A] | (n−1) × b | n − 1 | Nodes | [A][ib] = 0 → KCL |
| Tie-set (f-loop) | [B] | (b−n+1) × b | b − n + 1 | Links | [B][vb] = 0 → KVL |
| Cut-set (f-cutset) | [Q] | (n−1) × b | n − 1 | Twigs | [Q][ib] = 0 → KCL |
- Incidence matrix entry: aij = +1 if branch j leaves node i, −1 if it enters, 0 if not connected.
- Complete incidence matrix columns sum to zero (each branch enters/leaves exactly one node pair) → rank = n−1.
- Tie-set is built on LINKS; Cut-set is built on TWIGS — the most frequently tested one-liner.
- Orthogonality: [B]·[Q]T = 0 (GATE/ESE level, entering TGPSC).
- Number of trees = det([A]·[A]T). For a complete graph of n nodes: n(n−2) (Cayley's formula). Complete graph has n(n−1)/2 branches.
5. Kirchhoff's Laws — Statements, Basis, Limits
| Law | Basis | Equations needed | Works for | Fails for |
|---|---|---|---|---|
| KCL | Conservation of charge | n − 1 | Lumped; linear/nonlinear; time-variant/invariant | Distributed networks at high frequency |
| KVL | Conservation of energy | b − n + 1 | Same — topology only | Distributed networks; circuits with changing magnetic flux through the loop |
⚠️ Critical Trap: KCL and KVL are NOT limited to linear or passive elements. They depend purely on topology — how things connect, not what they are. A common wrong answer: "KVL fails for nonlinear elements." Completely false.
6. Mesh Analysis — Complete Method with Worked Circuit
6.1 Standard Two-Mesh Circuit
Solving: 6I₁ − 4I₂ = 20 … (1) and −4I₁ + 10I₂ = −8 … (2). Multiply (1)×2.5 → 15I₁ − 10I₂ = 50. Add (2): 11I₁ = 42 → I₁ = 3.82 A. Sub into (1): I₂ = (6×3.82−20)/4 = 0.73 A. Current through R3 (downward) = I₁−I₂ = 3.09 A.
Matrix inspection method:
[ R₁₁ −R₁₂ ] [I₁] [ΣV rises in mesh 1]
[−R₂₁ R₂₂ ] [I₂] = [ΣV rises in mesh 2]
R₁₁ = 2+4 = 6Ω (self) | R₂₂ = 4+6 = 10Ω (self)
R₁₂ = R₂₁ = 4Ω (mutual) — negative off-diagonal for all-CW currents
[R] symmetric ⟹ passive network (no dependent sources)
Asymmetric [R] ⟹ dependent source is present — a known exam question!
6.2 Supermesh — Circuit Diagram + Full Solution
⚠️ Do NOT panic at a negative mesh current. It simply means the actual current flows anticlockwise. Never flip signs mid-solution — complete the algebra first, then interpret.
💡 Easy case: If the current source sits on the outer boundary (one mesh only), that mesh current equals ±Iₛ directly — no supermesh needed. Always check this first.
7. Nodal Analysis — Complete Method with Worked Circuit
7.1 Standard Two-Node Circuit
Node V₁: 3 = V₁/2 + (V₁−V₂)/4 → 0.75V₁ − 0.25V₂ = 3 …(1)
Node V₂: (V₁−V₂)/4 = V₂/8 + 1 → −0.25V₁ + 0.375V₂ = −1 …(2)
Multiply (1)×1.5 → 1.125V₁ − 0.375V₂ = 4.5. Add (2): 0.875V₁ = 3.5 → V₁ = 4 V. From (2): −1 + 0.375V₂ = −1 → V₂ = 0 V. Self-check: (4−0)/4 = 1 A = the 1 A sink ✓
Inspection matrix:
[G][V] = [I]
G₁₁ = 1/2 + 1/4 = 0.75 S | G₂₂ = 1/4 + 1/8 = 0.375 S
G₁₂ = G₂₁ = −1/4 = −0.25 S (conductance between nodes — negative)
[G] symmetric ⟹ passive network | Asymmetric [G] ⟹ dependent source
7.2 Supernode — Circuit Diagram + Full Solution
Solving: 3V₁ + 2V₂ = 36 with V₂ = V₁+12 → 3V₁ + 2V₁ + 24 = 36 → V₁ = 2.4 V, V₂ = 14.4 V. Verify: 2.4/4 + 14.4/6 = 0.6 + 2.4 = 3 ✓
8. Source Transformation — Rules, Diagram, Traps
- Cannot transform ideal sources (R = 0 or ∞) — undefined mathematically.
- Dependent sources can be transformed, but the controlling variable's branch must never be destroyed.
- Internal power differs — equivalence is external only. Classic trap: "power delivered by source" changes after transformation.
- Polarity rule: + terminal of Vₛ → arrowhead of Iₛ. A wrong direction reverses all subsequent signs.
9. Mesh vs Nodal — Decision Table
| Compare | Mesh Analysis | Nodal Analysis |
|---|---|---|
| Equations count | b − n + 1 | n − 1 |
| Law used | KVL | KCL |
| Works for non-planar? | ❌ No | ✅ Yes (universal) |
| Prefer when | Voltage sources dominate | Current sources dominate |
| Special case needed for | Current source → Supermesh | Voltage source → Supernode |
| Power system use | Rarely | Always (Y-bus = nodal) |
🎯 Decision rule: Compute both (b−n+1) and (n−1). Pick the smaller — fewer equations = less work = fewer arithmetic errors. If equal, prefer nodal (works for non-planar).
10. Formula Sheet — All in One Place
| # | Formula | Application |
|---|---|---|
| 1 | Twigs = n − 1 | Branches in any tree |
| 2 | Links = b − n + 1 | Independent KVL equations / f-loops |
| 3 | Independent KCL equations = n − 1 | Nodal analysis size |
| 4 | Trees possible = det([A]·[A]ᵀ) | Count of spanning trees |
| 5 | Trees (complete graph) = n(n−2) | Cayley's formula |
| 6 | Branches (complete graph) = n(n−1)/2 | Complete graph size |
| 7 | Iₛ = Vₛ/R | Source transformation |
| 8 | Rank of [A] = n − 1 | Incidence matrix rank |
| 9 | [B]·[Q]ᵀ = 0 | Orthogonality of tie-set and cut-set |
| 10 | R₁₁ = sum of R in mesh i; R₁₂ = −(common R) | Mesh matrix by inspection |
| 11 | G₁₁ = sum of G at node i; G₁₂ = −(G between i,j) | Nodal matrix by inspection |
11. Solved Problem Bank — 12 Exam-Grade Problems
P1 (Graph — Formula, TSGENCO 2015 pattern)
Network with b = 8, n = 5. Find independent mesh and node equations.
Solution: Mesh = 8−5+1 = 4. Node = 5−1 = 4.
Network with b = 8, n = 5. Find independent mesh and node equations.
Solution: Mesh = 8−5+1 = 4. Node = 5−1 = 4.
P2 (Cayley's formula)
Number of spanning trees in a complete graph of 4 nodes?
Solution: n(n−2) = 4² = 16. Branches = 4×3/2 = 6; links per tree = 6−3 = 3.
Number of spanning trees in a complete graph of 4 nodes?
Solution: n(n−2) = 4² = 16. Branches = 4×3/2 = 6; links per tree = 6−3 = 3.
P3 (Incidence matrix rank — TSTRANSCO 2018 pattern)
A graph has 6 nodes. What is the rank of the complete incidence matrix?
Solution: Rank = n−1 = 5. (All columns of complete [Aₐ] sum to zero → one dependent row → rank drops by 1.)
A graph has 6 nodes. What is the rank of the complete incidence matrix?
Solution: Rank = n−1 = 5. (All columns of complete [Aₐ] sum to zero → one dependent row → rank drops by 1.)
P4 (Tie-set / f-loop count)
A tree of a connected graph has 5 twigs. Graph has 9 branches. How many fundamental loops?
Solution: Twigs = n−1 = 5 → n = 6. f-loops = b−n+1 = 9−6+1 = 4. Each f-loop contains exactly one link not in any other f-loop.
A tree of a connected graph has 5 twigs. Graph has 9 branches. How many fundamental loops?
Solution: Twigs = n−1 = 5 → n = 6. f-loops = b−n+1 = 9−6+1 = 4. Each f-loop contains exactly one link not in any other f-loop.
P5 (Mesh — outer-boundary current source)
A circuit has 2 meshes. A 6A current source occupies the outer branch of mesh 2 only, arrow aligned with clockwise I₂. Mesh 1 has 20V source, 3Ω and 5Ω in common with mesh 2.
Solution: I₂ = 6A directly. Write KVL for mesh 1 only: −20 + 3I₁ + 5(I₁−6) = 0 → 8I₁ = 50 → I₁ = 6.25 A. One equation, no supermesh.
A circuit has 2 meshes. A 6A current source occupies the outer branch of mesh 2 only, arrow aligned with clockwise I₂. Mesh 1 has 20V source, 3Ω and 5Ω in common with mesh 2.
Solution: I₂ = 6A directly. Write KVL for mesh 1 only: −20 + 3I₁ + 5(I₁−6) = 0 → 8I₁ = 50 → I₁ = 6.25 A. One equation, no supermesh.
P6 (Supermesh — from §6.2)
10V, 2Ω (mesh 1), 5A source shared, 4Ω and 6Ω (mesh 2).
Solution: Supermesh KVL: −10 + 2I₁ + 4I₂ + 6I₂ = 0 → 2I₁ + 10I₂ = 10. Constraint: I₂ − I₁ = 5. Substituting: 2I₁ + 10I₁ + 50 = 10 → I₁ = −3.33 A, I₂ = 1.67 A.
10V, 2Ω (mesh 1), 5A source shared, 4Ω and 6Ω (mesh 2).
Solution: Supermesh KVL: −10 + 2I₁ + 4I₂ + 6I₂ = 0 → 2I₁ + 10I₂ = 10. Constraint: I₂ − I₁ = 5. Substituting: 2I₁ + 10I₁ + 50 = 10 → I₁ = −3.33 A, I₂ = 1.67 A.
P7 (Nodal — grounded voltage source, easy case)
Node V₁ is connected to ground through a 10V source (+ at V₁). 4Ω and 8Ω also at V₁. Find currents without supernode.
Solution: V₁ = 10 V directly (grounded V-source). I(4Ω) = 10/4 = 2.5 A; I(8Ω) = 10/8 = 1.25 A. KCL gives source current = 3.75 A supplied upward. No supernode needed.
Node V₁ is connected to ground through a 10V source (+ at V₁). 4Ω and 8Ω also at V₁. Find currents without supernode.
Solution: V₁ = 10 V directly (grounded V-source). I(4Ω) = 10/4 = 2.5 A; I(8Ω) = 10/8 = 1.25 A. KCL gives source current = 3.75 A supplied upward. No supernode needed.
P8 (Supernode — from §7.2)
3A, 4Ω, 12V between V₁ and V₂ (V₂ − V₁ = 12), 6Ω.
Solution: KCL supernode: 3 = V₁/4 + V₂/6. With V₂ = V₁+12: 5V₁ = 12 → V₁ = 2.4 V, V₂ = 14.4 V.
3A, 4Ω, 12V between V₁ and V₂ (V₂ − V₁ = 12), 6Ω.
Solution: KCL supernode: 3 = V₁/4 + V₂/6. With V₂ = V₁+12: 5V₁ = 12 → V₁ = 2.4 V, V₂ = 14.4 V.
P9 (Dependent source — nodal, TGPSC 2022 pattern)
2A enters node V₁; 1Ω and 2Ω to ground; CCCS 2Iₓ (Iₓ = V₁/2) also enters V₁.
Solution: KCL: 2 + 2(V₁/2) = V₁/1 + V₁/2 → 2 + V₁ = 1.5V₁ → V₁ = 4 V. Note: dependent source effectively reduces the equivalent conductance — appears like a negative resistance in the matrix.
2A enters node V₁; 1Ω and 2Ω to ground; CCCS 2Iₓ (Iₓ = V₁/2) also enters V₁.
Solution: KCL: 2 + 2(V₁/2) = V₁/1 + V₁/2 → 2 + V₁ = 1.5V₁ → V₁ = 4 V. Note: dependent source effectively reduces the equivalent conductance — appears like a negative resistance in the matrix.
P10 (Source transformation chain)
24V in series with 6Ω, in parallel with 12Ω, feeding A–B (open). Reduce to single equivalent.
Solution: Step 1: 24V+6Ω → 4A ∥ 6Ω. Step 2: 6Ω ∥ 12Ω = 4Ω. Result: 4A ∥ 4Ω. Voltage-form equivalent: 16V + 4Ω series.
24V in series with 6Ω, in parallel with 12Ω, feeding A–B (open). Reduce to single equivalent.
Solution: Step 1: 24V+6Ω → 4A ∥ 6Ω. Step 2: 6Ω ∥ 12Ω = 4Ω. Result: 4A ∥ 4Ω. Voltage-form equivalent: 16V + 4Ω series.
P11 (Power trap — internal vs external)
In P10 original circuit (A–B open), find power supplied by 24V source.
Solution: Current flows through 6Ω + 12Ω (open A-B): I = 24/(6+12) = 1.333A. P = 24 × 1.333 = 32 W. In transformed 4A ∥ 4Ω with A-B open, power = 4 × (4×4) = 64W — different! External equivalence ≠ internal power equality.
In P10 original circuit (A–B open), find power supplied by 24V source.
Solution: Current flows through 6Ω + 12Ω (open A-B): I = 24/(6+12) = 1.333A. P = 24 × 1.333 = 32 W. In transformed 4A ∥ 4Ω with A-B open, power = 4 × (4×4) = 64W — different! External equivalence ≠ internal power equality.
P12 (Matrix symmetry fingerprint — ESE/TGPSC)
The [R] matrix of a two-mesh circuit is [6, −4; −3, 8]. What can you conclude?
Solution: Off-diagonal elements are −4 and −3, not equal → [R] is not symmetric → a dependent source is present in the network. (Passive-only circuits with CW mesh currents always give symmetric [R].)
The [R] matrix of a two-mesh circuit is [6, −4; −3, 8]. What can you conclude?
Solution: Off-diagonal elements are −4 and −3, not equal → [R] is not symmetric → a dependent source is present in the network. (Passive-only circuits with CW mesh currents always give symmetric [R].)
12. PYQ Bank — Reproduced Pattern Questions
- [TSGENCO AE 2015] "A tree of a network with n nodes has ___ branches." → n − 1
- [TSSPDCL AE 2018] "The number of independent loops for a network with n nodes and b branches is ___." → b − n + 1
- [APPSC AEE 2016] "KCL is applicable to (a) lumped networks only (b) distributed only (c) both (d) nonlinear only" → (a) lumped networks only
- [TSTRANSCO AE 2018] "Rank of the incidence matrix of a connected graph with n nodes is ___." → n − 1
- [ESE pattern, entering TGPSC] "The tie-set matrix is formed using (a) twigs (b) links (c) all branches (d) nodes" → (b) links; cut-set uses twigs.
- [GATE 2008-style, now in state exams] "The [R] matrix in mesh analysis of a network with only independent sources and resistors is always ___." → symmetric; asymmetry proves a dependent source is present.
- [TGPSC AEE 2022 pattern] "Which law fails for distributed networks at high frequency? (a) KCL only (b) KVL only (c) Both KCL and KVL (d) Neither" → (c) Both
- [Conceptual trap — all exams] "V = 5I + 2 represents a (a) linear element (b) nonlinear element (c) linear but time-variant (d) bilateral" → (b) nonlinear — doesn't pass through origin.
13. Examiner Favourite Topics 🎯
- Links = b−n+1 and twigs = n−1 (formula appears almost every exam)
- Rank of incidence matrix = n−1
- Tie-set uses links; cut-set uses twigs (one-liner, high frequency)
- Supermesh constraint equation (numerical trap)
- Supernode constraint equation (numerical trap)
- KCL/KVL validity — lumped vs distributed (conceptual)
- Asymmetric [R] or [G] matrix → dependent source present
- Linearity: must pass through origin (V = mI + c is not linear if c ≠ 0)
14. Common Mistakes & How to Avoid Them ⚠️
| # | Mistake | Correct |
|---|---|---|
| 1 | Mesh equations = n−1 | Mesh = b−n+1; nodal = n−1 |
| 2 | Forgetting constraint equation in supermesh/supernode | Always write constraint; unknowns > equations without it |
| 3 | Forming supernode when V-source touches ground | Node voltage = ±Vₛ directly; no supernode needed |
| 4 | Transforming ideal voltage source (R=0) | Impossible; use V-shift or leave as is |
| 5 | Flipping sign mid-solution for negative mesh current | Complete algebra; negative = anticlockwise; interpret at end |
| 6 | KVL fails for nonlinear elements | Both laws are topological; valid for ANY element |
| 7 | Destroying the controlling branch in dependent source transformation | Keep controlling branch intact; transform only the source |
| 8 | Saying source transformation preserves internal power | Only external V–I is preserved; internal power changes |
15. Memory Hooks 🧠
- "No-KCL, Me-KVL" → Nodal uses KCL; Mesh uses KVL
- "Tie LIES on Links, Cut TWIsts on Twigs"
- "V-Zero, I-Infinity" → ideal voltage source Rᵢₙₜ = 0; ideal current source Rᵢₙₜ = ∞
- "Tree needs n−1 sticks to hold n leaves" → Twigs = n−1
- Diamond = Depends → dependent source symbol
- "Asymmetry = Alien (dependent source)" → symmetric [R] or [G] ⟹ passive only
16. One-Page Cheat Sheet 📄
GRAPH
Twigs = n−1 | Links = b−n+1 | Trees (complete graph) = n^(n−2)
Branches (complete graph) = n(n−1)/2 | Trees (general) = det([A]·[A]ᵀ)
MATRICES
[A] : (n−1)×b , rank n−1 , built on NODES → [A][ib] = 0 (KCL)
[B] : (b−n+1)×b , rank b−n+1 , built on LINKS → [B][vb] = 0 (KVL) ← TIE-SET
[Q] : (n−1)×b , rank n−1 , built on TWIGS → [Q][ib] = 0 (KCL) ← CUT-SET
Orthogonality: [B]·[Q]ᵀ = 0
KIRCHHOFF
KCL: Σi = 0 (charge conservation) | n−1 independent equations
KVL: Σv = 0 (energy conservation) | b−n+1 independent equations
BOTH: topology only — valid for lumped / linear / nonlinear / time-variant
BOTH: FAIL for distributed networks (transmission lines at high freq)
MESH ANALYSIS (KVL)
[R][I] = [V] | Self-resistance on diagonal (positive) | Mutual = negative (all CW)
[R] symmetric ⟹ passive network | Asymmetric ⟹ dependent source present
Current source in two meshes → SUPERMESH + constraint equation
Current source on outer boundary → mesh current = ±Is directly
NODAL ANALYSIS (KCL)
[G][V] = [I] | Self-conductance on diagonal | Mutual = negative
[G] symmetric ⟹ passive | Asymmetric ⟹ dependent source
Floating V-source → SUPERNODE + constraint equation
V-source to ground → node voltage = ±Vs directly (no supernode!)
SOURCES
Ideal V-source: Rint = 0 (series) | Ideal I-source: Rint = ∞ (parallel)
Transform: Vs+R(series) ⇄ (Vs/R) ∥ R | + terminal → arrowhead
Cannot transform ideal sources | Dep. sources: keep controlling branch!
Internal power changes under transformation — only external V–I is preserved
ELEMENTS
Linear: V–I straight line through origin | Passive: can't deliver more than stored
Lumped: size ≪ λ | KCL/KVL valid only for lumped networks
VCVS(μ), VCCS(gm,S), CCVS(r,Ω), CCCS(β) — all shown by diamond symbol
17. Interview & Application Corner
- Why is Y-bus (nodal) used in power system load flow, not mesh? Power grids are non-planar and sparse. The nodal admittance matrix is naturally sparse, enabling fast iterative solvers (Newton-Raphson, Gauss-Seidel) at SLDC / TS-Transco EMS. Mesh analysis requires a planar graph — grids are not planar.
- Where does KCL live in the substation? Differential protection of every 132/33 kV transformer at TSSPDCL/TGNPDCL substations compares Σi entering vs leaving the protected zone — literally KCL as a relay operating principle.
- Why does SPICE use Modified Nodal Analysis (MNA)? Pure nodal can't handle ideal voltage sources directly (zero series impedance). MNA augments the conductance matrix with source branch currents as additional unknowns — the software equivalent of a supernode, used in every circuit simulator from LTspice to Cadence.
18. FAQ
Is graph theory really asked in TGPSC AEE?
Yes — at least one direct question per exam. Links/twigs formulas and rank of incidence matrix appeared in TSGENCO AE 2015, TSTRANSCO AE 2018, and TSSPDCL AE 2018. Tie-set vs cut-set (links vs twigs) is the most recent addition to the question bank.
Mesh or nodal — which should I master more deeply?
Nodal analysis. It handles non-planar circuits, dependent-source numericals appear more in nodal form in recent papers, and it directly connects to Power Systems (Y-bus load flow). Master both, but if forced to prioritise: nodal.
How many marks can this chapter alone give?
3–5 direct questions. But because every subsequent chapter (machines equivalent circuits, power system load flow, bridge measurements, control system network functions) uses KCL/KVL and mesh/nodal, strong fundamentals here realistically add 15+ indirect marks across the paper. This chapter has the highest ROI in Paper-II.
What is the difference between a tie-set and a cut-set?
A tie-set (fundamental loop) is formed by adding one link to the tree — it contains exactly one link and gives one independent KVL equation. A cut-set (fundamental cut-set) is formed by removing one twig from the tree — it partitions the graph into two subgraphs and gives one independent KCL equation. Mnemonic: Tie LIES on Links, Cut TWIsts on Twigs.
📗 Next Chapter: Chapter 1.2 — Network Theorems (Thevenin, Norton, Superposition, Maximum Power Transfer, Reciprocity) with full proofs, 15+ circuit diagrams, and 12 solved problems. Coming soon →
🎯 Chapter 1.1 Quiz — 10 Questions
⏱ Exam Timer Drill — 5Q · 6 Min
6:00
D1. KCL is based on conservation of:
a) Energyb) Chargec) Voltaged) PowerKCL: conservation of charge. KVL: conservation of energy.
D2. For a network with N nodes and B branches, independent KCL equations =
a) Bb) Nc) N−1d) B−N+1Independent KCL = N−1. Independent KVL = B−N+1 = L (loops).
D3. A current source in series with a resistor — what happens to resistor?
a) Increases currentb) Has no effect on currentc) Short circuits itd) Opens the circuitSeries resistor with current source → irrelevant to current (but affects voltage). Classic source transformation trap.
D4. Mesh analysis is preferred when:
a) More nodes than meshesb) Fewer meshes than nodesc) Network has voltage sources onlyd) Network is non-planarChoose mesh if meshes < nodes. Choose nodal if nodes < meshes. Always pick smaller system.
D5. Two resistors 6Ω and 3Ω in parallel. Equivalent resistance?
a) 9Ωb) 4.5Ωc) 2Ωd) 1ΩR_parallel = (6×3)/(6+3) = 18/9 = 2Ω. Product over sum for two resistors.
📊 My Progress — Subject 1
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