EEE U111 · Electrical Sciences · Module 2 · Lectures 4–8

Circuit Analysis: Methods & Theorems

Bobrow & Gupta (Asian ed.) §2.1–2.6 — nodal and mesh analysis; source transformation, Thevenin, Norton, maximum power transfer, superposition.

Start here — a fact worth knowing

Nothing in this module appears in any school syllabus — nodal analysis, mesh analysis, and Thevenin's theorem are new to the entire lecture hall. It is also the course's load-bearing module: transients (module 3), AC (module 4), and transistor circuits (module 5) all begin by simplifying a network with exactly these tools. The one skill it leans on hard: solving 2–3 simultaneous linear equations quickly and correctly.

First time through? Read the first-time lesson — same scope, taught slowly with the intuition for each method — then come back here to revise and drill. Two companions: Reading a circuit diagram (the node-spotting skill every method here starts from) and the circuit drills for reps.

1 · The game plan

Both methods pick unknowns that make one Kirchhoff law automatic and use the other to generate exactly enough equations. Nodal: unknowns = node voltages (KVL automatic), write KCL per node; happiest with current sources. Mesh: unknowns = clockwise loop currents (KCL automatic), write KVL per mesh; happiest with voltage sources. Equation counts: nodes − 1 vs. window panes; pick the smaller, but fluency beats optimality.

2 · Nodal analysis §2.1–2.2

Recipe: ground the busiest node; label the rest v₁, v₂, …; write KCL at each labelled node with every resistor current as (v_mine − v_yours)/R; solve. A voltage source to ground fixes that node (one unknown gone); a floating voltage source → supernode (merge the two nodes for KCL, add v_a − v_b = V).

Worked example 1Two-node nodal solve, with audit

6 A source into node 1; 2 Ω from node 1 to ground; 2 Ω between nodes 1 and 2; 4 Ω from node 2 to ground; 1 A source drawing current out of node 2.

KCL, node 1 (in = out): 6 = v₁/2 + (v₁ − v₂)/2 → 2v₁ − v₂ = 12.
KCL, node 2: (v₁ − v₂)/2 = v₂/4 + 1 → 2v₁ − 3v₂ = 4.
Solve: v₁ = 8 V, v₂ = 4 V. Branch currents by Ohm: 4 A, 2 A, 1 A — KCL holds at both nodes by inspection.
Unpack: the elimination

Subtract equation 2 from equation 1: (2v₁ − v₂) − (2v₁ − 3v₂) = 12 − 4 → 2v₂ = 8 → v₂ = 4; then 2v₁ = 12 + 4 = 16. The matching 2v₁ coefficients made subtraction the fast move — glance for that before grinding.
Power audit: 6 A source at 8 V delivers 48 W; 1 A source (4 V across, current entering its + side) absorbs 4 W; resistors v²/R: 32 + 8 + 4 = 44 W. Absorbed 48 W = delivered ✓.

3 · Mesh analysis §2.3

Recipe: assign every window-pane mesh a clockwise current; write KVL around each. A border element carries the difference: walking mesh 1, a shared R drops R(i₁ − i₂) — my current minus the neighbour's, from the walker's perspective. Current source on the outer edge → that mesh current is known; on a border → supermesh (merge panes for KVL, add i₁ − i₂ = I).

Worked example 2Two-mesh solve, with audit

Left mesh: 20 V source (pushing clockwise), 2 Ω; shared middle branch: 4 Ω. Right mesh: 3 Ω, and a 9 V source whose + terminal the clockwise walk enters first.

Mesh 1: 20 = 2i₁ + 4(i₁ − i₂) → 6i₁ − 4i₂ = 20.
Mesh 2: 4(i₂ − i₁) + 3i₂ + 9 = 0 → 7i₂ = 4i₁ − 9.
Solve: i₁ = 4 A, i₂ = 1 A; the shared 4 Ω carries 4 − 1 = 3 A downward.
Unpack: the elimination

From mesh 2: i₂ = (4i₁ − 9)/7. Substitute into mesh 1 and clear the 7: 140 = 42i₁ − 16i₁ + 36 = 26i₁ + 36 → i₁ = 104/26 = 4, then i₂ = (16−9)/7 = 1.
Power audit: 20 V source delivers 20 × 4 = 80 W. Absorbed: 2 Ω → 32 W; 4 Ω → 3² × 4 = 36 W; 3 Ω → 3 W; the 9 V source (current into its +) absorbs 9 W. Total 80 W ✓. A source absorbing = a battery charging — normal, report it.

4 · The equivalence toolkit §2.4–2.5

Source transformation: V in series with R ⇄ I = V/R in parallel with R — identical at the terminals, in both extremes and everywhere between.

Thevenin: any linear two-terminal network ≡ V_th in series with R_th. Find them:

V_th = open-circuit voltage at the terminals.
R_th = resistance looking in with independent sources killed (V → short, I → open) — only valid when there are no dependent sources.
Always available (and mandatory with dependent sources): R_th = V_oc/i_sc.
Unpack: why V_oc/i_sc works

Apply the claim to the equivalent itself: open-circuited it shows V_th; short-circuited it drives i_sc = V_th/R_th. So the ratio of the real network's two measurable extremes is R_th. Two measurements, no dissection — this is also how it's done on a lab bench.

Norton: the same equivalence as a current source: I_N = i_sc in parallel with R_th (one source transformation away from Thevenin).

Worked example 3Thevenin, Norton, and the best load

12 V source in series with 4 Ω to node A; 8 Ω from A to ground; terminals A–ground. Find the Thevenin and Norton equivalents, then the load extracting maximum power.

V_th: open-circuit divider → 12 × 8/(4+8) = 8 V.
R_th: kill the 12 V (short) → 4 ∥ 8 = 8/3 Ω. Cross-check: i_sc = 12/4 = 3 A (the short bypasses the 8 Ω), and 8/(8/3) = 3 A ✓. Norton: 3 A ∥ 8/3 Ω.
Max power transfer: R_L = R_th = 8/3 Ω, giving P_max = V_th²/4R_th = 64/(32/3) = 6 W. (Check: load current 8/(16/3) = 1.5 A; 1.5² × 8/3 = 6 W ✓.)
Unpack: where R_L = R_th comes from

P = V_th²R_L/(R_th+R_L)². Differentiate with the quotient rule: numerator of dP/dR_L ∝ (R_th+R_L)² − R_L·2(R_th+R_L) = (R_th+R_L)(R_th − R_L), which is zero at R_L = R_th. Your daily-drill quotient rule, earning money.

5 · Superposition §2.6

In a linear circuit, any voltage or current = sum of contributions from each independent source acting alone, others killed (V → short, I → open). Mini-example: 12 V source + 4 Ω into node A, 3 A source into A, 8 Ω to ground. Alone: v′ = 12 × 8/12 = 8 V; v″ = 3 × (4∥8) = 8 V; total v_A = 16 V (nodal check: (16−12)/4 + 16/8 = 1 + 2 = 3 A ✓).

Two hard rules: dependent sources stay on in every partial circuit; and power never superposes (quadratic — here 8 Ω truly dissipates 16²/8 = 32 W, not 8 + 8 = 16 W). Superpose v and i only; compute p from totals.

Classic traps — the module's mark-losers

1) Shared-branch signs in mesh: from mesh k's walk it's always (i_k − neighbour), never the other way. 2) Killing sources wrongly: dead voltage source = short, dead current source = open — students swap these under time pressure. 3) Killing a dependent source: never — not in R_th, not in superposition; with dependent sources present, find R_th via V_oc/i_sc, or by a test source when there is no independent source (V_oc = i_sc = 0). And write the controller equation before the KCL/KVL, or you will have one unknown too many. 4) Superposing power: only v and i superpose. 5) "Maximum power" ≠ "maximum efficiency": at matched load, efficiency is exactly 50% — half of everything dies in R_th. 6) Mixed mesh directions: all clockwise, always; a negative answer just means the real current circulates the other way.

6 · Dependent sources: the exam pattern §1.8, §2.1–2.6

Past mid-sem papers put a diamond into nearly every Module 2 question, so this section is four exam-shaped solves. The whole method in two lines: (1) treat the diamond as an ordinary source of its kind, then write one controller equation expressing its controlling v_x or i_x in your unknowns; (2) never kill it — not for R_th, not in superposition. First time seeing this? The lesson §5 teaches it slowly.

Finding R_th	Allowed when
Kill independent sources, reduce	No dependent sources at all
R_th = V_oc/I_sc	At least one independent source (example 4)
Test source: kill independents, keep diamonds, push 1 A in, R_th = v_T/1 A	Always; the only route with no independent source (example 5)

Worked example 4Norton with a VCCS, via V_oc and I_sc

Find the Thevenin and Norton equivalents at a–b, and the maximum power a resistive load across a–b could draw.

Controller first. v_x sits across the 6 Ω with + on the source side, so v_x = 18 − v₁. The diamond pushes 0.5v_x = 0.5(18 − v₁) into node v₁.
V_oc. With a–b open, the 2 Ω carries nothing, so v_a = v₁. KCL at v₁, in = out: (18 − v₁)/6 + 0.5(18 − v₁) = v₁/3. Both left terms share the factor (18 − v₁), and 1/6 + 1/2 = 2/3, so (2/3)(18 − v₁) = v₁/3. Multiply by 3: 2(18 − v₁) = v₁ → 36 = 3v₁ → V_oc = v₁ = 12 V. (Here v_x = 6 V and the diamond pushes 3 A.)
I_sc. Short a–b: the 2 Ω now runs from v₁ to ground and carries v₁/2, which is I_sc. The diamond stays on, so KCL gains one outflow: (2/3)(18 − v₁) = v₁/3 + v₁/2. Multiply by 6: 4(18 − v₁) = 2v₁ + 3v₁ → 72 = 9v₁ → v₁ = 8 V, so I_sc = 8/2 = 4 A.
R_th = V_oc/I_sc = 12/4 = 3 Ω. Thevenin: 12 V in series with 3 Ω. Norton: 4 A in parallel with 3 Ω.
Max power: R_L = R_th = 3 Ω, P_max = V_th²/4R_th = 144/12 = 12 W.

The mark-loser: "kill the 18 V, reduce" gives 6 ∥ 3 + 2 = 4 Ω — wrong, because it silently deleted the diamond.

Worked example 5R_th by test source — no independent source anywhere

Find the Thevenin equivalent seen at a–b.

Choose the route. Nothing independent drives this network, so V_oc = 0 and I_sc = 0 — their ratio is 0/0, useless. Test source: push 1 A into a (and out of b), find v_a.
The 2 Ω is the only path from a into the network, so it carries the whole 1 A: v_a = v_c + 2 × 1.
KCL at c (diamond as a normal source): the 1 A arriving leaves down the 6 Ω and down the diamond: 1 = i_x + 0.5i_x = 1.5i_x → i_x = 2/3 A. The controller is the 6 Ω's own current, so v_c = 6 × 2/3 = 4 V.
v_a = 4 + 2 = 6 V, so R_th = 6 V / 1 A = 6 Ω. The Thevenin equivalent is a bare 6 Ω — no source, because the network has none.
Sanity check: the diamond always carries half the 6 Ω's current at the same voltage, so it behaves like a 12 Ω resistor; 6 ∥ 12 = 4, plus 2 gives 6 Ω ✓. Ignoring the diamond gives 8 Ω ✗.

Worked example 6Superposition with the diamond kept on

Using superposition, find v_a and i_x.

① 12 V alone (6 A → open; diamond stays, controlled by this circuit's i_x′). KCL at a: i_x′ = v′/4 + 0.5i_x′, i.e. 0.5i_x′ = v′/4. Controller: i_x′ = (12 − v′)/2, so (12 − v′)/4 = v′/4 → v′ = 6 V, i_x′ = 3 A.
② 6 A alone (12 V → wire, so the 2 Ω's left end is at 0 V). Controller: i_x″ = (0 − v″)/2. KCL at a: i_x″ + 6 = v″/4 + 0.5i_x″ → 0.5i_x″ + 6 = v″/4 → −v″/4 + 6 = v″/4 → v″ = 12 V, i_x″ = −6 A. (The diamond's value here is 0.5 × (−6) = −3 A: it pushes 3 A up. Negative is fine.)
Add: v_a = 6 + 12 = 18 V, i_x = 3 + (−6) = −3 A — 3 A actually flows from a back into the 12 V source.
Check on the full circuit: in at a: i_x + 6 = −3 + 6 = 3 A; out: 18/4 + 0.5(−3) = 4.5 − 1.5 = 3 A ✓.

The mark-loser: killing the diamond in each partial circuit gives 8 + 8 = 16 V. Independent sources take turns; the diamond is always on.

Worked example 7Working backwards from the load's power

The load R_L = 6 Ω dissipates 24 W. Find the magnitude of V_s and the current it supplies.

Start where the number is — at the load, and walk towards the source. P = v²/R gives v_b² = 24 × 6 = 144, so v_b = 12 V (magnitude; −12 V would give the same power, which is why the paper asks for a magnitude).
Currents at b: i_L = 12/6 = 2 A, the 12 Ω takes 12/12 = 1 A, so by KCL the 2 Ω feeds 3 A into b.
KVL round the right loop: the diamond must supply the 2 Ω's drop plus v_b: 3i₁ = 2 × 3 + 12 = 18 V.
Controller equation: 3i₁ = 18 → i₁ = 6 A — the current in the left loop.
Left loop: V_s = 6 × (3 + 6) = 54 V, supplying 6 A.
Audit: the diamond delivers 18 × 3 = 54 W = 18 W (2 Ω) + 12 W (12 Ω) + 24 W (load) ✓; the 54 V source delivers 54 × 6 = 324 W = 108 W + 216 W ✓. The diamond's power comes from a supply the diagram doesn't draw — that's what "dependent source" models.

7 · Your minimal prerequisite kit

Simultaneous linear equations, 2–3 unknowns, fast and correct — the module's engine. If elimination feels slow, drill five systems tonight; every exam problem here contains one.
All of Module 1, especially the sign convention and the divider formulas — Thevenin problems are divider problems wearing a coat.
Power forms p = vi = i²R = v²/R, for the audit.
The quotient rule (max-power derivation) — covered by the MATH daily drill.

8 · What to practise

Skill	Where	How many
Nodal analysis (incl. one supernode)	Bobrow §2.1–2.2 problems + tutorial sheet	5–6, auditing each
Mesh analysis (incl. one supermesh)	§2.3 problems + tutorial sheet	5–6
Source transformation chains; Thevenin & Norton	§2.4–2.5 problems	6–8, at least two with a dependent source (V_oc/i_sc route)
Superposition; max power transfer	§2.6 problems	4–5
Dependent sources: V_oc/I_sc, test source, superposition with the diamond on, working back from power	§2.4–2.6 problems with dependent sources + the Dependent sources drill	5–6, one of each shape in §6

Same habit as Module 1: tutorial sheet first, textbook to top up, and every solved circuit gets the 30-second power audit — it is the answer key you carry into a closed-book exam.