Circuit analysis, taught from zero

1 · Why you need a method, not cleverness

Module 1 solved circuits by spotting things: "it's one loop, so KVL gives the current." That works until the circuit has three loops and five nodes, and every path you stare at tangles into another. The fix is what engineering always does when cleverness stops scaling: a procedure that works every time, without insight.

Both procedures in this module do the same thing under the hood — they choose a small set of unknowns that automatically respect one of Kirchhoff's laws, then use the other law to generate exactly enough equations. Solve the simultaneous equations; every voltage and current in the circuit falls out.

2 · Nodal analysis: give every junction a height §2.1–2.2

The idea in one image: voltage is like height, so pick one node as sea level — call it the reference, or ground, 0 V — and label every other node with its height above sea level: v₁, v₂, … Those are your only unknowns. Any element's voltage is then just a difference of two heights, which is why KVL is automatically satisfied — walk any loop of heights and you return to your starting height by arithmetic alone.

The recipe:

1. Choose ground (pick the node with the most connections — fewer terms later).
2. Label the other node voltages.
3. Write KCL at each labelled node, expressing every resistor current with Ohm's law: the current flowing from node a to node b through R is (v_a − v_b)/R — my height minus yours, over the resistance.
4. Solve the simultaneous equations.

Let's work one properly. Here it is — and notice the ground symbol has already been placed on the bottom rail, so that whole rail is our 0 V reference and only two node voltages are unknown:

Now write KCL at each labelled node. Read every term straight off the picture:

Node 1 — in = out: 6 = v₁/2 + (v₁ − v₂)/2. (Into the node: the 6 A. Out: down to ground, and across to node 2.)

Node 2 — in = out: (v₁ − v₂)/2 = v₂/4 + 1. (In: the current arriving from node 1. Out: down to ground, and the 1 A the source removes.)

Clear fractions (×2 and ×4) and tidy: 2v₁ − v₂ = 12 and 2v₁ − 3v₂ = 4. Subtract: 2v₂ = 8 → v₂ = 4 V, then v₁ = 8 V. Every current now falls out by Ohm's law: 4 A down the first 2 Ω, 2 A across, 1 A down the 4 Ω — and KCL holds at both nodes by inspection. (The notes page reruns this with the full power audit.)

One wrinkle: voltage sources in a nodal problem. If a source connects a node to ground, it's a gift — that node's voltage is simply known, one unknown gone. If it floats between two non-ground nodes, the standard move is the supernode: treat the two nodes as one island for KCL, and add the source's equation v_a − v_b = V to the system. Know the idea; drill it on the tutorial sheet.

3 · Mesh analysis: currents that go in rings §2.3

The dual method. A mesh is a minimal "window pane" loop of the circuit. Assign each mesh its own circulating current — all clockwise, always, no exceptions (consistency is the whole game) — and those are your unknowns. KCL is now automatic: a loop current enters and leaves every node it touches. You write KVL once around each mesh.

The one subtlety worth slowing down for: an element on the border between two meshes carries both loop currents at once — its actual current is the difference. Walking around mesh 1, the shared resistor's drop is R(i₁ − i₂); walking mesh 2 it's R(i₂ − i₁) — my current minus the neighbour's, always from the perspective of the mesh you're walking.

Here is the worked example — two window panes side by side, sharing the middle branch. Both mesh currents are drawn clockwise, because they always are:

Mesh 1 (walk clockwise): rise 20, drop 2i₁, drop 4(i₁ − i₂): 20 = 6i₁ − 4i₂.

Mesh 2 (walk clockwise): drop 4(i₂ − i₁), drop 3i₂, drop 9: 4(i₂ − i₁) + 3i₂ + 9 = 0 → 7i₂ = 4i₁ − 9.

Substitute and solve: i₁ = 4 A, i₂ = 1 A (algebra unpacked on the notes page), so the shared 4 Ω carries 4 − 1 = 3 A downward. The 9 V source absorbs power — current is being pushed into its + terminal, a battery being charged, exactly like Module 1's example 1.

The wrinkle here is the mirror of nodal's: a current source on a mesh border fixes the difference of two mesh currents (the supermesh idea — merge the two panes for KVL and add i₁ − i₂ = I). A current source on an outer edge is a gift: that mesh current is known outright.

4 · The theorems: replacing circuits with smaller ones §2.4–2.6

Everything so far solves a circuit as given. Lectures 6–8 add a different power: rewriting a circuit into a simpler one that behaves identically. Four rewrites and one bookkeeping principle.

Source transformation — the warm-up rewrite

A voltage source V in series with R is indistinguishable, from its two terminals, from a current source I = V/R in parallel with the same R. Check the two extremes: open-circuited, both show voltage V = IR; short-circuited, both push I = V/R. Same behaviour at the ends, and (being linear) everywhere between. Use it to massage a circuit toward whichever method you prefer.

Thevenin's theorem — the big one

Here is the module's most-used result. Take any linear two-terminal network — dozens of sources and resistors, arbitrary mess — and as far as the outside world can tell, it is exactly equivalent to one voltage source V_th in series with one resistor R_th.

Why you should care: because circuits are usually "a big network driving one interesting element." Thevenin shrinks the big network to two components, once, and then every question about the interesting element becomes a one-loop Module 1 problem. Change the load, reuse the same equivalent — that's the whole workflow of amplifier design later.

Finding the two numbers:

• V_th = the open-circuit voltage across the terminals (remove the load, compute the dangling voltage).
• R_th = the resistance seen looking into the terminals with all independent sources killed: voltage sources → shorts (a dead battery is a wire), current sources → opens (a dead current source pushes nothing — a gap). Series/parallel reduce what remains.
• Cross-check (and the required route when dependent sources are present, since they never get killed): short the terminals, find i_sc, and use R_th = V_th/i_sc.

Worked slowly, with pictures. The network: a 12 V source in series with 4 Ω reaching node a; an 8 Ω from a down to ground; the terminals we care about are a–b.

Finding V_th. Leave a–b open. Nothing is connected there, so no current flows out of node a — which means the 12 V simply divides across the 4 Ω and 8 Ω in series. (That "no current flows into an open branch" step is the picture from Reading circuits §5.) So V_th = 12 × 8/(4+8) = 8 V.

Finding R_th. Kill the independent source — a voltage source becomes a wire — and look back in through the terminals:

The cross-check. Short a to b instead: now the 8 Ω has a wire across it, so it's bypassed entirely (Reading circuits §5 again), leaving i_sc = 12/4 = 3 A. Then R_th = V_th/i_sc = 8/3 Ω ✓ — agrees, so both numbers are trustworthy.

Norton's theorem is the same result wearing the other uniform: a current source I_N = i_sc in parallel with the same R_th. Thevenin ⇄ Norton is just one source transformation. Here: 3 A in parallel with 8/3 Ω.

Maximum power transfer — a calculus payoff

Connect a load R_L to a Thevenin equivalent. Which R_L extracts the most power? Too small → big current but tiny load voltage; too big → the opposite. The sweet spot, found by writing P = V_th²R_L/(R_th+R_L)² and setting dP/dR_L = 0 (the notes page unpacks the differentiation — your MATH drill in action):

For our example: R_L = 8/3 Ω, P_max = 64/(32/3) = 6 W. One honesty note the exam likes: at matched load, half the power dies in R_th — maximum power ≠ maximum efficiency (efficiency there is exactly 50%).

Superposition — one source at a time

Linear circuits obey a beautiful accounting rule: the response (any voltage or current) equals the sum of the responses to each independent source acting alone, with the others killed (same killing rules: V→short, I→open). Complex circuit, several sources → several easy circuits, added up.

Mini-example: a 12 V source in series with 4 Ω feeds node a; a 3 A source also feeds node a; an 8 Ω runs from a to ground. One circuit becomes two easy ones:

Panel ②, 12 V alone: the 3 A branch is an open gap, so no current leaves node a that way and the 12 V just divides across 4 Ω and 8 Ω — v′ = 12 × 8/12 = 8 V. Panel ③, 3 A alone: with the 12 V shorted to a wire, the 4 Ω now runs from a to ground alongside the 8 Ω, so the 3 A pushes into 4 ∥ 8 = 8/3 Ω, giving v″ = 3 × 8/3 = 8 V. Add them: v_a = 8 + 8 = 16 V.

Confirm with one nodal equation on the original: (12 − v)/4 + 3 = v/8. Multiply by 8: 2(12 − v) + 24 = v → 48 = 3v → v = 16 V ✓.

Two rules that save marks: dependent sources are never killed — they stay on in every partial circuit; and power does not superpose (it's quadratic: here the true power in the 8 Ω is 16²/8 = 32 W, nothing like 8²/8 + 8²/8 = 16 W). Superpose voltages and currents only; compute power at the end, from totals.

5 · Dependent sources in every method §1.8, §2.1–2.6

Here is a fact about the past mid-sem papers that should shape your revision: almost every Module 2 question has a diamond in it. A Norton equivalent with a "2.5v_x" source, superposition with a "4i" source, a mesh problem with a current-controlled source on a shared branch. The methods above don't change at all — there is exactly one new habit, plus one new way of finding R_th. That is the whole of this section.

Recall from Module 1 what the diamond is: a source whose value is a formula in some other quantity of the circuit — 2i_x, 0.5v_x. It is not an energy input you can switch off; it is a rule the circuit obeys (it models the inside of a transistor or op-amp, which is why the course cares).

The one new habit: write it like a source, then add its controller equation

In nodal or mesh analysis, treat the diamond exactly like an ordinary source of the same kind — a current diamond contributes its current to a KCL equation, a voltage diamond contributes its voltage to a KVL equation. Then write one extra equation: the controlling quantity (i_x or v_x) expressed in terms of your unknowns (node voltages or mesh currents). Substitute it in, and you are back to an ordinary system with the ordinary number of unknowns.

Watch it on the smallest possible circuit:

1. KCL at a, diamond treated as a normal source: in = out gives i_x + 2i_x = v_a/6.
2. Controller equation: i_x is the current through the 2 Ω from the source's top (10 V) to node a, so i_x = (10 − v_a)/2.
3. Substitute and solve: 3(10 − v_a)/2 = v_a/6. Multiply both sides by 6: 9(10 − v_a) = v_a, so 90 = 10v_a and v_a = 9 V. Then i_x = 0.5 A and the diamond pushes 1 A. Check KCL: in 0.5 + 1 = 1.5 A, out 9/6 = 1.5 A ✓.

Mesh analysis works the same way: a voltage diamond such as 3i goes into a KVL walk as a rise or drop like any battery, and the controller equation writes that i in mesh currents (if it sits on a shared branch it is a difference of two mesh currents).

Supernode and supermesh with a diamond. Nothing new — combine the two ideas. A voltage diamond floating between two non-ground nodes makes a supernode whose constraint is v_a − v_b = (the diamond's formula); a current diamond on a border between two meshes makes a supermesh whose constraint is i₁ − i₂ = (the diamond's formula). Then add the controller equation as usual. Count equations before you start: one per unknown, plus one controller equation per diamond.

Thevenin and Norton: three routes to R_th

V_th (open-circuit voltage) and I_N (short-circuit current) are found exactly as before — the diamond stays in the circuit and you solve with nodal or mesh plus its controller equation. The question is R_th, and there are three routes:

Why route 1 fails with a diamond: its current or voltage depends on what happens elsewhere, so it is neither a wire nor a gap — you can't replace it with either. You must keep it and let the equations decide. Why route 3 is needed: a network made only of resistors and diamonds has nothing driving it, so V_oc = 0 and I_sc = 0, and 0/0 tells you nothing. So you drive it yourself with a test source and see how it responds — the ratio of your push to its response is its resistance. (Why 1 A? Any value works, because everything in the circuit scales with it; 1 A just makes the division trivial.)

Run it. KCL at a: the 1 A test current leaves through the 6 Ω and through the 3 Ω branch. The 3 Ω has v_a at its top and 0.5v_a at its bottom (the diamond's + terminal), so it carries (v_a − 0.5v_a)/3:

Had you ignored the diamond and treated the 3 Ω as going straight to ground, you'd have said 6 ∥ 3 = 2 Ω — wrong. The diamond "props up" the bottom of the 3 Ω, so less current flows through it and the network looks more resistive. The notes page has two exam-level versions of this (worked examples 4 and 5).

Superposition with a diamond on board

Superposition sums the effects of the independent sources, one at a time. The diamond is not one of them — it is part of the circuit's wiring rules — so it stays on in every partial circuit, with its formula now written in that partial circuit's own controlling variable (i_x′ in the first, i_x″ in the second). Each partial circuit then needs its controller equation, which is why exam superposition questions with a diamond are really "two small nodal problems." A diamond's value can even come out negative in one partial circuit — that just means it pushes the other way there. Worked example 6 on the notes page shows the whole thing, including what goes wrong if you kill the diamond.

Power of a dependent source

Exactly the Module 1 rule, p = vi with the passive sign convention: if current enters the diamond's + terminal it absorbs vi; if current leaves its + terminal it delivers vi. For a current diamond you need its terminal voltage (from your node voltages); for a voltage diamond you need its current. In the first example of this section: the diamond pushes 1 A up out of its top, and its top is at 9 V above ground, so it delivers 9 × 1 = 9 W. Audit: the 10 V source delivers 10 × 0.5 = 5 W; the resistors absorb 0.5² × 2 + 9²/6 = 0.5 + 13.5 = 14 W = 5 + 9 ✓. Diamonds can deliver power — they model transistors, which draw it from a supply the diagram doesn't show.

6 · You're ready — what to do next

That's lectures 4–8: two systematic solvers and five rewriting tools. This module rewards reps more than any other — the methods only feel mechanical after you've run them a dozen times:

• Get the reps on the circuit drills page — fresh nodal, mesh and Thevenin problems with drawn circuits and a worked method behind every answer, including a Dependent sources topic (Norton with a VCCS, test-source R_th).
• Work the notes page examples — full solves with the power audit at the end of each; examples 4–7 are the dependent-source exam pattern.
• Tutorial sheet first, always — supernode and supermesh drill lives there, and surprise tests draw from current material.
• Textbook: 8–10 problems across §2.1–2.3 (nodal + mesh), 8–10 across §2.4–2.6 (theorems) — including at least two Thevenin problems with a dependent source, done via V_oc/i_sc.