EEE U111 · Electrical Sciences · Mid-sem prep

Circuit-analysis ladders: from a tiny circuit to the past paper

The past-paper questions look advanced because they stack several small ideas on top of each other. Each ladder below takes them apart: every rung is a small circuit with clean numbers that adds exactly one new idea, and the top rung is the real Oct 2024 question.

Start here

What looks "advanced" in these papers is mostly stacking. Oct 2024 Q2 is three ideas you can learn one at a time: write the controller equation, find Voc and Isc separately, and divide. Q4 is superposition plus one rule (the dependent source stays on). None of the rungs below needs anything beyond Module 2.

The target is the Oct 2024 paper. It was set in your semester slot and is the closest model of your exam.

Cold check · about 20 minutes

Do this before reading any solutions. It tells you which rung to start on, so you don't spend time on steps you can already do.

  1. Ladder 1, rung 1 (one loop with a dependent source): find i. 5 min.
  2. Ladder 2, rung 1 (two sources, superposition): find v. 5 min.
  3. Ladder 3, rung 1 (working back from load power): find Vs. 4 min.
  4. Open Oct 2024 Q2 and Q4, and for each write only the first moves: which method, which unknowns, and the controller equation. Don't solve. 6 min.
Where to start

Climb until you get stuck, then drop down one rung. In each ladder, find the first rung you couldn't do cold and start on the rung below it. That rung should feel easy. Redo it quickly, then climb. If rung 1 was fine and your first moves for Q2 and Q4 matched the past-papers page, skip straight to the top rungs.

Skip for now (this week) Mar 2025 Q1(a) (the Norton resistance comes out negative, −20/19 Ω, which is correct but unusual) and Mar 2025 Q4 (AC, and parts (c)–(d) are flawed as printed) aren't worth your time before 6 Oct. Everything on this page builds towards the Oct 2024 paper.

Ladder 1 · Dependent sources → Oct 2024 Q2 (Norton)

The one habit this ladder builds: a dependent source is written like any source, and then you add one extra equation (the controller equation) that says what its controlling variable equals in your unknowns. The lesson's dependent-sources section explains the idea; the rungs below practise it.

Rung 1New on this rung: the controller equation

🎧 Watch first (3 min): the controller-equation habit, narrated.

One loop. The diamond is a voltage source whose value is 0.5 × (the voltage across the 4 Ω).

Find the loop current i and the voltage vx.

Try it first: write KVL round the loop, then one more equation for vx.

Answer

i = 3 A, vx = 12 V. The diamond is at 0.5 × 12 = 6 V.

Worked solution
  1. KVL clockwise, starting at the bottom-left: up through the source (+24), then drops through 2 Ω (2i), 4 Ω (vx) and the diamond. On the bottom wire the current flows right to left, so it enters the diamond's + terminal, which makes the diamond a drop of 0.5vx: 24 − 2i − vx − 0.5vx = 0.
  2. That is one equation with two unknowns, so write the controller equation. vx is the voltage across the 4 Ω, with + at the top where the current enters: vx = 4i.
  3. Substitute: 24 − 2i − 4i − 2i = 0 ⇒ 8i = 24 ⇒ i = 3 A, so vx = 12 V.
  4. Check: the drops 6 + 12 + 6 = 24 V ✓.

Ignoring the diamond would give 24/6 = 4 A. Here the diamond behaves like an extra 2 Ω (it drops 2i), which is why the current is smaller.

Rung 2New on this rung: controller equation in nodal analysis

🎧 Watch first (3 min): the controller-equation habit, narrated.

One unknown node, a. The diamond pushes 0.25vx amps up into a, where vx is the voltage across the 4 Ω.

Find the node voltage va.

Try it first: KCL at a (in = out), then write vx in terms of va.

Answer

va = 18 V (and vx = 6 V, so the diamond pushes 1.5 A).

Worked solution
  1. KCL at a: in through the 4 Ω, (24 − va)/4, plus in from the diamond, 0.25vx; out through the 6 Ω, va/6.
  2. Controller equation: vx is the voltage across the 4 Ω, + on the left (24 V side): vx = 24 − va.
  3. Substitute: (24 − va)/4 + (24 − va)/4 = va/6, i.e. (24 − va)/2 = va/6. Multiply by 6: 72 − 3va = va, so va = 18 V.
  4. Check: vx = 6 V, so 1.5 A comes through the 4 Ω and 0.25 × 6 = 1.5 A from the diamond. In: 3 A. Out: 18/6 = 3 A ✓.

Ignoring the diamond gives a plain divider, 24 × 6/10 = 14.4 V. That is wrong, because the diamond adds current into a.

Rung 3New on this rung: the test source

🎧 Watch first (3 min): the test-source method, narrated.

No independent source at all. The diamond's value is 8i volts, where i is the current down through the 4 Ω.

Find the Thevenin resistance Rth seen from terminals a–b.

Try it first: with nothing to drive it, Voc = 0 and Isc = 0, and 0/0 tells you nothing. So push in a test current of 1 A and find the voltage it produces.

Answer

Rth = 4 Ω. (Treating the diamond as dead would give 6 ∥ 4 = 2.4 Ω, which is wrong.)

Worked solution
  1. Why a test source: with no independent source, nothing drives the circuit, so Voc = 0 and Isc = 0. Their ratio is 0/0. Instead, connect a 1 A current source pushing into terminal a and call the voltage it creates v. Then Rth = v/1.
  2. KCL at a: the 1 A leaves down through the 6 Ω and down the right branch: 1 = v/6 + i.
  3. KVL down the right branch, including the controller: the current i enters the diamond's + terminal, so the diamond is a drop of 8i: v = 4i + 8i = 12i, so i = v/12.
  4. Solve: 1 = v/6 + v/12 = v/4, so v = 4 V and Rth = 4 Ω.

Sense check: the right branch behaves like 4 + 8 = 12 Ω, and 6 ∥ 12 = 4 Ω ✓. If you "kill" the diamond by shorting it, you get 6 ∥ 4 = 2.4 Ω, the classic trap.

Rung 4New on this rung: Voc and Isc, two separate circuits

An independent source and a dependent one. The diamond sits in the path to the terminal, like the one in Oct 2024 Q2: vm − vn = 0.5va, where va is the voltage across the 6 Ω.

Find the Thevenin equivalent (Vth, Rth) at terminals a–b, and the Norton current.

Try it first: solve the circuit twice, once with a–b open and once with a–b shorted. Write the controller equation first: va = vn.

Answer

Vth = Voc = 12 V, IN = Isc = 2.4 A, Rth = 12/2.4 = 5 Ω. (Killing the diamond would give 3 ∥ 6 + 2 = 4 Ω, which is wrong.)

Worked solution
  1. Controller equation: va = vn, so vm = vn + 0.5vn = 1.5vn.
  2. Open circuit (a–b open): no current can flow through the 2 Ω, so none flows through the diamond either. Node n is then a plain divider: vn = 12 × 6/(3 + 6) = 8 V. With no current, the 2 Ω drops nothing, so Voc = vm = 1.5 × 8 = 12 V.
  3. Short circuit (a joined to b, so a is at 0 V): the 2 Ω now carries vm/2 = 1.5vn/2 = 0.75vn. That current flows out of node n through the diamond. KCL at n: (12 − vn)/3 = vn/6 + 0.75vn. Multiply by 12: 48 − 4vn = 2vn + 9vn, so 15vn = 48 and vn = 3.2 V. Then Isc = 0.75 × 3.2 = 2.4 A.
  4. Divide: Rth = 12/2.4 = 5 Ω. Norton: 2.4 A in parallel with 5 Ω.

This is Q2's method exactly. Q2 just has one more path (the 1 Ω and 1 V branch) feeding the terminal.

Top rungThe past paper: Oct 2024 Q2, 18 marks

As set · verbatim from the paperQ.2 find the Norton' s equivalent current source and resistance at the terminals 1-11 of the network shown Fig.3. [18 Marks]
Fig. 3, redrawn. The diamond is a voltage-controlled voltage source: VM − VA = 2Va, where Va is the voltage across the lower 2 Ω. The bottom wire (1′) is the reference.

What each rung gave you here: the controller equation Va = VN (rungs 1–2); the diamond in the path to the terminal (rung 4); Voc and Isc as two separate circuits, then divide (rung 4); and a test source to check RN (rung 3).

Answer

IN = 2 A, RN = 1 Ω (Voc = 2 V).

🎧 Narrated walkthrough of this exact question (about 7.7 min) · full written solution: past papers, Oct 2024 Q2.

Ladder 2 · Superposition → Oct 2024 Q4

Superposition turns one circuit with several sources into several easy circuits with one source each. There are only two rules for switching sources off: a dead voltage source becomes a short (a wire), and a dead current source becomes an open (the branch is removed). Dependent sources are never switched off.

Rung 1New on this rung: one source at a time, and the two kill rules

Two independent sources: 12 V and 3 A.

Use superposition to find v, the voltage across the 6 Ω.

Try it first: draw two partial circuits, one with each source switched off, and find each part of v.

Answer

v = 6 + 3 = 9 V.

Worked solution
  1. 12 V alone (the 3 A becomes an open, so that branch disappears): the 6 Ω and 3 Ω are in parallel, 6 × 3/(6 + 3) = 2 Ω, in series with the 2 Ω. Divider: v′ = 12 × 2/(2 + 2) = 6 V.
  2. 3 A alone (the 12 V becomes a short, so the 2 Ω now goes straight to the bottom wire): the 3 A sees 2 ∥ 6 ∥ 3. In conductances: 1/2 + 1/6 + 1/3 = 1, so the combination is 1 Ω, and v″ = 3 × 1 = 3 V.
  3. Add: v = 6 + 3 = 9 V.
  4. Check with one nodal equation: (12 − 9)/2 + 3 = 4.5 A in, 9/6 + 9/3 = 4.5 A out ✓.

Rung 2New on this rung: the dependent source stays on in every partial circuit

The same circuit, with the 3 Ω replaced by a diamond that pulls 0.5i down out of node a, where i is the current down the 6 Ω.

Use superposition to find v and i.

Try it first: still two partial circuits (one per independent source), but the diamond appears in both of them.

Answer

v = 8 + 4 = 12 V and i = 2 A. (Switching the diamond off in each part gives 9 + 4.5 = 13.5 V, which is wrong.)

Worked solution
  1. Controller equation (true in every partial circuit): i = v/6, so the diamond pulls 0.5v/6 = v/12 out of node a.
  2. 12 V alone (3 A open, diamond on): KCL at a: (12 − v′)/2 = v′/6 + v′/12 = v′/4. Multiply by 4: 24 − 2v′ = v′, so v′ = 8 V.
  3. 3 A alone (12 V shorted, diamond on): KCL at a: 3 = v″/2 + v″/6 + v″/12 = 9v″/12, so v″ = 4 V.
  4. Add, then use the controller: v = 12 V, so i = 12/6 = 2 A. Check: in, (12 − 12)/2 + 3 = 3 A; out, 2 + 0.5 × 2 = 3 A ✓.

Rung 3New on this rung: redrawing when a dead current source opens a branch

Three independent sources. The 4 Ω is in series with the 3 A source.

(a) Use superposition to find va. (b) Find the voltage at b, the top of the 3 A source.

Try it first: three partial circuits. Before any arithmetic, redraw each one. When the 3 A is switched off, what happens to the 4 Ω?

Answer

(a) va = 8 + 3 + 6 = 17 V. (b) vb = 17 + 3 × 4 = 29 V.

Worked solution
  1. 12 V alone (9 V shorted, 3 A open): the open current source leaves the 4 Ω dangling with no current, so the whole right branch disappears. What is left: 12 V, 3 Ω, then 6 Ω to the bottom wire (the shorted 9 V). Divider: v′ = 12 × 6/(3 + 6) = 8 V.
  2. 9 V alone (12 V shorted, 3 A open): the right branch is gone again. The 9 V drives through 6 Ω, and va is taken across the 3 Ω (to the shorted 12 V): v″ = 9 × 3/(3 + 6) = 3 V.
  3. 3 A alone (both voltage sources shorted): all 3 A flows through the 4 Ω into a, whatever its value, then splits between 3 Ω and 6 Ω to the bottom wire: 3 ∥ 6 = 2 Ω, so v‴ = 3 × 2 = 6 V.
  4. Add: va = 8 + 3 + 6 = 17 V. Check with KCL at a: (12 − 17)/3 + (9 − 17)/6 + 3 = −5/3 − 4/3 + 3 = 0 ✓.
  5. (b) The 3 A flows through the 4 Ω from b to a, so b is higher by 3 × 4 = 12 V: vb = 17 + 12 = 29 V.

The lesson of this rung: a resistor in series with a current source never changes the rest of the circuit (the source forces its current anyway). It only changes the voltage across the source itself.

Top rungThe past paper: Oct 2024 Q4, 18 marks

As set · verbatim from the paperQ.4. Using the superposition theorem, determine V1, the voltage across the 3-ohm resistor, in the circuit shown in Fig.5. [18 Marks]
Fig. 5, redrawn. The diamond is a current-controlled voltage source of value 4i, where i is the current down through the 3 Ω.

What each rung gave you here: three independent sources means three partial circuits (rungs 1 and 3); the 4i diamond stays on in every one, with its controller i = V1/3 (rung 2); and each partial circuit needs redrawing before any arithmetic (rung 3).

Answer

V1 = 30 − 48 + 12 = −6 V (so i = −2 A).

🎧 Narrated walkthrough of this exact question (about 7 min) · full written solution: past papers, Oct 2024 Q4.

Ladder 3 · Working backwards from power → Oct 2024 Q1(b)

When the question gives you a number at the load (a power or a current) and asks about the source, don't set up equations. Start at the load and walk back towards the source using Ohm's law and KCL, one element at a time.

Rung 1New on this rung: walking from the load back to the source

The 2 Ω load dissipates 8 W.

Find Vs and the current it supplies.

Try it first: from P = I²R, find the load current, then move left one element at a time.

Answer

Vs = 12 V, supplying 3 A.

Worked solution
  1. Load: IL2 × 2 = 8 ⇒ IL = 2 A, so vq = 2 × 2 = 4 V.
  2. The 1 Ω carries the same 2 A, so vp = 4 + 1 × 2 = 6 V.
  3. The 6 Ω carries 6/6 = 1 A. KCL at p: the 2 Ω must bring in 1 + 2 = 3 A.
  4. The 2 Ω drops 3 × 2 = 6 V, so Vs = 6 + 6 = 12 V, supplying 3 A.

Rung 2New on this rung: crossing a dependent-source "bridge"

Two halves that share only the bottom wire. The diamond pushes 2ix, where ix is the current in the left 4 Ω. The 3 Ω load dissipates 12 W.

Find ix, Vs, and the current Vs supplies.

Try it first: walk back through the right half until you reach the diamond, then use its value to cross over to the left half.

Answer

ix = 2 A, Vs = 14 V, supplying 2 A.

Worked solution
  1. Load: IL2 × 3 = 12 ⇒ IL = 2 A, so the load voltage is 6 V.
  2. The 1 Ω carries 2 A, so the node before it sits at 6 + 2 = 8 V. The right-hand 4 Ω then carries 8/4 = 2 A.
  3. KCL at the top of the diamond: it must supply 2 + 2 = 4 A. So 2ix = 4 ⇒ ix = 2 A. That is the bridge crossed.
  4. Left half: the 2 A flows through the series 4 Ω (a drop of 8 V), then into 4 ∥ 12 = 3 Ω (a drop of 2 × 3 = 6 V). So Vs = 8 + 6 = 14 V, supplying 2 A.

This has Q1(b)'s shape exactly. Q1(b) just has an extra resistor in the right half, and its controller is a voltage (Vx) instead of a current.

Top rungThe past paper: Oct 2024 Q1(b), 12 marks

As set · verbatim from the paperQ1. (b) For the network shown in Fig. 2, find the magnitude of Vo and the current supplied by it, given that the power loss in RL = 2Ω resistor is 18W. [12 Marks]
Fig. 2, redrawn. The dependent current source's value 2Vx is set by the voltage across the left 5 Ω.
Answer

|Vo| = 8.55 V, supplying 1.35 A (with Vx = 6.75 V).

Full written solution: past papers, Oct 2024 Q1.

After the ladders