EEE U111 · Electrical Sciences · Module 1 · Lectures 1–3

Basic Elements & Laws

Bobrow & Gupta (Asian ed.) §1.1–1.8 — sources, KCL & KVL, power and the sign convention, R, L, C, and dependent sources.

Start here — a fact worth knowing

Most of this module's content is 12th-physics territory (current electricity, capacitors), so nobody in the hall is ahead on substance. What's genuinely new — the passive sign convention, ideal sources as promises, dependent sources — is new to everyone. The marks in this module come from habits, not background: sign discipline and the power-balance check.

Seeing circuits for the first time (or 12th physics feels distant)? Read the first-time lesson first — same scope, taught slowly — then come back here to revise and drill. If the diagrams themselves are the problem, Reading a circuit diagram fixes that in 45 minutes. For reps: the circuit drills.

1 · The variables, power, and the sign convention

Current i = dq/dt (amperes), with a chosen arrow direction. Voltage v (volts) is always a difference between two points, marked + / −. Instantaneous power:

p(t) = v(t) · i(t)

The passive sign convention — the module's one rule to rule them all

If the current arrow enters the + terminal of an element, p = vi is power absorbed. Positive result → really absorbing; negative result → actually delivering (and nothing is wrong). Corollary used as an answer-check in every example below: in any circuit, Σ absorbed = Σ delivered at every instant.

2 · Sources: promises, not fixed quantities

An ideal voltage source fixes its terminal voltage for any current; an ideal current source fixes its current for any voltage. In both cases the other quantity is decided by the rest of the circuit. A dependent source (diamond symbol) is a source whose value is a formula in some other circuit variable — four flavours (VCVS, VCCS, CCVS, CCCS), read off the label: what it produces × what controls it. They exist because they model transistors and op-amps (lectures 23–26); handling rule in §6 below.

3 · KCL and KVL §1.3–1.4

KCL (charge conservation): at any node, Σ current in = Σ current out. KVL (energy conservation): around any closed loop, Σ rises = Σ drops. Sign recipe for KVL: fix a walking direction, record each element with the sign of the terminal you enter (enter at − → rise, enter at + → drop), and never switch schemes mid-loop.

Worked example 1One loop, two sources — with the full power audit

A 12 V source drives a series loop containing a 2 Ω resistor, a 4 Ω resistor, and a 3 V source connected opposing the 12 V one (current enters its + terminal).

The 3 V source's + faces right, so the clockwise current enters it at +: it opposes the 12 V and, as the audit below shows, absorbs power.

KVL clockwise (with current i clockwise): rise 12, drop 2i, drop 4i, drop 3. So 12 = 2i + 4i + 3 → 6i = 9 → i = 1.5 A.
Element voltages: 2 × 1.5 = 3 V and 4 × 1.5 = 6 V. Loop check: 3 + 6 + 3 = 12 ✓.
Power audit. Delivered: 12 V source, current leaves its + terminal → 12 × 1.5 = 18 W. Absorbed: resistors i²R = 4.5 W and 9 W; the 3 V source absorbs 3 × 1.5 = 4.5 W. Total absorbed 4.5 + 9 + 4.5 = 18 W = delivered ✓.
Unpack: why does a source absorb power here?

Current flows into the 3 V source's + terminal, so by the sign convention it absorbs p = vi > 0. Physically: the 12 V source is pushing current backwards through it — this is exactly what happens when a charger charges a battery. "Source" names the element, not its role in this particular circuit.

The audit costs 30 seconds and catches nearly every sign error. Run it on every solved circuit.

4 · Series, parallel, and the two dividers §1.5

Series (same current): R_eq = R₁ + R₂; voltage divider v₁ = v · R₁/(R₁+R₂). Parallel (same voltage): R_eq = R₁R₂/(R₁+R₂); current divider i₁ = i · R₂/(R₁+R₂) — the other resistor on top: more current takes the easier path.

Worked example 2Current source into a parallel pair

A 6 A current source feeds 3 Ω and 6 Ω in parallel. Find each branch current, the source voltage, and audit the power.

Both resistors span the same two nodes, so both see the source's voltage v — which the circuit sets, not the source.

Equivalent resistance: (3 × 6)/(3 + 6) = 2 Ω, so the source voltage is v = 6 × 2 = 12 V (a current source's voltage — set by the circuit).
Branch currents by Ohm: 12/3 = 4 A and 12/6 = 2 A. KCL check: 4 + 2 = 6 A in = out ✓.
Unpack: the current-divider shortcut

Directly: i₃ = 6 × 6/(3+6) = 4 A — opposite resistor on top. It's just Ohm's law twice: both branches share v = i·R_eq, so branch 1 carries v/R₁ = i·R_eq/R₁, and R_eq/R₁ = R₂/(R₁+R₂).
Power audit: source delivers 12 × 6 = 72 W; resistors absorb v²/R = 144/3 = 48 W and 144/6 = 24 W; 48 + 24 = 72 W ✓.

5 · Capacitors and inductors §1.6–1.7

Capacitor: i = C dvdt, E = ½Cv² Inductor: v = L didt, E = ½Li²

Perfect duals (swap v↔i, C↔L). The four facts used constantly:

DC steady state: capacitor → open circuit; inductor → short circuit (plain wire).
Continuity: capacitor voltage cannot jump; inductor current cannot jump. (These become the initial conditions of every transient problem in module 3.)

Steady-state mini-example: 10 V source in series with 2 Ω, an inductor (L = 0.5 H), and 3 Ω, with a capacitor (C = 100 µF) connected across the 3 Ω. At steady state the inductor is a wire and the capacitor an open, so i = 10/(2+3) = 2 A and v_C = drop across 3 Ω = 6 V.

Unpack: the stored energies

E_L = ½ × 0.5 × 2² = 1 J; E_C = ½ × (100×10⁻⁶) × 6² = 1.8 mJ. Watch the prefixes: µF × V² gives µJ-scale numbers — writing 1.8 J here is the classic prefix slip.

6 · Dependent sources in action §1.8

The handling rule: write KCL/KVL treating the diamond as a normal source, but immediately express its controlling variable in terms of your unknowns — it is a formula, not a new unknown.

Worked example 3A series loop with a VCVS

A 12 V source drives a series loop through R₁ = 2 Ω, R₂ = 4 Ω, and a dependent voltage source of value 3v₁ (absorbing orientation), where v₁ is the voltage across R₁.

The diamond's value is a formula in v₁, which is measured back at R₁ — substitute it before writing KVL, not after.

One loop → one unknown current i. Express the controller at once: v₁ = 2i, so the diamond's value is 3v₁ = 6i.
KVL: 12 = 2i + 4i + 6i = 12i → i = 1 A, v₁ = 2 V, dependent source = 6 V.
Power audit: independent source delivers 12 W; absorbed: R₁ 2 W, R₂ 4 W, dependent source 6 × 1 = 6 W; total 12 W ✓.
Unpack: can a dependent source deliver instead?

Yes — routinely (that's amplification). Its role falls out of the solved signs, exactly like the 3 V source in example 1. Never assume a direction for its power; compute it from the sign convention.

Classic traps — the module's mark-losers

1) Forcing every power positive. Negative absorbed power = delivering; report it, don't "fix" it. 2) Switching KVL sign schemes mid-loop. Pick enter-at-sign, keep it for the whole walk. 3) Misreading topology: series means same current, parallel means same voltage — decide by connection, not by how the diagram is drawn. 4) Making the voltage source fix its current (or a current source its voltage) — the circuit decides those. 5) Treating a dependent source as an extra unknown instead of substituting its formula immediately. 6) Prefix slips: keep µ, m, k straight — mA × kΩ = V is safe; µF with anything needs care (see §5's unpack).

7 · Your minimal prerequisite kit

Everything this module assumes. If a line is shaky, patch just that line:

Ohm's law v = iR and the three power forms p = vi = i²R = v²/R (12th physics).
Series/parallel resistance and both divider formulas (§4 above is the refresher).
Solving 2–3 simultaneous linear equations, fast — barely needed this module, but lectures 4–8 (nodal/mesh) run on it. Front-run it now: 15 minutes of practice, twice this week.
Derivatives of polynomials and exponentials (for C dv/dt, L di/dt) — the MATH daily drill already builds exactly this.
SI prefixes p, n, µ, m, k, M — instant conversion, both directions.

8 · What to practise

Tutorial sheets: sheets 1–2 with hints and verified solutions, and the tutorial practice drill. Sheet 1's graph, average, rating and wire questions lean on the Module 1 extras lesson, which is outside §1.1–1.8.

Skill	Where	How many
KCL/KVL on single- and two-loop circuits, with power audit	Bobrow ch. 1 problems (§1.3–1.4) + tutorial sheet	6–8, auditing every one
Series/parallel reduction, both dividers	§1.5 problems	4–6 (mixed reductions)
L & C: element equations, steady-state, stored energy	§1.6–1.7 problems	4–5
Circuits with one dependent source	§1.8 problems + tutorial sheet	at least 3 — this is the new skill

Tutorial-sheet problems first (they come with discussion, and surprise tests draw from current material); textbook problems to top up. Check once whether your copy of Bobrow prints answers to end-of-chapter problems — if not, verify your work with the power audit instead: it's a built-in answer key.