Circuits from zero

1 · Two quantities run everything: current and voltage

Strip away every symbol and a circuit is just this: charged particles being pushed around closed loops of wire. Two numbers describe the situation at any point, and every single thing in this course is about one or both of them.

Current i — the flow itself. Stand at a cross-section of wire and count charge passing per second: that rate is the current, in amperes. 1 A = 1 coulomb/second. Formally i = dq/dt — a derivative, but the meter-reading intuition is what you'll use. Current has a direction: we draw an arrow and count flow along the arrow as positive. (Convention counts the flow direction of positive charge; the electrons actually drift the other way, and it never matters.)

Voltage v — the push. Voltage is energy per unit charge, in volts: 1 V = 1 joule/coulomb. The crucial habit, from day one: voltage is always between two points — a difference, like height. "The voltage at A" is sloppy; "the voltage of A above B" is exact. We mark the two points + and − and say the + side is v volts above the − side.

The water analogy, used with care: current is the flow rate in a pipe, voltage is the pressure difference driving it, and a battery is the pump. It's good for intuition and it fails politely later (nothing "spills out" of a cut wire) — use it, don't marry it.

2 · The cast of characters: sources and the resistor

A circuit diagram is a set of elements connected by ideal wires (zero resistance — a wire is just "the same point, stretched"). Module 1's cast has five members. First the three you half-know:

The ideal voltage source holds a fixed voltage between its terminals — say 12 V — no matter what current flows through it. It promises a pressure difference, and lets the rest of the circuit decide the flow. A good battery approximates one.

The ideal current source is the mirror image: it forces a fixed current through itself — say 2 A — no matter what voltage appears across it. Less familiar from school, equally important here.

The resistor converts electrical energy to heat, and obeys the law you know: v = iR — voltage across is proportional to current through. The constant R is resistance in ohms (Ω). Two consequences you also know, restated in course language:

• Series (same current through both): resistances add. R_eq = R₁ + R₂. The voltage splits in proportion to resistance — the voltage divider: v₁ = v · R₁/(R₁+R₂).
• Parallel (same voltage across both): R_eq = R₁R₂/(R₁+R₂). The current splits inversely to resistance — the current divider: i₁ = i · R₂/(R₁+R₂) (note: the other resistor on top — more current takes the easier path).

3 · Power, and the sign convention that runs the course

Push a charge through a pressure difference and energy changes hands. The rate of that energy transfer is power:

Fine — but energy transferred which way? Is the element consuming energy or supplying it? This is decided by a bookkeeping rule so central the course is unplayable without it, the passive sign convention:

Why this matters: in every circuit, energy is conserved, so total power absorbed = total power delivered, at every instant. That gives you a free, complete answer-check on any solved circuit — add up all the powers with correct signs and the total must be zero. The worked examples on the notes page run this check every time; make it your habit too.

One more vocabulary item from the handout: p(t) = v(t)·i(t) is called instantaneous power because both v and i may vary with time; for now (DC) they're constants, and the distinction becomes interesting in the AC module.

4 · The two laws every circuit obeys §1.3–1.4

Everything in lectures 1–8 is these two laws plus Ohm's law, applied systematically. Both are conservation laws wearing circuit clothes.

KCL — charge doesn't pile up

A node is a junction where elements meet. Charge can't accumulate at a node (there's nowhere for it to go — a node has no volume), so Kirchhoff's Current Law says: at every node, at every instant,

That's it. Three wires meet; 3 A comes in on one, 4 A on another; then 7 A must leave on the third. It's the traffic rule of junctions: cars in = cars out.

KVL — a round trip costs nothing

Voltage is like height. Walk any closed loop in a circuit — climbing up through sources, dropping down through resistors — and you must return to your starting height. Kirchhoff's Voltage Law: around any closed loop,

The practical recipe (and where the sign errors breed — slow down here): pick a loop, pick a walking direction, and walk it once, writing each element's voltage with the sign of the terminal you enter first: enter at − and leave at + is a rise (+); enter at + is a drop (−). The signed sum equals zero. Any consistent scheme works; what kills marks is switching schemes mid-loop.

A narrated walk: a 12 V source pushing current clockwise through a 2 Ω then a 4 Ω resistor in one loop. The current is some i, so the resistors drop 2i and 4i. Walking clockwise: up 12 (through the source, − to +), down 2i, down 4i, and we're home: 12 = 2i + 4i = 6i, so i = 2 A. The drops are 4 V and 8 V — and 4 + 8 = 12, the loop closes. KVL solved the circuit in one line.

5 · The calculus elements: capacitor and inductor §1.6–1.7

Resistors relate v and i by a number. The other two passive elements relate them by a derivative — they have memory, and they store energy instead of burning it.

The capacitor is two plates that store charge: q = Cv, with capacitance C in farads. Differentiate and remember i = dq/dt:

Read the equation like a sentence: current flows only while the voltage is changing. Steady voltage → zero current. Two consequences you'll use constantly:

• In DC steady state a capacitor is an open circuit — once everything settles, no current flows through it.
• A capacitor's voltage cannot jump — an instant jump would need infinite current. (Its current may jump.)

The inductor is the mirror image — a coil that resists changes in current through magnetic effects:

Voltage appears only while the current is changing. So: in DC steady state an inductor is a plain wire (a short circuit), and an inductor's current cannot jump (its voltage may). Notice the perfect duality — swap v↔i and C↔L and each capacitor fact becomes the inductor fact. Learn one column, get the other free.

These two elements are why the course needed your calculus: when module 3 asks what happens in the moments after a switch flips, the answer is a differential equation, and d/dt fluency (the MATH daily drill) becomes circuit fluency.

6 · Dependent sources — the new idea in this module §1.8

Everything above had a whiff of 12th physics. This one is genuinely new, to everyone: a dependent (controlled) source is a source whose value is set by a voltage or current elsewhere in the circuit. Its symbol is a diamond, and its label is a formula, like "3v₁" or "0.5i_x", where v₁ or i_x is measured somewhere else on the diagram.

Why would anyone invent that? Because it's how amplification is modelled. A transistor is, roughly, a device where a small quantity at one place controls a big flow at another — a valve, where a light touch on the handle governs a torrent. When the course reaches BJTs and op-amps, their circuit models are made of exactly these diamonds. Learn dependent sources now and lecture 23 becomes bookkeeping; skip them and it becomes magic.

There are four flavours, named by what controls × what's produced — voltage-controlled voltage source (VCVS), voltage-controlled current source (VCCS), current-controlled voltage source (CCVS), current-controlled current source (CCCS). You don't memorise the acronyms; you read them off the label: what kind of quantity does the formula produce, and what kind does it depend on?

How to handle one in analysis — the only rule you need: treat it as a normal source while writing KCL/KVL, but immediately express its controlling variable in terms of your circuit unknowns. It is not a new unknown; it's a formula wearing a source costume. Here is the example:

Series loop → one current i. The controlling variable, expressed in unknowns at once: v₁ = 2i, so the diamond's value is 6i. KVL: 12 = 2i + 4i + 6i = 12i → i = 1 A. The dependent source behaves like a 6 V drop — but only because the solution made it so. (Full version with the power check: notes page, worked example 3.)

7 · You're ready — what to do next

That's lectures 1–3 in full: current and voltage, sources, the sign convention, KCL/KVL, series/parallel machinery, the two storage elements, and dependent sources. Turn it into marks:

• Get the reps on the circuit drills page — generated dividers, KCL/KVL and power-sign problems with drawn circuits and a worked method behind every answer.
• Work the notes page — its three worked examples do the full exam-style bookkeeping (including the power-balance check every time), and its traps section names the exact sign errors that cost marks.
• Do the tutorial-sheet problems the week they're set — with surprise open-book tests worth 10%, staying current is the exam strategy.
• Textbook problems: end-of-chapter problems tied to §1.2–1.8 — 8–10 across KCL/KVL, dividers, and at least two with a dependent source. (Check once whether your Bobrow copy prints answers; if not, tutorial problems first.)