Electric Circuits¶
Syllabus mapping
PY-11-03 — Electricity and Magnetism. Covers: electric current; conventional current vs electron flow; DC vs AC; resistance and Ohm's law; ohmic vs non-ohmic conductors; electrical power; series and parallel circuits, including ammeter/voltmeter placement and combination-circuit problem solving.
Wording above is paraphrased for teaching use — cross-check against the official syllabus PDF (see Resources) before treating any line as verbatim NESA text.
You need to know: how to calculate current, resistance and power, the difference between conventional current and electron flow, how series and parallel circuits behave differently, and how to solve combination-circuit problems.
Assumed knowledge check¶
Before this page makes sense, you should be comfortable with:
- Algebra — rearranging multi-step equations
- Electrostatics — charge, potential difference
Core¶
Current¶
Current is the rate of flow of charge — coulombs per second (amperes).
Conventional current vs electron flow¶
Conventional current is defined as the direction positive charge would flow — from positive terminal to negative terminal through an external circuit. This is a historical convention, fixed before anyone knew what was actually moving inside a wire.
Electron flow is what's actually happening in a metal conductor: negatively charged electrons move from the negative terminal to the positive terminal — the opposite direction to conventional current. Unless a question specifically asks about electron flow, use conventional current — it's what circuit diagrams and formulas assume.
DC vs AC¶
- DC (direct current): current flows in one direction only (e.g. a battery)
- AC (alternating current): current periodically reverses direction (e.g. mains power)
Resistance and Ohm's law¶
- Ohmic conductors: resistance stays constant regardless of voltage/current — a graph of \(V\) against \(I\) is a straight line through the origin.
- Non-ohmic conductors: resistance changes with voltage/current — the \(V\)–\(I\) graph is curved. A filament globe is a classic example: as current flows, the filament heats up and its resistance increases, so the graph curves rather than staying straight.
Power¶
Both are the same quantity — pick whichever form matches the variables you already have, using \(V=IR\) to substitute between them if needed.
Worked example
A \(12\ \Omega\) resistor carries a current of \(2\ \text{A}\). Find the power dissipated.
Series circuits¶
Current is the same everywhere in a series circuit (only one path, no branching) — but voltage divides across each component.
- Ammeter placement: in series with the component (so all the current passes through it).
- Voltmeter placement: in parallel across the component (measuring the potential difference between its two ends).
Parallel circuits¶
Voltage is the same across each branch in a parallel circuit — but current divides between branches (more current through the lower-resistance branch).
🖼️ Diagram needed: series circuit, parallel circuit, and a combination circuit, with consistent circuit-symbol style and correct ammeter/voltmeter placement shown. TODO — placeholder until sourced/drawn.
Worked example
Two resistors, \(6\ \Omega\) and \(3\ \Omega\), are connected in parallel. Find the combined resistance.
Notice the combined parallel resistance (\(2\ \Omega\)) is less than either individual resistor — adding a parallel path always makes it easier for current to flow overall.
Practical¶
Suggested prac: combination circuit measurement and error analysis — build a circuit with both series and parallel sections, measure voltage/current at multiple points, compare against calculated predictions, and discuss sources of discrepancy (meter internal resistance, wire resistance, measurement uncertainty — see Working Scientifically). TODO — link this school's actual prac instructions once written.
Advanced
Full combination circuits — a mix of series and parallel sections in one circuit. The systematic approach: reduce parallel sections to their single equivalent resistance first, then treat the whole thing as an equivalent series circuit; work back outward once you need to find current/voltage at a specific point inside a reduced section.
Non-ohmic component problems — since a single resistance value doesn't apply, you generally need to read values directly off a given \(V\)–\(I\) graph rather than compute with a fixed \(R\).
Extension
Beyond the Physics 11–12 syllabus — won't appear in the HSC, included for interest / depth study inspiration.
📎 Depth study idea
Ohm's law (\(V=IR\)) is an empirical relationship — it works extremely well for many materials, but it's not a fundamental law of physics the way Newton's laws are. It emerges from a microscopic picture: charge carriers (electrons in a metal) drift through the material at a low average speed, colliding with the lattice, with current related to that drift velocity by \(I = nqv_dA\) (charge carrier density × charge × drift velocity × cross-sectional area — not on your data sheet, genuinely beyond Year 11 scope). A depth study could investigate why some materials are strongly non-ohmic (like a filament globe, or a diode) by connecting their resistance's dependence on temperature or voltage back to this microscopic picture, rather than treating "ohmic vs non-ohmic" as just a shape-of-graph classification.
Video/visual resources¶
- 🖥️ PhET Simulation — Circuit Construction Kit — TODO: confirm current PhET link (excellent fit for this whole page)
- 🎥 Khan Academy — TODO: source a series/parallel circuits explainer
- 🎥 Physics High — TODO: check for a NSW-syllabus-aligned circuits video
Check yourself¶
-
A charge of \(15\ \text{C}\) flows past a point in a circuit over \(5\ \text{s}\). Find the current.
Answer
\(I = \dfrac{q}{t} = \dfrac{15}{5} = 3\ \text{A}\)
-
Three resistors (\(4\ \Omega\), \(4\ \Omega\), \(4\ \Omega\)) are connected in series to a \(12\ \text{V}\) supply. Find the current in the circuit.
Answer
\(R_{series} = 4+4+4 = 12\ \Omega\)
\(I = \dfrac{V}{R} = \dfrac{12}{12} = 1\ \text{A}\)
-
A \(60\ \Omega\) resistor and a \(30\ \Omega\) resistor are connected in parallel across a \(12\ \text{V}\) supply. Find the current through each resistor, and the total current drawn from the supply.
Answer
Voltage is the same across each parallel branch: \(12\ \text{V}\) across both.
\(I_{60} = \dfrac{12}{60} = 0.2\ \text{A}\), \(I_{30} = \dfrac{12}{30} = 0.4\ \text{A}\)
Total current: \(I_{parallel} = 0.2+0.4 = 0.6\ \text{A}\)