Are the circuit equations on the data sheet?

Yes. NESA's sample data sheet for the new (2025) syllabus includes I = q ÷ t, V = IR, P = VI, P = I^2R and the series and parallel rules. P = V^2 ÷ R is not on it, but you can get it from P = VI and V = IR. Check which sheet your school gives you for Year 11 exams.

What is the difference between conventional current and electron flow?

Conventional current is the direction positive charge would flow: from the positive terminal, around the circuit, to the negative terminal. Electron flow is the real motion of electrons in a metal, which is the opposite way. Circuit rules use conventional current.

Why does adding a resistor in parallel lower the total resistance?

Each new branch is another path for charge, so more current flows for the same voltage. More current at the same voltage means less total resistance, which is why Rparallel is always smaller than the smallest branch.

What makes a conductor ohmic?

Its resistance stays the same as the voltage changes (at a constant temperature), so current is proportional to voltage and the I–V graph is a straight line through the origin. A metal resistor kept cool is ohmic; a filament lamp is not.

Why are household appliances connected in parallel?

Each appliance then gets the full mains voltage and can be switched on or off without affecting the others. In series they would share the voltage, and one switched off would break the circuit for all of them.

In a circuit, conventional current flows clockwise. Which way do the electrons in the wires move?

Back and forth with no overall direction

They don't move; only energy moves

The current through a filament lamp is measured as the voltage across it is increased. Which statement is correct?

Its resistance rises as the current rises, because the filament gets hotter.

Its resistance stays constant, so its I–V graph is a straight line.

Its resistance falls as the current rises.

It has no resistance once it is glowing.

Why is a voltmeter connected in parallel with a component?

It measures the energy per coulomb between the two ends of the component, so it must connect to both ends.

So that all of the circuit's current goes through it.

Because it has a very small resistance.

Because it measures the charge flowing through the component.

Year 11 Physics · Electricity and magnetism · Electric circuits · 14 min read

Electric circuits

Every circuit question in Year 11 Physics comes down to three quantities: current (how much charge flows each second), voltage (how much energy each coulomb of charge carries between two points) and resistance (how hard it is to push charge through). V = IR ties them together, and the series and parallel rules tell you how they share out around a circuit.

This guide covers current and charge, conventional current, Ohm's law and ohmic conductors, power and energy, series and parallel circuits, where meters go and how to solve combination circuits. Then come the common mistakes, a quick quiz and a full worked example.

What you need to know

Use I = q ÷ t to calculate electric current, and relate the direction of conventional current to the way charge actually flows.
Explain why an external voltage drives electrons through a metal conductor connected to it, and compare alternating current (AC) with direct current (DC).
Explain why temperature, length, cross-sectional area and material each affect a conductor's resistance.
Analyse how voltage, current and resistance relate with V = IR, and examine how current varies with voltage for ohmic and non-ohmic conductors in a practical.
Relate energy transformations to power, and use P = W ÷ Δt = ΔE ÷ Δt, P = VI and P = I²R (with conservation of energy) to solve problems, including energy lost in circuits.
Compare, draw and analyse series and parallel circuits, and explain why voltmeters go in parallel and ammeters in series.
Use the series rules (same current everywhere; V_series = V₁ + V₂ + …; R_series = R₁ + R₂ + …) and the parallel rules (1 ÷ R_parallel = 1 ÷ R₁ + 1 ÷ R₂ + …; I_parallel = I₁ + I₂ + …).
Solve for voltage, current and resistance in combination circuits, and in a practical compare measured values with calculated ones and account for the differences.

The equations you will use:

I = q ÷ tCurrent in amperes (A) = charge in coulombs (C) ÷ time in seconds. Convert minutes to seconds first.

V = IRVoltage (V), current (A), resistance (Ω). Use the voltage across that resistor and the current through it.

P = W ÷ Δt = ΔE ÷ ΔtPower is the rate of energy transfer, in watts (J s⁻¹).

P = VI P = I²RElectrical power. I²R is the heating in a resistor or a wire: the energy 'lost' in transmission lines.

P = V² ÷ R not on the data sheetNot on the sheet, but it follows from P = VI and V = IR. Handy for parallel branches, where V is known.

R_series = R₁ + R₂ + … V_series = V₁ + V₂ + …Series: the same current flows through every component; the voltages add up to the supply voltage.

1 ÷ R_parallel = 1 ÷ R₁ + 1 ÷ R₂ + … I_parallel = I₁ + I₂ + …Parallel: every branch has the same voltage; the branch currents add up to the total. Invert at the end.

Dot points: PY11-EM-18, PY11-EM-19, PY11-EM-20, PY11-EM-21, PY11-EM-22, PY11-EM-23, PY11-EM-24, PY11-EM-25, PY11-EM-26, PY11-EM-27, PY11-EM-28, PY11-EM-29, PY11-EM-30, PY11-EM-31, PY11-EM-32, PY11-EM-33, PY11-EM-34, PY11-EM-35, PY11-EM-36, PY11-EM-37, PY11-EM-38, PY11-EM-39 (2025 Physics syllabus, Year 11), in our own words.

Mini lesson

Current, charge and which way it flows

Current is the rate charge flows past a point: I = q ÷ t. One ampere is one coulomb per second. In a metal wire the moving charges are free electrons.

A battery or power supply sets up a voltage across the circuit, which creates an electric field inside the wire. That field pushes on the free electrons, so they drift from the negative terminal, through the circuit, to the positive terminal. Conventional current is defined the other way, from + to −, as if positive charges were moving. Circuit diagrams and the rules in Year 11 and Year 12 use conventional current.

In direct current (DC, from a battery) the charges drift one way all the time. In alternating current (AC, from a power point) the supply voltage reverses polarity over and over (50 full cycles a second in Australia), so the charges jiggle back and forth. Energy is still delivered either way.

Conventional current (blue) goes from + to − around the outside circuit. The electrons (orange) actually drift the other way.

Resistance and Ohm's law

Resistance is how much a component opposes current: R = V ÷ I, in ohms (Ω). It comes from the free electrons colliding with the vibrating ions of the metal. That explains the four factors:

Length: a longer wire means more collisions along the way, so R is proportional to length.
Cross-sectional area: a thicker wire gives more paths side by side, so R is inversely proportional to area.
Material: materials differ in how many free electrons they have and how often those electrons collide, so copper resists far less than nichrome of the same size.
Temperature: hotter ions vibrate more, so electrons collide more often and a metal's resistance rises.

Ohmic and non-ohmic conductors

A conductor is ohmic if its resistance stays constant, so current is proportional to voltage: double the voltage, double the current. Its I–V graph is a straight line through the origin, and with I on the y-axis the gradient is 1 ÷ R.

A filament lamp is non-ohmic. As the current grows, the filament heats up, its resistance rises, and the graph bends over. For a non-ohmic component, R = V ÷ I still gives the resistance at one point on the graph, but it is not the gradient, and it changes from point to point.

Ohmic: a straight line through the origin (constant R). Filament lamp: the curve flattens because the hot filament's resistance rises.

Power and energy

Power is how fast energy is transformed: P = ΔE ÷ Δt, in watts. In a circuit the battery's chemical energy becomes electrical potential energy of the charges, which the components turn into heat, light or motion. Electrically, P = VI: the energy each coulomb carries (V) times the coulombs per second (I).

Combine P = VI with V = IR and you get P = I²R. This is why power lines lose energy as heat: the loss depends on the square of the current, so halving the current cuts the loss to a quarter. Conservation of energy then tells you the power delivered to the load is the power supplied minus the I²R lost in the wires.

Series and parallel

In a series circuit there is one path. The same current goes through every component (charge isn't used up, only energy is), the voltages across the components add up to the supply voltage (energy per coulomb is conserved around the loop), and the resistances add: R_series = R₁ + R₂ + ….

In a parallel circuit each component has its own branch between the same two points, so every branch has the full branch voltage. The branch currents add up to the total current, and the total resistance comes from 1 ÷ R_parallel = 1 ÷ R₁ + 1 ÷ R₂ + …. Adding a branch gives the current another path, so the total resistance goes down: it is always less than the smallest branch.

Series: one path, same current, voltages add. Parallel: separate branches, same voltage, currents add.

Meters, and solving combination circuits

An ammeter measures the current through a component, so it goes in series with it, and it has a very small resistance so it doesn't reduce the current it is measuring. A voltmeter measures the voltage between two points, so it connects in parallel across the component, and it has a very large resistance so almost no current is diverted through it.

For a combination circuit, work from the inside out and then back: (1) replace each parallel group with its equivalent resistance, (2) add the series resistances to get the total, (3) find the total current with I = V ÷ R_total, (4) work back out, using V = IR for each series part and sharing the remaining voltage across each parallel branch. Check at the end that the branch currents add up to the total.

The ammeter sits in the loop so all the current goes through it. The voltmeter connects across the resistor's two ends.

Common mistakes

Thinking current is used up, so it is smaller after each component in a series circuit.Charge is conserved: the current is the same at every point in a series circuit. It is energy (voltage) that is shared out between the components.
Adding parallel resistances like series ones, or forgetting to flip 1 ÷ R_parallel at the end.Work out 1 ÷ R₁ + 1 ÷ R₂, then take the reciprocal. Check: the answer must be smaller than the smallest branch.
Drawing the ammeter across a component, or the voltmeter in the main loop.Ammeter in series (low resistance), voltmeter in parallel (high resistance).
Treating conventional current and electron flow as the same direction.Conventional current goes + to − outside the battery. Electrons drift − to +. Say which one you mean.
Putting minutes, mA or kΩ straight into the equations.Convert to seconds, amperes and ohms before you substitute: 250 mA = 0.250 A, 2.2 kΩ = 2200 Ω.
Using the supply voltage with one resistor's resistance in V = IR.Match the three values to the same component: the voltage across it, the current through it, its resistance (or use the totals for the whole circuit).

Worked example

A 12.0 V battery is connected to resistor R₁ = 5.0 Ω in series with a parallel pair, R₂ = 20 Ω and R₃ = 30 Ω. Assume the battery and wires have no resistance.

Calculate the current through R₃ and the power used in R₁.

(5 marks)

I'll think out loud, the way you should in the exam, using KNOW-UD: Knowns, Not known, Organise, Work it out, Units, Direction.

Prediction first. A parallel pair is always less than its smaller resistor, so 20 Ω and 30 Ω together are less than 20 Ω, about 12 Ω. The total is then about 17 Ω, so the current is about 12 ÷ 17 ≈ 0.7 A. Most of the 12 V sits on the bigger part (the pair, about 8 V), and the 30 Ω branch takes the smaller share of the current, so I expect about 0.3 A through R₃. For R₁: about 0.7² × 5 ≈ 2.5 W.

K · Knowns

I copy the circuit with every value on it. R₂ and R₃ join at the same two junctions, so they are in parallel; R₁ carries the whole current, so it is in series with the pair.

V = 12.0 V
R₁ = 5.0 Ω (series), R₂ = 20 Ω and R₃ = 30 Ω (parallel)

Knowns on the sketch: 12.0 V, R₁ = 5.0 Ω in series with R₂ = 20 Ω and R₃ = 30 Ω in parallel. Unknowns: I₃ and P₁.

N · Not known

Two answers: a current in amperes and a power in watts. On the way I'll need the total resistance, the total current and the voltage across the pair.

I₃ = ? A, P₁ = ? W

O · Organise

I shrink the parallel pair to one resistor, add R₁ for the total, find the total current, then work back out. All of the current goes through R₁, so P₁ = I²R₁.

1 ÷ R_p = 1 ÷ R₂ + 1 ÷ R₃, R_T = R₁ + R_p
I_T = V ÷ R_T, V_p = I_TR_p, I₃ = V_p ÷ R₃
P₁ = I_T²R₁

W · Work it out

I keep the full calculator values for the total current and only round at the end. I label every line so the marker can follow which resistor each number belongs to.

1 ÷ R_p = 1/20 + 1/30 = 0.08333 Ω⁻¹ → R_p = 12.0 Ω
R_T = 5.0 + 12.0 = 17.0 Ω
I_T = 12.0 ÷ 17.0 = 0.70588 A
V_p = 0.70588 × 12.0 = 8.4706 V (and V₁ = 0.70588 × 5.0 = 3.5294 V)
I₃ = 8.4706 ÷ 30 = 0.28235 A
P₁ = 0.70588² × 5.0 = 2.4913 W

U · Units

Volts over ohms gives amperes, and amperes squared times ohms gives watts. The resistances are given to two significant figures, so I give two.

I₃ ≈ 0.28 A
P₁ ≈ 2.5 W

D · Direction

Current and power here are scalars, so instead of a direction I do a sense check with the circuit rules. Voltages: 3.53 V + 8.47 V = 12.0 V, the supply. Currents: I₂ = 8.4706 ÷ 20 = 0.42353 A, and 0.42353 + 0.28235 = 0.70588 A, the total. Both rules hold.

V₁ + V_p = 12.0 V ✓ I₂ + I₃ = I_T ✓

The current through R₃ is 0.28 A, and R₁ uses 2.5 W.

Check against the prediction. I predicted about 0.3 A through R₃ and about 2.5 W in R₁. I got 0.28 A and 2.5 W, and the voltages and currents add up correctly around the circuit.

Exam advice

What markers look for

Show the formula, the substitution and the answer with a unit for every step. In a multi-step circuit problem, one slip early on costs only that step if your working is clear, because markers can follow your numbers through.General advice from NESA HSC marking feedback, 2021 to 2025, summarised
Label every quantity with the component it belongs to (V₁, I₃, R_p). HSC markers have asked students to identify every symbol they use in a calculation, and in a combination circuit unlabelled numbers are very hard to follow.General advice from NESA HSC marking feedback, 2021 to 2025, summarised
Convert to SI units before substituting: mA to A, kΩ to Ω, minutes and hours to seconds, kW to W. HSC markers listed missed conversions as a common error across several questions.NESA HSC marking feedback, 2024, summarised
Year 11 circuit rules don't go away. In 2024, after an electromagnetism question, HSC markers advised students to review the circuit laws (how current and voltage share out around a circuit), so make the series and parallel rules automatic now.NESA HSC marking feedback, 2024, summarised
When a practical asks why measured and calculated values differ, name the real cause: the battery's internal resistance, the resistance of the connecting wires and meters, or the resistors' tolerance (often ±5%). 'Human error' earns nothing.CortexHSC teaching notes

Summaries in our own words. Read NESA's own feedback with the exam packs: 2024.

Where this leads in Year 12

Circuits underpin Year 12 Electromagnetism. Current in a wire feels a force in a magnetic field (F = lIB sin θ), which drives electric motors, and a spinning motor's back emf changes the current it draws. Transformers use P = VI and energy conservation (V_pI_p = V_sI_s), and the I²R heating from this topic is one reason real transformers and power lines lose energy.

FAQ

Are the circuit equations on the data sheet?: Yes. NESA's sample data sheet for the new (2025) syllabus includes I = q ÷ t, V = IR, P = VI, P = I²R and the series and parallel rules. P = V² ÷ R is not on it, but you can get it from P = VI and V = IR. Check which sheet your school gives you for Year 11 exams.
What is the difference between conventional current and electron flow?: Conventional current is the direction positive charge would flow: from the positive terminal, around the circuit, to the negative terminal. Electron flow is the real motion of electrons in a metal, which is the opposite way. Circuit rules use conventional current.
Why does adding a resistor in parallel lower the total resistance?: Each new branch is another path for charge, so more current flows for the same voltage. More current at the same voltage means less total resistance, which is why R_parallel is always smaller than the smallest branch.
What makes a conductor ohmic?: Its resistance stays the same as the voltage changes (at a constant temperature), so current is proportional to voltage and the I–V graph is a straight line through the origin. A metal resistor kept cool is ohmic; a filament lamp is not.
Why are household appliances connected in parallel?: Each appliance then gets the full mains voltage and can be switched on or off without affecting the others. In series they would share the voltage, and one switched off would break the circuit for all of them.

Related guides

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