Topic guide 8 · Unit 4 Topic 2 · Machines, mechanisms and control
Electrical circuits, power and efficiency
Nearly every circuits question comes down to three moves. Find the total resistance, use $V = IR$ to get the current, then work back to whatever was asked. Circuits is new to the external exam for 2026 and has no past questions, so the worked examples here are Tangent Tuition’s own.
Tick off each skill once you can do it without help.
0
past QCAA circuits questions, 2020–2025
1
formula you have to recall
5
syllabus dot points, none examined before
Skills in this topic
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Part 1
What the exam asks about circuits
There is no past QCAA question on circuits. A search of all six question and response books and all six multiple choice books from 2020 to 2025, for resistor, resistance, ohm, voltage, current, amp, battery, electric and circuit, finds no genuine circuits question. Every match is something else: soil and air resistance, chemical resistance in a plastics table, high-voltage cable insulation in a plastics question, logic circuits and motors.
Circuits is new to the external exam for 2026. The QCAA revision report (July 2024) lists “calculations involving series and parallel circuits” as introduced subject matter in Unit 4. Its clarified list adds “formulas for electrical as well as mechanical power and efficiency”, and the v1.4 syllabus (March 2026) then added “× 100%” to the efficiency formulas.
You have met some of this before. Unit 2 covers $V = IR$, $P = VI$ and $E = Pt$, and series and parallel circuits with two resistors. Unit 4 extends this to three or more resistors, the total current sums and efficiency.
Every formula but one is printed in the formula book, so these questions test whether you can apply the formulas, not whether you remember them. Every answer can also be checked with arithmetic, which is not true of the other new topic, heat treatment.
What the syllabus says
Unit 4 Topic 2, Machines, mechanisms and control, has five dot points on circuits:
•Calculate to solve problems involving basic series circuits, using the formulas $R_{total} = R_1 + R_2 + R_3 + R_4 + \dots$, $V_{total} = V_1 + V_2 + V_3 + V_4 + \dots$ and $V = IR$.
•Calculate to solve problems involving electrical power, using the formulas $P = VI$ and $E = Pt$.
•Calculate to solve problems involving electrical power efficiency, using the formula η = power output ÷ power input × 100% $= \dfrac{P_{out} }{P_{in} } \times 100\%$.
•Calculate to solve problems involving electrical energy efficiency, using the formula η = energy output ÷ energy input × 100% $= \dfrac{E_{out} }{E_{in} } \times 100\%$.
The word “basic” appears in both circuit dot points. That suggests pure series or pure parallel arrangements rather than combined networks, with manageable numbers. It is a prediction, because no question has been set yet.
•$I_{total} = \dfrac{V}{R_1} + \dfrac{V}{R_2} + \dots$ for a parallel circuit. The syllabus names it but the book does not print it. It is the one formula you have to recall, and Part 4 shows how to use it.
•$P = I^2R$ and $P = \dfrac{V^2}{R}$. Both follow from $P = VI$ and $V = IR$, and either can save a step.
•That $E = Pt$ needs the time in seconds to give joules.
•That in parallel, $R_T$ is always smaller than the smallest branch. This is the check that catches a wrong reciprocal.
Apart from that one formula, everything this topic needs is in the book.
Part 2
Series or parallel: what is shared
A circuit moves charge around a loop. Voltage ($V$, in volts) is the push that drives it. Current ($I$, in amps) is the rate of flow. Resistance ($R$, in ohms) opposes the flow. Ohm’s law ties the three together, and it is the most useful relationship in this topic:
$$V = IR$$
A series circuit has one path. The same current flows through every component, and the supply voltage divides between them. A parallel circuit has several branches. Every branch gets the full supply voltage, and the branch currents add to give the total current.
Series: one path through 4, 6 and 2 Ω. $R_T = 12\ \Omega$, the same 1 A flows everywhere, and the voltage drops add to the 12 V supply.Parallel: branches of 4, 6 and 12 Ω across the same 12 V. The branch currents of 3, 2 and 1 A add to 6 A, and $R_T = 2\ \Omega$.
These two circuits are the worked examples in Parts 3 and 4.
Series: one path
•Current. The same through every component.
•Voltage. Divides between the components. The voltages add to $V_T$.
•Total resistance. The resistances add, so $R_T$ is always larger than any one resistor.
•Adding a component increases $R_T$ and decreases the current.
•If one component fails, everything stops.
Parallel: branches
•Current. Divides between the branches. The branch currents add to $I_T$.
•Voltage. The full supply voltage across every branch.
•Total resistance. The reciprocals add, so $R_T$ is always smaller than the smallest branch.
•Adding a branch decreases $R_T$ and increases the total current.
•If one branch fails, the other branches keep working.
Game plan: Any circuit
1Decide whether there is one path (series) or several branches (parallel) before you write anything.
2Find the total resistance.
3Use $V = IR$ to get the total current.
4Work back to whatever the question asks for: a branch current, the voltage across one component, a power, an energy or an efficiency.
5Run a check. Series voltages add to the supply, and parallel currents add to the total.
Try it yourself
Switch between series and parallel
Change the supply and the three resistors, then switch the connection. Watch what each resistor shares with the others, and which one takes the most power.
Part 3
Solving a series circuit
In series the resistances add, so the total is always larger than any one resistor. Adding a resistor in series always increases the total resistance. The voltages add in the same way:
$$R_T = R_1 + R_2 + R_3 + \dots$$
$$V_T = V_1 + V_2 + V_3 + \dots$$
The current is the same everywhere, so once you have it from $V = IR$ you can find the voltage across each resistor.
Worked example
Worked example: a series circuit
This is a Tangent Tuition example, not a past QCAA question. No circuits question has been set yet. Try it before you open the solution.
Worked example 1
Tangent Tuition example · Series circuit
A 12 V supply is connected in series to resistors of 4 Ω, 6 Ω and 2 Ω. Determine the current in the circuit and the voltage across each resistor.
Total resistance.
$R_T = 4 + 6 + 2 = 12\ \Omega$
Current, from Ohm’s law. In series this is the current everywhere.
The total is always smaller than the smallest branch. Adding a branch lowers the total resistance and raises the total current.
The reciprocal step is where parallel questions go wrong
For resistors of 4, 6 and 12 Ω, $\dfrac{1}{R_T} = \dfrac{1}{4} + \dfrac{1}{6} + \dfrac{1}{12} = \dfrac{6}{12} = 0.5$, so $R_T = 1 \div 0.5 = 2\ \Omega$.
The mistake is stopping at 0.5 and writing it down as the answer. The sum gives you $\dfrac{1}{R_T}$, not $R_T$, so you still have to invert it.
To check, compare with the smallest branch. Here that is 4 Ω and the answer is 2 Ω. Parallel resistance is always less than the smallest branch. If your answer is bigger than any single resistor, you skipped the inversion or added the resistances as if they were in series.
The one formula you have to recall
The syllabus lists three formulas for parallel circuits, and the book prints two of them. This one is named in the Unit 4 subject matter and is absent from the formula book:
Learn it, because it gives you the total current without finding $R_T$. Every branch has the full supply voltage across it, so each branch current is $V \div R$, and the branch currents add. It takes one line and no reciprocals.
The reciprocal step is where parallel questions go wrong, and this route avoids it. Use it first when only the current is wanted. When you need $R_T$ as well, use it as a cross-check. If $V \div R_T$ and the branch-current sum disagree, look at your inversion.
Worked example
Worked example: a parallel circuit
This is a Tangent Tuition example, not a past QCAA question. No circuits question has been set yet. Try it before you open the solution.
Worked example 2
Tangent Tuition example · Parallel circuit
The same 12 V supply is connected in parallel to resistors of 4 Ω, 6 Ω and 12 Ω. Determine the total resistance, the current in each branch and the total current.
Branch currents. Every branch has the full 12 V across it:
$I_1 = \dfrac{12}{4} = 3\ \text{A}$
$I_2 = \dfrac{12}{6} = 2\ \text{A}$
$I_3 = \dfrac{12}{12} = 1\ \text{A}$
Total current.
$I_T = 3 + 2 + 1 = 6\ \text{A}$
Check. $I_T = V \div R_T = 12 \div 2 = 6$ A. The two routes agree, which confirms both the reciprocal step and the branch currents.
Answer
$R_T = 2\ \Omega$ · 3 A, 2 A and 1 A · $I_T = 6\ \text{A}$
The same supply, two very different circuits
The supply is the same and two of the three resistors are the same, but the parallel circuit draws six times the current of the series one. It dissipates $12 \times 6 = 72$ W against the series circuit’s $12 \times 1 = 12$ W.
On the same supply, the same resistors in parallel always give a lower total resistance, more current and more power than in series.
Part 5
Power, energy and efficiency in a circuit
Electrical power is $P = VI$, in watts. Combined with Ohm’s law it also gives $P = I^2R$ and $P = \dfrac{V^2}{R}$, but the book prints only $P = VI$.
Energy is power multiplied by time, $E = Pt$, in joules. The time must be in seconds.
Efficiency works exactly as it does for machines. Use $\eta = \dfrac{P_{out} }{P_{in} } \times 100\%$ with powers or $\eta = \dfrac{E_{out} }{E_{in} } \times 100\%$ with energies. For a motor, the input is electrical and the useful output is mechanical, and the difference is lost as heat.
Game plan: Power, energy and efficiency
1$P = VI$ gives the power a device draws, in watts.
2$E = Pt$ gives the energy it uses, in joules, with $t$ in seconds.
3$\eta = \dfrac{P_{out} }{P_{in} } \times 100\%$ compares the useful output power with the input power.
4$\eta = \dfrac{E_{out} }{E_{in} } \times 100\%$ does the same with energies, and gives the same answer for the same device over the same time.
5Whatever energy is missing from the output has become heat.
Unit traps
Unit trap
Where it bites
Minutes in $E = Pt$
Eight minutes is 480 seconds. Using 8 gives an answer 60 times too small.
kW against W
The prefix table is in the formula book. 1.2 kW is 1200 W.
kWh against J
A kilowatt-hour is 3.6 million joules. Use joules unless the question asks for something else.
mA against A
400 mA is 0.4 A.
Percentage against decimal
Multiply by 100 and write the % sign. The v1.4 formulas include “× 100%”.
Worked example
Worked example: motor efficiency
This is a Tangent Tuition example, not a past QCAA question. No circuits question has been set yet. Try it before you open the solution.
Worked example 3
Tangent Tuition example · Motor efficiency
A 24 V battery drives a motor that draws 3.5 A for 8 minutes. The motor delivers 55 W of useful mechanical power. Determine its efficiency and the energy lost as heat.
Check. $\dfrac{26\,400}{40\,320} \times 100 = 65.48\%$, the same efficiency by the energy route. The two efficiency formulas must agree.
The question states no precision, so the efficiency is given to the nearest whole unit, 65%. Keep 65.48% for the check.
Answer
65% · 13 920 J
Part 6
Earning the marks without a mark guide
No QCAA marking guide has marked a circuits question, so there is nothing to quote. The predicted dot points below are modelled on the 2020 and 2021 calculation guides. The wording has since shifted to “determines …” in 2022 and 2023 and “calculates …” in 2024 and 2025, but the structure of one mark per step has stayed the same.
Predicted dot points (a prediction, not a quote)
Predicted dot point
What would earn it
provides correct formula and substituted values for total resistance
The formula written out with the numbers in.
provides correct working to give correct total resistance
The arithmetic.
provides correct formula and substituted values for current
$V = IR$ rearranged, with the numbers in.
provides correct working to give correct answer with correct unit provided
The number, the unit and the stated precision.
Two conventions worth assuming
Every criteria statement on this page is derived, not quoted. Two conventions from the mechanical guides are still worth assuming. Each step usually carries a mark (the formula-and-substitution line had its own mark in the 2020, 2021 and 2023 guides, less often since), and follow-through applies down a chain.
A topic with no past questions
Two things make this one less daunting than it sounds.
First, every formula but one is supplied. Learn $I_{total} = \dfrac{V}{R_1} + \dfrac{V}{R_2} + \dots$, and everything else is in the book. The work is recognising which arrangement you are looking at and running the steps in the right order.
Second, the answers are checkable. Series voltages must add to the supply. Parallel currents must add to the total. $I_T$ must equal $V \div R_T$. The two efficiency formulas must agree. If you build those checks in, you can mark your own work.
Self-checks to run inside the question
Checkpoint
How to run it
Series or parallel?
One path or several. Decide before you write anything.
Did I invert?
The reciprocal sum gives $\dfrac{1}{R_T}$, not $R_T$.
Is $R_T$ sensible?
In series, bigger than every resistor. In parallel, smaller than the smallest.
Do they add up?
Series voltages add to the supply. Parallel currents add to the total.
Seconds?
$E = Pt$ needs seconds. Convert minutes first.
Which is the output?
For a motor, electrical in and mechanical out. Efficiency is always below 100%.
Unit written?
Ohms, amps, volts, watts, joules or per cent.
What changes for 2026: this whole topic
Series and parallel calculations are new to Unit 4. The revision report lists them as introduced, and it lists the electrical power and efficiency formulas as clarified. $V = IR$, $P = VI$ and $E = Pt$ were already in the old formula book but were never examined. The series and parallel sums and the two electrical efficiency formulas are new to the book, and “× 100%” was added in v1.4 (March 2026).
Because the formulas are supplied, I expect these items to test application rather than recall, and the dot points say the arrangements will be basic. My prediction for the most likely shapes is a straightforward series or parallel calculation worth three to five marks, and a motor efficiency chain that could sit alongside the mechanical power questions.
Game plan
Your game plan for circuits
1Decide series or parallel: one path, or several branches.
2Find the total resistance. In parallel, invert the reciprocal sum and check that $R_T$ is below the smallest branch.
3Find the total current with $V = IR$. In parallel you can go straight there with $I_{total} = \dfrac{V}{R_1} + \dfrac{V}{R_2} + \dots$
4Work back to the component the question asks about.
5For power and energy, use $P = VI$ and then $E = Pt$, with the time in seconds.
6Run the checks, then write every final answer to the nearest whole unit with its unit.
The efficiency routine is the same one you use for machines. Logic control is a different topic, with gates instead of numbers, and it has its own game plan, Game plan 6.
Four questions written from the syllabus dot points, with every number checked. No past example exists, so the mark values follow comparable mechanical calculation questions and the mark notes are predictions. Final answers are to the nearest whole unit unless the question says otherwise.
(B) is the difference of the two, and it also equals the smaller resistor, which is where a guess lands. (C) is the average. (D) adds them as if the circuit were in series. It is bigger than both resistors, so the smaller-than-the-smallest-branch check catches it straight away.
Answer
(A) 2 Ω
Question 2 of 4
Skill 2: Series circuits · Simple familiar · [4 marks]
A 24 V supply is connected in series to resistors of 5 Ω, 7 Ω and 12 Ω.
a) Calculate the total resistance of the circuit. [1 mark]
b) Calculate the current flowing in the circuit. [2 marks]
c) Calculate the voltage across the 12 Ω resistor. [1 mark]
(a)
$R_T = 5 + 7 + 12 = 24\ \Omega$
1 mark · correct working to give the total resistance.
A cordless drill runs from an 18 V battery pack. The motor draws a current of 2.5 A and is operated continuously for 6 minutes. Over that period the drill delivers 30 W of useful mechanical power.
Determine the efficiency of the drill and the energy lost as heat during the 6 minutes.
You need four quantities before the final answer: the input power, the input energy, the output energy and the difference. Convert the time to seconds before anything else.
$P_{in} = VI = 18 \times 2.5 = 45\ \text{W}$
2 marks · formula and substituted values for the input power, then the input power.
Check. $\dfrac{10\,800}{16\,200} \times 100 = 66.67\%$, the same as the power route. Run both routes, because their agreement checks the whole question.
Watch the time. Using $t = 6$ gives $E_{in} = 270$ J and $E_{out} = 180$ J, sixty times too small, but the efficiency still comes out at 66.67% because the error cancels in the ratio. Only a unit check on the energies catches it.
Answer
67% · 5400 J
Sources: Engineering 2025 v1.4 general senior syllabus, Unit 4 Topic 2 · Engineering formula and data book 2025 v1.4 · QCAA Engineering 2025 general senior syllabus revision report (July 2024) · QCAA Engineering external assessment question and response books, multiple choice question books and marking guides 2020 to 2025, searched in full (no circuits question exists in the past papers). All numerical answers verified computationally.
