A transformer changes voltage, not the amount of energy available
A transformer is easy to draw and surprisingly easy to misunderstand. A pupil sees two coils wound around an iron core, hears the words step-up and step-down, and starts memorising which side has more turns. That is only the beginning. A strong answer explains why the alternating input produces a changing magnetic field, how this changing field induces an e.m.f. in the second coil, and why a larger output voltage does not mean energy has appeared from nowhere. The useful examination skill is to keep voltage, current and power distinct while checking which side is primary and which is secondary.
Imagine a school display that receives a small alternating input and shows a different alternating output. If the secondary coil has twice as many turns as the primary coil, an ideal model gives twice the voltage. Yet the power available does not double. In the ideal model, the output current is half the input current. Real transformers also lose some energy. Every conclusion in this chapter begins with the question: which quantity is changing and which conservation rule still holds?
This volume goes beyond the overview of Vol 0271 on induction, generators and transformers and continues from Vol 0331 on the alternating generator waveform. The first established the mechanism, the second established the alternating input, and this chapter makes the numerical consequences rigorous. All figures and problems are original teaching illustrations, not official or predicted examination questions.
What the 2027 SEC syllabus actually asks
The Singapore Examinations and Assessment Board’s 2027 G3 school-candidate syllabus directory, updated 26 October 2026, lists standalone Physics as K323, with 6091 shown as the older reference code. The official K323 syllabus, Topic 19, includes the structure and operation of a simple iron-cored transformer, the ideal relationships Vp/Vs = Np/Ns and VpIp = VsIs, energy losses in cables, and the advantages of high-voltage transmission. These are explicit learning outcomes, not guesses about a future paper.
The standalone K323 examination comprises Paper 1 multiple choice, 40 marks in one hour and 30% weighting; Paper 2 structured and free response, 80 marks in one hour 45 minutes and 50% weighting; and Paper 3 practical, 40 marks in one hour 50 minutes and 20% weighting. A topic can be assessed through explanation, numerical work or interpretation of unfamiliar information. The directory also lists different combined Science syllabuses, including K326 and K327; do not assume every detail of standalone K323 is compulsory in each combined course. Follow your own entered subject and examination year.
The canonical learning route is the G3 SEC EMS Learner’s Guide. For the necessary arithmetic, use the Mathematics Hub and PSLE Learning Guide. A fraction is not a weakness to conceal inside a Physics formula; it is a foundational skill to repair.
The mechanism: one changing field, two coils
A simple transformer has a primary coil connected to the alternating input, a secondary coil supplying the output, and a common iron core that helps magnetic flux link the coils. The alternating current in the primary creates a changing magnetic field. That changing field passes through the secondary and induces an alternating e.m.f. there. It is the change that matters. A steady direct-current input, after the brief switching transient, does not create the continuously changing magnetic field needed for ordinary transformer operation.
The coils are not normally electrically connected to each other in this simple description. Energy is transferred through the changing electromagnetic field. If a diagram puts the input on the left, that is merely a drawing convention; a later question may place the primary on the right. Label by function, not position. The side connected to the source is primary. The side delivering the transformed output is secondary.
The turn count matters because each turn experiences the changing magnetic flux. Under the idealised assumptions used in the syllabus equation, the voltage ratio follows the turns ratio. A bigger secondary turn count gives a larger secondary voltage; a smaller secondary count gives a smaller secondary voltage. The rule uses actual turn counts or their ratio, not wire length, coil thickness, the position on the page or a guess from the size of the symbol.
Equation one: voltage follows the turn ratio
Write the equation exactly as a matched ratio: Vp/Vs = Np/Ns. Rearrange it to Vs = Vp × Ns/Np. This form makes the cause clear: secondary voltage equals the primary voltage multiplied by secondary turns relative to primary turns. For 100 primary turns, 400 secondary turns and 6 V alternating input, Vs = 6 × 400/100 = 24 V. The result is a step-up transformer because output voltage exceeds input voltage.
The same hardware with 400 primary turns and 100 secondary turns is a step-down arrangement for the same input: Vs = 6 × 100/400 = 1.5 V. Never place the larger number in the numerator simply because you expect an impressive answer. First write Np and Ns beside the correct sides of the diagram. If both turn counts are multiplied by the same number, their ratio remains unchanged; the ideal voltage ratio remains unchanged as well.
In an examination answer, use an equal-sign chain that readers can check. State the ratio, substitute, rearrange, calculate and label the voltage. A result without the substitution hides whether the student has reversed the sides. A result expressed only as ‘four times bigger’ may not meet a question asking for a numerical output in volts. Read what the question requests, then stop after supplying the required information.
Equation two: power balance determines current in the ideal model
The ideal relationship VpIp = VsIs says that input power equals output power, with voltage times current measured in watts. If the voltage is raised by a factor of four, current is reduced by a factor of four for the same transferred power. With input 12 V at 2 A, the ideal input power is 24 W. If secondary voltage is 48 V, output current is 24/48 = 0.5 A. The transformer did not create 96 W; it traded a larger voltage for a smaller current.
Rearrange carefully: Is = VpIp/Vs. If you need input current instead, Ip = VsIs/Vp. It is tempting to use the turn ratio twice in the same direction, saying both voltage and current increase together. That contradicts the ideal power relationship. Think in power terms before doing the second calculation. There are two quantities on each side of the balance, and they must multiply to the same result in the ideal model.
Real machines have losses, so actual output power is lower than input power. The ideal equations apply when the question explicitly assumes an ideal transformer or asks for a simple syllabus model. If a problem provides efficiency, use it to relate actual output power to input power. Do not silently assume 100% efficiency and then claim that a real transformer releases no heat.
High voltage helps transmission because cables heat less at lower current
A cable with resistance R dissipates electrical power as heat at a rate I²R. For the same transmitted power, increasing transmission voltage reduces current. Because loss depends on the square of current, halving current reduces resistive loss to a quarter if resistance is unchanged. This is the central reason for raising voltage along transmission lines and reducing it later for suitable uses. The conservation of energy is still respected: less energy is wasted in the conductors.
Consider the simplified task of transmitting 24,000 W. At 240 V, current would be 100 A in a basic P = VI model. At 2,400 V, it would be 10 A. For an illustrative cable resistance of 2 ohms, the respective I²R values would be 20,000 W and 200 W. That large first loss reveals that a constant delivered-power assumption and such a lossy cable would need a fuller system model; the numbers are best used to show the square-law comparison, not to promise exact measured delivery at either setting.
A careful pupil says ‘higher voltage allows lower current for the same transferred power, reducing heating loss’ instead of ‘high voltage makes electricity stronger’. It does not remove all losses. Nor does it make high voltage harmless. Mains and transmission systems are not suitable for pupil experimentation. Use diagrams, classroom models and teacher-provided data rather than attempting electrical construction or access to infrastructure.
A procedure that survives unfamiliar exam questions
Begin with a labelled inventory: primary voltage Vp, secondary voltage Vs, primary turns Np, secondary turns Ns, input current Ip and output current Is. Put units beside electrical quantities. Decide whether the model is ideal. Then calculate only the requested unknown. The quickest reliable route is often to solve the turn-ratio equation for voltage, then use power balance if the task also asks for current. Finally check whether the direction of change makes physical sense.
If Np is larger than Ns, Vs should be lower than Vp. In the ideal model Is should then be larger than Ip for the same power. That qualitative expectation is an error detector, not a substitute for calculation. If your computed figures violate both rules, recheck the fraction orientation. If only one violates the expectation, look for a mistaken unit, missing decimal or failure to distinguish input from output.
Some questions provide a turn ratio in words such as primary to secondary = 5:2. That means Np/Ns = 5/2. It is not Ns/Np. Rewriting verbal ratios explicitly prevents mistakes before arithmetic begins. More involved problems may contain two transformer stages; apply each stage in sequence and keep its primary and secondary labels local to that stage.
Common errors that cost otherwise avoidable marks
The first is treating step-up as a promise of more power. The second is swapping the turn counts and then rationalising the resulting answer. The third is using a direct-current supply as though it drove an ordinary transformer continuously. The fourth is reading a coil’s position instead of its electrical role. The fifth is putting volts where amperes belong after a power calculation. Each error has a distinct correction: conservation, labelling, time variation, functional identification and unit checking.
Another error is reporting that an iron core creates energy. The core provides a magnetic path and improves flux linkage; it is not a fuel. If the question asks why transformer operation relies on a changing field, the material of the core alone is not the answer. A student must name the changing magnetic field and the induced secondary e.m.f. When the task asks about high-voltage transmission, explain the current and cable-heating chain rather than repeating only the turns ratio.
Twenty-four original cases: work before reading
These questions cover identification, proportional reasoning, electrical power, cable losses and evidence evaluation. They do not reproduce SEAB items. Write each solution on paper first, compare the reasoning, then correct the first wrong step. The deliberately varied situations encourage transfer rather than memorisation.
Identify the primary
A diagram connects a 9 V a.c. source to a coil on the right; a measuring instrument sits beside the left coil.
Exam check: The right coil is primary because it receives the input. The left coil is secondary because it supplies the output. Do not let page layout decide the labels. Write the two roles before counting turns; every subsequent equation depends on them. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
A clear step-up
The primary has 60 turns at 4 V; the secondary has 180 turns.
Exam check: Vs = 4 × 180/60 = 12 V. Output voltage is three times input, so the arrangement steps up. The calculation does not predict output current until input current or power is known. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
A clear step-down
The primary has 800 turns at 240 V; the secondary has 100 turns.
Exam check: Vs = 240 × 100/800 = 30 V. The output is one eighth of the input. Writing a result of 1,920 V would reveal that the ratio was inverted. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Reverse turns
An ideal transformer converts 20 V to 5 V and has 1,000 primary turns.
Exam check: Use Np/Ns = Vp/Vs = 20/5 = 4, so Ns = 1,000/4 = 250 turns. Start from the whole ratio; do not guess that a lower voltage must mean exactly half as many turns. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Unknown primary turns
A 6 V input gives 18 V output, and the secondary has 450 turns.
Exam check: Np/Ns = 6/18 = 1/3, hence Np = 150 turns. The result shows fewer turns on the primary than secondary, which is consistent with the threefold voltage increase. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Equal turns
Two ideal coils each have 350 turns; input is 15 V a.c.
Exam check: The ideal voltage ratio is 1, so the output is 15 V. A transformer need not change voltage merely because it has two coils; the turn ratio tells us whether voltage changes. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Both turn counts doubled
A primary and secondary have 100 and 400 turns; a redesign gives 200 and 800.
Exam check: Both arrangements have Ns/Np = 4. Under the ideal model, the output-to-input voltage ratio remains four. Doubling both counts is not equivalent to doubling the ratio. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Turn ratio given in words
The stated primary-to-secondary turns ratio is 7:2 and Vp is 42 V.
Exam check: Vs = 42 × 2/7 = 12 V. Copy the order of the verbal ratio faithfully. ‘Primary to secondary’ means Np:Ns, not the reverse. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Input current and stepped-up voltage
An ideal 10 V input draws 3 A; the output is 30 V.
Exam check: Input power is 10 × 3 = 30 W. Is = 30/30 = 1 A. The current falls because the voltage rises while the ideal transferred power remains 30 W. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Reverse current
An ideal transformer supplies 6 V at 4 A from a 24 V input.
Exam check: Output power is 6 × 4 = 24 W. Input current Ip = 24/24 = 1 A. The lower output voltage allows a higher output current without adding power. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Two-stage voltage transformation
A stage converts 20 V to 100 V; a second stage has 300 primary and 60 secondary turns.
Exam check: The second stage gives 100 × 60/300 = 20 V. The output returns to 20 V in the ideal voltage model. Re-label the second-stage primary and secondary rather than reusing first-stage coil numbers. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Check impossible power
A response claims 12 V at 5 A input produces 48 V at 3 A output in an ideal transformer.
Exam check: The alleged input power is 60 W and output power is 144 W. They cannot both describe an ideal passive transformer. An output at 48 V would require 60/48 = 1.25 A if no losses occur. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Current ratio from turns
Np is 120, Ns is 480, and Ip is 8 A in an ideal transformer.
Exam check: Voltage increases fourfold, so Is = 8/4 = 2 A. We need no actual voltage figures. The power-balance relation together with the turn ratio is sufficient. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Explain the direct-current trap
A pupil connects a steady d.c. source to a schematic transformer and expects an ordinary continuous secondary output.
Exam check: A steady current creates a magnetic field that is not continually changing. After switching transients, the simple induction mechanism does not supply a sustained output as alternating input would. This is a conceptual answer, not a prompt to test household circuits. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Lenz’s reasoning
An examiner asks why induction does not give unlimited energy for no mechanical effort.
Exam check: Induced effects oppose the change producing them. Sustained generator operation requires input work; the transformer transfers the received energy rather than creating a new supply. This links the transformer chapter back to generator mechanisms. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Cable current comparison
The same ideal transmitted power is carried at 200 V and 2,000 V.
Exam check: The second voltage is tenfold, so the required current is one tenth as large at the same power. This explains why high-voltage transmission is advantageous before any cable resistance is specified. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Squared heating effect
Current in a particular resistance decreases from 10 A to 5 A.
Exam check: Since loss is I²R, the new loss is (5/10)² = one quarter of the original. A 50% reduction in current is a 75% reduction in resistive heating loss for fixed resistance. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Cable resistance doubles
Current stays 4 A but cable resistance changes from 3 ohms to 6 ohms.
Exam check: Loss changes from 4² × 3 = 48 W to 4² × 6 = 96 W. For unchanged current, resistive loss is proportional to resistance, not its square. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Loss at a lower current
A cable resistance is 0.5 ohms; compare 12 A and 3 A.
Exam check: At 12 A, loss is 144 × 0.5 = 72 W. At 3 A, it is 9 × 0.5 = 4.5 W. Lowering current by a factor of four lowers the loss by a factor of sixteen. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Why a high-voltage line is not a classroom project
A student proposes checking a transmission principle using a wall socket and improvised wire.
Exam check: That is not an appropriate investigation. The principle can be understood through school-approved equipment, published data and symbolic calculations. A written conclusion does not require handling hazardous electrical voltages. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
A real output lower than input
A real transformer receives 50 W and delivers 44 W to its load.
Exam check: The missing 6 W is converted into other forms, commonly including heat and other losses. Efficiency is 44/50 = 88%. It would be wrong to apply the ideal power equality to this measured situation without qualification. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Efficiency and required input
A device supplies 72 W with 90% efficiency.
Exam check: Input power is output/0.90 = 80 W. Total loss is 8 W. The efficiency figure is extra information beyond the ideal transformer syllabus relation, so show why you are using it. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Turn count versus measured voltage
A nominal 1:2 turns ratio produces a measured voltage ratio below 2 under load.
Exam check: The ideal turn equation is a model, not a promise that every real measurement has no deviation. Resistive drop and other losses may affect terminal voltages. The correct response distinguishes the ideal prediction from the observation. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
What to say about the iron core
The question asks for the core’s role rather than the voltage equation.
Exam check: A suitable iron core provides a magnetic path that helps changing magnetic flux link both windings. The core supports induction; it does not generate energy, and its mere presence without a changing field does not sustain induced secondary voltage. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
The output exceeds the input in volts
A friend sees 6 V becoming 48 V and says the transformer has multiplied power eightfold.
Exam check: Voltage grew eightfold, but ideal output current would be one eighth of input current. Equal products V × I preserve ideal power. Without both current readings, the claim of power multiplication is ungrounded. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
No current given
A problem gives Np, Ns and Vp, then asks for secondary current.
Exam check: The turn data let you find Vs, but current still requires an input-current or power constraint, or other sufficient information. A precise answer identifies the missing quantity instead of inventing a load. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Graph meets transformer
A 50 Hz alternating generator drives a stationary ideal transformer.
Exam check: The changing input lets the transformer operate. The transformer changes voltage magnitude according to turns but does not by itself convert a 50 Hz input into a different frequency in the simple model. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Reading a unit error
A worked answer labels output current as ’12 V’ after dividing power by voltage.
Exam check: Power in watts divided by voltage in volts gives current in amperes. The numerical arithmetic may be correct but the unit is not. State 12 A if the calculation truly yields twelve amperes. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
Transmission is a chain, not one machine
A diagram shows generation, a step-up device, a transmission cable and a step-down device.
Exam check: Explain the sequence: electrical output from generation is transformed to higher voltage for lower-current transport, then transformed to appropriate lower voltage for distribution or use. Actual networks have equipment, protection and losses omitted from this simplified diagram. A complete answer identifies the model assumptions and tests the direction of change against conservation of energy.
How to turn this into a reliable revision method
Start with five minutes of retrieval, not a page of copied notes. Draw two boxes marked primary and secondary, then fill in Vp, Vs, Np, Ns, Ip and Is from memory. Say the physical mechanism aloud before writing either equation. If the order is wrong, return to the changing-field explanation. If the ratio is wrong, practise four short examples with reversed primary and secondary positions. If the numerical answer is wrong, check fraction operations and units before blaming a lack of Physics knowledge.
Next do three mixed questions without being told which equation to use. One should ask for a missing number of turns, one for current after a voltage transformation, and one for a cable-loss comparison. Leave enough space to explain the model. A strong learner can choose the relationship from the context; a weaker habit waits for the equation to be supplied by the heading. After correcting mistakes, rewrite only the line where the reasoning first failed.
Finish with a sixty-second explanation to someone else: ‘The a.c. supply makes a changing field. The core links it to the secondary coil. Turns control voltage ratio. In an ideal transformer power stays the same, so voltage and current change inversely. High-voltage transmission lowers current and cable heating.’ If any sentence feels like a slogan you cannot justify, return to the corresponding numerical case. The goal is understanding that travels into an unfamiliar paper.
A compact independent test
Without looking back, calculate the output voltage for 24 V input, 400 primary turns and 100 secondary turns. Then calculate the ideal secondary current if the input current is 0.5 A. Finally explain, in no more than three sentences, why carrying the same power at ten times the voltage helps with cable heating. Solutions: Vs = 24 × 100/400 = 6 V; input power is 12 W, so Is = 12/6 = 2 A; transmission current falls to one tenth, and I²R heating falls to one hundredth for fixed resistance. Check the direction of each result, not just the last digit.
A second test asks what additional information you need if only the turn ratio is given and the question requests power loss in an actual cable. You need enough information about current and cable resistance, or an equivalent specification. A transformer turn ratio alone cannot establish a numerical loss. A capable learner recognises the boundary of the model and says what evidence would close the gap.
Where to go next
For the overlapping mathematics of rates, expenses and graphs, continue to the Mathematics Hub. For earlier foundations in electrical power and kilowatt-hours, use Vol 0167; for a previous mixed-induction scenario see Vol 0272. Strong preparation connects those owners without turning this page into a duplicate. A later EMS scenario will ask which conclusions a school can justify with measured power, an explicit tariff and a sensible design proposal.
The steady rule is simple: write the source and destination, preserve the power constraint, and explain the loss mechanism. That is what turns a remembered formula into examination performance.
