Quick answer: To explain lower power loss in a transmission cable, hold useful delivered power constant, use P = VI to show that higher receiving voltage permits lower current, then use P_loss = I²R to show how heating falls in a cable of fixed resistance. Always name the load, line and sending end. Do not confuse a loss of power with energy lost over time.
This Learner’s Guide continues the water-turbine energy route with a new question: what happens after a generator makes electrical energy and it must travel through resistive wires? The worked examples are invented and labelled as models. The central skill is to write a complete scientific explanation and reconcile its numbers, not memorise a dramatic claim about high-voltage electricity.
- Verify the new 2027 SEC Physics code K323 and use the official syllabus
- Distinguish load voltage from source voltage
- Find current from a specified useful receiving power
- Calculate cable heating and source power consistently
- Show the square-law effect of changing current
- Separate physical explanation, mathematical calculation and justified claims
- Practise worked questions and check the energy ledger
Contents
- Start with the result, not a slogan
- Why this chapter follows the turbine
- The official 2027 examination frame
- A compact model you can hold in your head
- Four quantities and their units
- Read P = VI carefully
- Use Ohm’s law on the cable
- Derive the line-heating relationship
- Work the fixed-load comparison without shortcuts
- What is actually reduced?
- Why high voltage is helpful, not free
- Keep induction and transmission separate
- The resistance is the full path, not just the visible wire
- The role of temperature
- A fair comparison in a data-based question
- How to read an unfamiliar diagram
- A four-line examination explanation
- Mistake: source voltage equals load voltage
- Mistake: loss percentage uses a mysterious denominator
- Mistake: percent reduction is percentage points
- Mistake: voltage is energy
- Mistake: a model number becomes a real-world claim
- Do not confuse a.c. and d.c. applications
- Reading the command word explain
- Graph reading that keeps its axes honest
- Checking a time interval
- The meaning of efficiency in this model
- Why the safest response is often narrower
- An examination-day working order
- A seven-day route to durable understanding
- Worked practice with answers
- Frequently asked questions
- Official sources and next learning route
Start with the result, not a slogan
A power line is not an invisible tube that passes electrical energy without cost. If a current flows through a cable with resistance, some energy is transferred to the surroundings by heating. A higher transmission voltage can allow a smaller current for the same useful transmitted power, which reduces resistive loss. The essential condition is the same useful power requirement; without it, a comparison of two voltages is incomplete. A learner who writes only “high voltage means less energy lost” has missed the reason. A stronger response explains that power transfer is linked to voltage and current, while heating in a cable rises with the square of current. Keep those two relationships distinct, then join them. This is a conceptual G3 Physics guide, not advice to modify household mains wiring, and all numerical systems below are fictional low-power classroom models.
Why this chapter follows the turbine
The preceding Vol 0725 followed falling water through mechanical rotation to induced e.m.f. The mathematical reconstruction in Vol 0726 checked uneven time intervals, and Vol 0728 tested an exhibition’s claims. Now imagine the generated electricity must reach a model load at the other end of a wire. The new question is not whether the turbine has produced electricity. It is how much electrical energy arrives and what is lost along the way. Treat this as a separate evidence problem: source, cable and receiving load each deserve their own entries. The familiar transformer principles already have a broader foundation in Vol 0335; this chapter owns the narrower task of interpreting a fixed-load line-loss case.
The official 2027 examination frame
The 2027 SEAB G3 subject list, last updated 26 October 2026, assigns Physics K323, replacing the earlier reference code 6091. The K323 syllabus explicitly covers a.c. generators, iron-cored transformers, ideal-transformer voltage/current equations, energy loss in cables and the advantage of high-voltage transmission. The same syllabus requires candidates to handle data, explain patterns and use evidence in unfamiliar situations. Do not describe this fictional cable case as a past-year paper or a required practical apparatus task. The 2027 SEC, beginning in 2027, retains G3-level examination standards and publishes its actual syllabus and assessment scheme through SEAB. Keep official assessment facts separate from the examples designed for learning here.
A compact model you can hold in your head
Draw three boxes in words: sending end, line, receiving load. At the sending end electrical power enters the cable. The cable has a total resistance that includes whatever conductors form the full current path. At the receiving end the useful load receives electrical power. If the model ignores other losses, power entering the line equals useful load power plus cable heating power. This is not a magical formula that makes all other losses disappear in real equipment. It is a declared simplifying assumption that makes the example soluble. Before doing arithmetic, put a name next to each voltage: source voltage, voltage drop across the cable and voltage across the load. A familiar mistake is to call them all V and then use the right equation with the wrong number.
Four quantities and their units
Current, I, describes charge flow and is measured in amperes (A). Potential difference, V, is measured in volts (V). Resistance, R, is measured in ohms (Ω). Power, P, is the rate of energy transfer and is measured in watts (W). Energy is measured in joules (J), with one watt equal to one joule per second. A numerical result without the correct unit may look plausible but conceal a conceptual error: 50 W is a rate, while 50 J is an amount of energy. When a question provides a time, multiply the relevant power by time in seconds to find energy. If a graph shows watts at several instants, do not add the readings and call the sum joules. First determine the elapsed intervals or the intended method of finding energy.
Read P = VI carefully
Electrical power at a specified pair of terminals is the product of the potential difference across those terminals and the current through the relevant path: P = VI. A receiver operating at 120 V and drawing 10 A takes 1200 W of useful electrical power in this simple model. If the same receiving load power is supplied at 240 V, the necessary current is 5 A. The numerical comparison therefore halves the current. The sentence that explains the mechanism must hold the received power constant; otherwise someone could raise both voltage and current and increase rather than decrease cable heating. The symbol V must refer to the receiving-load voltage when this calculation describes the receiver, not a vaguely stated voltage somewhere along the route.
Use Ohm’s law on the cable
A conducting cable with model resistance R has a voltage drop V_drop = IR. That drop is not the same as the useful potential difference across the load. In the first case, current 10 A through a combined line resistance of 2 Ω produces a 20 V line drop. The load still has 120 V across it in the stated scenario, so the sending-end voltage must be 140 V if the elements are represented by this simple series model. At 5 A the same line drops 10 V, and the source need only provide 250 V for a 240 V receiver. Those values show why changing the receiving voltage changes other quantities too. Never use 120 V to calculate the cable drop; the cable is the component with resistance 2 Ω.
Derive the line-heating relationship
Power converted to thermal energy in the line is P_loss = V_drop I. With V_drop = IR, substitution gives P_loss = I × IR = I²R. This does not require advanced mathematics. It is a precise use of two physical laws in the same section of the circuit. If the current doubles while line resistance stays unchanged, I² becomes four times as large, so the heating rate becomes four times as large. If the current halves, heating becomes one quarter. A common mistake is to claim that heating halves whenever current halves. The relationship is nonlinear. Write down what stays constant before using a proportional comparison. In some real cables resistance changes with temperature; the examination-style model assumes the given value remains constant unless stated otherwise.
Work the fixed-load comparison without shortcuts
Suppose a fictional demonstration load must receive 1200 W and the entire line has resistance 2 Ω. Case A supplies the receiver at 120 V, giving I = 1200/120 = 10 A. Cable heating is 10² × 2 = 200 W. Cable drop is 10 × 2 = 20 V, so the sending voltage is 140 V and sending power is 140 × 10 = 1400 W. Case B supplies the receiver at 240 V, giving I = 5 A. Heating is 5² × 2 = 50 W. The drop is 10 V; sending voltage is 250 V; sending power is 250 × 5 = 1250 W. Both cases conserve energy within the stated model: 1400 = 1200 + 200 and 1250 = 1200 + 50. All figures are invented for practice and are not measurements from a real grid.
What is actually reduced?
The amount of useful receiving power is 1200 W in both cases. The line heating falls from 200 W to 50 W. The absolute reduction is 150 W; the percentage reduction in line heating is 150/200 × 100% = 75%. It would be false to say that the whole electricity bill, the generator input or the total environmental impact falls by 75%. The 75% applies only to the line-heating rate in this model, with a fixed useful receiving load and line resistance. The proportion of source power lost in the line also changes, but its denominator is different: 200/1400 ≈ 14.3% in A versus 50/1250 = 4% in B. A strong answer labels both percentages instead of presenting a single impressive number without its base.
Why high voltage is helpful, not free
If useful power P = VI is held fixed, raising V lowers the current I = P/V. Because heating in the line is I²R, the loss falls substantially as current decreases. A transformer can change an a.c. voltage: the ratio of primary and secondary voltages equals the ratio of corresponding coil turns for an ideal transformer, Vp/Vs = Np/Ns, while VpIp = VsIs in the ideal model. Actual equipment is not lossless. Higher voltages also require suitable insulation, clearance, protection and specialist design. The school lesson is a principle of electricity transmission, not a suggestion to raise voltage in a home project. Only qualified personnel should handle mains or distribution equipment.
Keep induction and transmission separate
A generator and a transformer do different jobs. A generator converts mechanical energy into electrical energy through electromagnetic induction, typically producing an alternating e.m.f. in the simple a.c. generator example. A transformer uses a changing magnetic field linking coils to change the a.c. voltage. A cable delivers electrical energy to another location while resistance produces heating. Those are three distinct stages. A transformer does not create additional energy because its output voltage is larger. If output voltage doubles in the ideal equation, output current adjusts so the power balance still holds. Repeating the word induction for every component is not an explanation. Say which component is acting, what is changing physically, and what quantity the candidate is asked to calculate.
The resistance is the full path, not just the visible wire
In a classroom sketch, one long straight line is often drawn between source and lamp. In a real closed circuit, current needs a complete path. If a problem says “total line resistance is 2 Ω,” use 2 Ω once. If it gives an outward cable and a return cable separately, add the stated resistances when that is what the circuit requires. If it instead provides resistance per unit length, multiply by the appropriate complete length only when the units and circuit description justify it. Double-counting a total resistance is as serious as leaving out the return conductor. Look for the words total, each, loop, one-way and return. They tell you which numerical model has actually been supplied.
The role of temperature
Wire resistance is sometimes treated as a fixed numerical constant in a calculation, but conductor resistance can change as temperature changes. Heating may therefore affect both the amount of energy transferred to surroundings and the resistance of a metal line. The problem may give a constant effective resistance deliberately, allowing a simpler comparison. A disciplined learner uses that assumption rather than inventing a new temperature coefficient or dismissing the result as useless. In an evaluation answer, a suitable limitation is “the model assumes line resistance remains 2 Ω throughout.” That is specific and supported. A weak limitation is “the experiment could be inaccurate,” which identifies neither the source of uncertainty nor its effect.
A fair comparison in a data-based question
A comparison is fair when its intended explanatory difference is isolated. Here the useful receiving power and model line resistance are fixed, while receiving voltage is changed. Current and power loss then follow from the equations. If a second table has a different load power or a different resistance, the results cannot be attributed only to voltage. Write a short conditions sentence before comparing values: “For a fixed 1200 W receiving load and a 2 Ω line, the higher-voltage case draws less current and has lower resistive heating.” This does more scientific work than a long introduction. It tells the examiner exactly why the numerical conclusion is valid for the stated setup and stops the reader from generalising it beyond those conditions.
How to read an unfamiliar diagram
Start by locating the receiving load. Mark the voltage that is actually across it and the current through it. Then locate each cable resistance and note whether it is in series with that load or represents an equivalent line resistance. Next identify the meter location: a voltmeter across a cable reads the cable drop; a voltmeter across the load reads load voltage. An ammeter in the same simple series path reads the same current. Finally look for a transformer and label primary and secondary sides. A neat sketch with three named voltage drops can prevent most substitution errors. It is possible to have the correct formula memorised and still insert the wrong V because the diagram has not been read carefully.
A four-line examination explanation
For a question asking why electricity is transmitted at high voltage, a concise response can be built in four linked steps. First, a larger voltage allows a smaller current for a given power transfer (P = VI). Second, a transmission cable has resistance. Third, its rate of resistive heating is I²R; lowering current reduces this loss. Fourth, therefore less electrical energy is wasted as heating in the cables over a given time. If the question asks specifically about a transformer, add that a step-up transformer increases alternating voltage and decreases current in the ideal energy-balance model. Avoid claiming that the transformer itself “provides extra energy.” Each line should earn its place by adding a new part of the reasoning.
Mistake: source voltage equals load voltage
The most subtle numerical trap in this guide is treating a load voltage as though it were a supply voltage even when there is a cable drop. In Case A, 120 V appears across the load but 140 V is required at the source. Writing sending power = 120 × 10 = 1200 W silently removes the 200 W cable loss. That would break the model’s energy balance. A useful correction strategy is to check whether the source power exceeds the delivered power by exactly the calculated line loss. If it does not, inspect the voltage labels before reworking the arithmetic. This single check catches a surprising number of errors more quickly than recalculating the entire table from the beginning.
Mistake: loss percentage uses a mysterious denominator
“The line lost 14.3%” should make a careful reader ask: 14.3% of what? In Case A, the source provides 1400 W; 200 W is heating, giving 14.3% of sending power. But heating compared with received useful power is 200/1200 = 16.7%, a different ratio. Both calculations can be arithmetically correct while answering different questions. The relevant denominator belongs to the wording: of source input, of delivered output, of last year’s loss, or of the first trial. When students lose marks on percentages, it is often not multiplication but the wrong reference quantity. Write the full fraction with labels before simplifying it.
Mistake: percent reduction is percentage points
Case A loses 14.3% of its source power, while Case B loses 4.0%. The difference in the loss shares is about 10.3 percentage points, not 10.3% as a relative change. The loss power itself falls from 200 W to 50 W, a 75% reduction. Those statements are not contradictory: they describe different quantities. If a graph plots the loss share as a percentage, describe changes along the vertical scale in percentage points when subtracting two percentages. If a sentence describes how much smaller one measured power is than another, calculate the relative percentage with the first value as the base. Precision in language protects otherwise sound science.
Mistake: voltage is energy
Voltage is measured in volts. Energy is measured in joules or kilowatt-hours, depending on context. Raising voltage is not the same as creating energy. In an ideal transformer, higher voltage is accompanied by lower current at the same power. In a resistive line, changing current changes heating, and the source may then need to provide less total power to deliver the specified useful load. Keep those effects separate. A student might write “the high-voltage line contains more electricity,” but this vague phrase cannot be checked. Replace it with an answerable claim about voltage, current, line resistance, power loss or received energy. The objective is not to sound technical. It is to say exactly what has changed.
Mistake: a model number becomes a real-world claim
The numbers 120 V, 240 V, 2 Ω and 1200 W in this chapter form an invented teaching case. They have not been taken from a Singapore grid, a school installation or a practical measurement. Even when the mathematical method is correct, those values cannot support claims about public infrastructure. In a public science exhibition, label them “illustrative fixed-load model” and explain the assumptions. An audience may reasonably ask whether the same percentage is expected under different cable materials, distances or loads. A good answer is that the physical tendency is useful, while a particular numerical saving depends on the system being examined. Trust improves when a writer says both what is known and what has not been measured.
Do not confuse a.c. and d.c. applications
The 2027 G3 Physics syllabus includes a.c. generators and transformers. An ordinary idealised iron-cored transformer depends on changing magnetic flux, which is why it is discussed in an a.c. context. A simple direct-current supply held steady does not continuously induce the same transformer output in the secondary coil. Students should not imply that any battery can be connected directly to the primary of a conventional transformer to obtain a steady stepped-up d.c. voltage. This chapter’s arithmetic about current and resistive line losses can be applied to a stated circuit model, but the discussion of transformer voltage changes needs the appropriate alternating or time-varying input. Distinguish the general relationship from the particular device.
Reading the command word explain
If the command is “state,” a short valid fact may suffice. If it is “explain,” an answer needs a cause-and-effect chain. If it is “calculate,” show the relevant substitution with correct unit. If it is “compare,” quote both figures and say what changes while identifying controlled conditions. If it is “suggest,” propose something physically plausible within the stated constraints without presenting speculation as observation. The K323 assessment objectives include interpreting information, making reasoned explanations, and drawing inferences, so the same chapter supports several types of response. Practise rewriting a formula-only answer into words. The best examination writing makes the physical mechanism visible, not just the arithmetic.
Graph reading that keeps its axes honest
Imagine a graph of line-heating power in watts against current in amperes for a fixed 2 Ω line. The relation P_loss = 2I² curves upward. At 0 A, loss is 0 W; at 2 A, 8 W; at 4 A, 32 W; at 6 A, 72 W. The graph is not a straight line through the origin because loss is proportional to the square of current rather than current itself. Doubling current from 2 A to 4 A quadruples loss from 8 W to 32 W. A reader must label the vertical axis W and horizontal axis A and must not call the curve a voltage–current characteristic. Graph shape is evidence of the model supplied, not a permission to extrapolate beyond the physical assumptions of a real cable.
Checking a time interval
Power tells you how quickly energy is transferred; time tells you how much has accumulated. If Case A’s cable heating remains 200 W for 30 seconds, the modelled thermal energy transferred is 200 × 30 = 6000 J. Case B’s 50 W over the same interval gives 1500 J. Over one minute the values become 12,000 J and 3000 J. The 75% reduction is retained because the duration is the same. If the trials last different durations, direct comparison of joules would be unfair unless time is considered. A line can have lower heating power but release more total heat if it runs long enough. State whether a question asks for power or energy and keep a matching time basis when comparing results.
The meaning of efficiency in this model
For this deliberately simple three-part model, a transmission efficiency can be defined as useful power received divided by sending power, multiplied by 100%. Case A gives 1200/1400 ≈ 85.7%; Case B gives 1200/1250 = 96%. These are efficiencies of the line-plus-load accounting boundary, not turbine efficiency, not transformer efficiency, and not total power-station efficiency. They apply under the assigned fixed resistance, fixed receiving power and other idealisations. Before using “efficient,” name the boundary: what is the input and what is the useful output? Without a boundary, a seemingly impressive efficiency number can mislead. That is particularly important when an unseen question combines several energy-transfer stages.
Why the safest response is often narrower
Physics questions sometimes contain everyday language such as “better,” “green” or “less wasteful.” Convert such words into measurable quantities. Here “less wasteful” can mean less cable-heating power for the specified receiving power. It does not automatically mean lower installation cost, lower environmental impact across equipment manufacture, or a safer system. Voltage itself creates hazards; an appropriate real design involves standards and specialist protection. A scientifically honest conclusion is narrower and more useful than a universal promise. This habit also serves English comprehension and Mathematics real-world modelling: quote the precise evidence, explain its scope and avoid the unsupported final leap.
An examination-day working order
On a timed question, spend the first moments labelling the receiving load, source and cable. Underline whether a power is “delivered,” “transmitted,” “generated” or “lost.” Write down the given line resistance and whether it is total. Select P = VI for the load, V_drop = IR for the line, and P_loss = I²R for heating. Check sending power = load power + loss only if the simplified series model says there are no other losses. Include units in every intermediate line. Round at the end according to the question’s instruction. Then answer the command word in one clear sentence. This order is a practical defence against reading the numbers correctly but solving the wrong physical problem.
A seven-day route to durable understanding
On day one, explain the source–line–load model without equations. On day two, practise changing I while R remains fixed and predict the effect on I²R. On day three, use P = VI to hold receiving power constant while voltage changes. On day four, label separate source, line and load voltages on three fresh sketches. On day five, calculate energy over different durations and discover why watts and joules cannot be interchanged. On day six, write a four-line explanation of high-voltage transmission without consulting notes. On day seven, complete a timed unseen question and audit every denominator and unit. Space the checks rather than re-reading the same paragraph many times; retrieval and correction are more useful than familiarity alone.
Worked practice with answers
The following short problems move from single calculations to bounded conclusions. Solve each without looking at the answer first; the correction sentence is part of the learning.
Practice 1: A simple cable
Question: The total line resistance is 3 Ω and current is 4 A. Find cable loss.
Worked answer: P_loss = I²R = 4² × 3 = 48 W.
Why it matters: The number is a rate of energy transfer, not 48 J. No receiving load was specified, so do not invent its voltage.
Practice 2: Halving current
Question: A 6 A line has 72 W heating loss at fixed resistance. What happens at 3 A?
Worked answer: The current ratio is 3/6 = 1/2. Heating becomes (1/2)² × 72 = 18 W.
Why it matters: The loss falls by 54 W, or 75%, not 50%. The square-law comparison assumes resistance is unchanged.
Practice 3: Finding a cable drop
Question: A modelled total cable resistance is 1.5 Ω and line current is 8 A. Find the potential difference across the cable.
Worked answer: V_drop = IR = 8 × 1.5 = 12 V.
Why it matters: That drop is not the receiving voltage. It is the difference between source and receiving-end voltages in this model.
Practice 4: Receiving power
Question: A 600 W receiving device is specified to operate at 150 V. Find its current.
Worked answer: I = P/V = 600/150 = 4 A.
Why it matters: Use the voltage across the receiving device. Do not substitute a source voltage unless no line drop is present or the question says to.
Practice 5: Sending power
Question: A receiver takes 900 W while the cable converts energy to heat at 36 W. Assume no other losses.
Worked answer: P_send = 900 + 36 = 936 W.
Why it matters: This works because the problem states an accounting boundary without other losses. Real systems may have additional losses.
Practice 6: Full case reconciliation
Question: A 600 W load receives power at 100 V through 2 Ω total line resistance. Find current, drop and sending power.
Worked answer: I = 6 A; V_drop = 12 V; sending voltage = 112 V; P_send = 112 × 6 = 672 W.
Why it matters: Cross-check with P_loss = 6² × 2 = 72 W; 600 + 72 = 672 W.
Practice 7: Line efficiency
Question: Using the preceding case, determine the line-model efficiency.
Worked answer: Efficiency = 600/672 × 100% ≈ 89.3%.
Why it matters: Name the boundaries: received load power divided by sending power. Do not attach that percentage to the generator.
Practice 8: Percentage reference
Question: A line loses 40 W in one trial and 25 W in another. By what percent did heating decrease?
Worked answer: (40 − 25)/40 × 100% = 37.5%.
Why it matters: The first trial’s loss is the denominator because the question asks for the decrease relative to it.
Practice 9: Total energy
Question: A fixed line-heating rate of 18 W lasts for 45 seconds. Calculate thermal energy transferred.
Worked answer: E = Pt = 18 × 45 = 810 J.
Why it matters: A second answer of 810 W is dimensionally wrong. Check the time is in seconds.
Practice 10: Different test times
Question: Trial A loses 50 W for 12 s; B loses 40 W for 20 s. Compare energy lost.
Worked answer: A loses 600 J; B loses 800 J.
Why it matters: B’s smaller power still results in more total lost energy because it lasts longer.
Practice 11: Resistance assumption
Question: The line resistance doubles while current is unchanged. State the effect on heating.
Worked answer: P_loss = I²R; doubling R doubles the heating power.
Why it matters: Do not apply a square to resistance; the square belongs to current.
Practice 12: Fixed receiving load
Question: A 1000 W receiver operates at either 100 V or 200 V with the same total line resistance. Compare line currents.
Worked answer: I_A = 10 A and I_B = 5 A; the latter is half.
Why it matters: At fixed resistance the latter cable-heating rate is one-quarter of the former. This does not say the whole system’s energy use is quartered.
Practice 13: The full return path
Question: Two conductors each have resistance 0.8 Ω. They form the complete line to and from a receiving load.
Worked answer: R_total = 0.8 + 0.8 = 1.6 Ω.
Why it matters: If another question instead states “total line resistance 0.8 Ω,” do not double it.
Practice 14: Detect an impossible ledger
Question: A table claims 700 W sent, 700 W delivered and 60 W lost as line heating, with no other energy source.
Worked answer: Those three numbers are inconsistent: 700 W cannot equal 700 W + 60 W.
Why it matters: Identify the precise conflict instead of assuming the display is correct because its labels look polished.
Practice 15: Transformer ratio
Question: For an ideal transformer, primary coil 200 turns at 100 V and secondary coil 800 turns. Find secondary voltage.
Worked answer: Vs = 100 × 800/200 = 400 V.
Why it matters: This is an ideal ratio of voltages. It does not prove extra power is produced, and it assumes appropriate changing magnetic flux.
Practice 16: Transforming current
Question: If an ideal transformer takes 2 A at 100 V and delivers 400 V, find the ideal output current.
Worked answer: Power is 100 × 2 = 200 W; output current is 200/400 = 0.5 A.
Why it matters: The output voltage is four times as great but the current one quarter as large in the ideal model.
Practice 17: Compare honest claims
Question: A label reads “At 240 V the model line produces 75% less cable heat than at 120 V.” Is it acceptable?
Worked answer: Only if it states the comparison also assumes the same receiving power and line resistance.
Why it matters: If those conditions are absent, the statement sounds universal and exceeds the evidence.
Practice 18: Test a counterexample
Question: At 240 V a different receiving load draws 20 A. Can you assume lower heating than the 120 V, 10 A case?
Worked answer: No. At the same 2 Ω line resistance, 20 A produces 800 W heating versus 200 W at 10 A.
Why it matters: Higher voltage alone does not guarantee less heating when power and current are allowed to change.
Frequently asked questions
Why not transmit with the smallest possible current?
Smaller current reduces resistive heating for a given line resistance, but real transmission equipment must balance electrical, insulation, safety and economic constraints. The textbook principle is not an instruction for field design.
Does a transformer make electricity?
No. An ideal transformer trades voltage against current while maintaining power in the model. Real transformers have losses.
What is the best first formula?
Find what the question has specified. For a known receiving power and voltage, use P = VI to determine current. For cable heating, use I²R.
How can I check a complex answer quickly?
Compare source power with useful receiving power plus stated cable loss. Check units, then check whether the percentage denominator matches the words.
Are these figures actual Singapore transmission values?
No. The values are fabricated for a school-level mathematics and physics exercise. Use current official sources for examination requirements, not these examples for grid engineering.
What if my answer says loss goes down with higher voltage?
Add the fixed-power condition and the causal current-and-square-law reasoning. A statement without the controlling condition is incomplete.
Does more voltage always mean less lost energy?
No. At fixed useful power and unchanged line resistance it allows lower current and hence lower heating, but different loads, durations and resistance can change the result.
Where do I find the exact SEC syllabus?
Use SEAB’s 2027 G3 SEC school-candidate list and the K323 Physics syllabus linked above. These remain authoritative for assessment formats and content.
Is this an official SEAB practical task?
No. This is an invented model used to practise required physics explanations, calculations and evidence reading.
How should a student revise this alongside Mathematics?
Work the calculations again, then check the word-problem translation route if converting statements into equations is still difficult.
Official sources and next learning route
Official examination sources: SEAB SEC overview · 2027 G3 subject list · Physics K323 (2027) · Mathematics K310. For earlier foundations, use the PSLE Science explanation route and the after-PSLE secondary transition route.
Once the physical line-loss ledger is reliable, the next lesson is to test its ratios, percentages and thresholds as a Mathematics real-world problem. A useful explanation is one that can be checked: define the load, name the voltage, calculate the current, identify the loss and stop where the evidence stops.
