G3 SEC Physics Variable-Load Cable Evidence

Quick answer: When current changes over time, calculate cable heating separately for each interval, multiply each power by its duration, and add the energies. In this invented 24 V source model with a fixed 2 Ω line, currents of 2 A for one hour, 4 A for two hours and 6 A for one hour cause 144 Wh of line heating. The correct average line-heating power is 36 W, not 32 W.

This guide moves beyond the previous fixed-delivered-power comparison. It teaches G3 Physics learners how to track a changing demand profile, keep source and receiver quantities separate, and explain exactly which conclusions the assumptions allow. Every example is fictional and text-only; it is not advice about real electrical wiring.

2027 SEC G3 Physics: K323 (prior reference code 6091)
Find line-heating power using I²R, then energy using time
Check source power, receiver power and the energy ledger
Explain peak, average and a time-weighted result
Reject invalid rows and unsupported public claims
Practise with worked examples, independent questions and a four-week plan

The new problem: demand changes
Know which quantities are being held constant
Current is a rate of charge flow
Why the cable heats up
A complete four-hour demand profile
From power to energy
Understand the source voltage ledger
Construct source, cable and receiver power
Calculate the full-day energy ledger
Average and peak power answer different questions
Why the average current is not enough
Why the total-energy fraction is 37.5%
Separate power capacity from energy use
Test a 5 A fourth scenario
Recognise an impossible proposed row
Compare fixed source and fixed receiver carefully
What the model does not include
How to write the Physics explanation
What a graph can show without a picture
Do not confuse an observation with an assumption
A practical skill without a dangerous practical build
A twelve-minute exam-style method
Why this fits the official 2027 G3 SEC
Four weeks from foundation to independent transfer
Parents: the question that reveals real understanding
Independent practice and worked answers
Frequently asked questions
Trusted references and related learning routes

The new problem: demand changes

The previous learner route held the delivered load fixed while comparing cable currents. This chapter changes that assumption. A small fictional teaching model runs for four hours, but the receiver takes different currents at different times. The question is no longer, “Which fixed operating point has the lower cable loss?” It is, “How much energy does the line waste across the entire operating day, and what can be inferred about the receiver?” A student who misses the change of assumption can perform flawless arithmetic and still write a false conclusion. Here we will make the condition visible before using a formula. The numbers describe a fictional educational circuit, not a distribution network or a safe experimental build. No student should work on mains electricity or attempt a physical installation from this article.

Know which quantities are being held constant

In this example, the model has a 24 V source and a line represented by a fixed resistance of 2 Ω. Those are the two fixed quantities. The line current changes between operating periods because the fictional receiver changes its effective demand. We do not hold receiver power constant. This difference is central. Holding delivered power constant, holding the source voltage constant and holding receiver resistance constant are three different problems. They may use the same formulas, but they do not have the same answers. At the top of your working, write the assumptions in full words. Then decide which power belongs to the source, which is dissipated in the line, and which reaches the receiver. A formula cannot decide your model for you.

Current is a rate of charge flow

Current is measured in amperes. It tells us the rate of charge passing a point in the circuit. It does not directly state the amount of electrical energy transferred or the cost of running an appliance. A source voltage, a resistance and a load are needed to describe this model’s energy transfers. When a pupil says “six amperes of electricity were used”, the sentence hides the actual physical quantity. Say instead that the current was 6 A for the stated interval. Next, identify the relevant potential difference and duration. This habit matters in SEC structured questions: physical quantities must be named and linked, not merely replaced by numerals. It also prevents confusing a peak current with the total current-time history.

Why the cable heats up

A resistive cable transfers electrical energy to thermal energy while current passes through it. Under the stipulated steady-condition model, the line-heating power is P = I²R. The square is not decoration. When the current doubles while R stays fixed, the line-heating rate becomes four times as large. This is why a time interval with high current may contribute disproportionately to total loss. It does not establish that an actual cable with changing temperature would retain exactly the same resistance; that is a simplification of this educational calculation. In an examination answer, identify the formula, substitute the given line resistance, and attach watts to the power. If the question asks for energy, watts alone cannot be the final unit.

A complete four-hour demand profile

Use this invented table: during hour 1 the current is 2 A; during hours 2 and 3 it is 4 A; during hour 4 it is 6 A. The durations are therefore 1 h, 2 h and 1 h. The line resistance is always modelled as 2 Ω. Calculate line-heating power separately for each interval: 2² × 2 = 8 W; 4² × 2 = 32 W; 6² × 2 = 72 W. Notice that the last interval is the highest-power interval, but it is not the longest. The first aim is to distinguish the rows without averaging them too soon. A single “typical current” conceals the relationship between squaring and time.

From power to energy

Power describes the rate of energy transfer; energy describes how much is transferred over a stated duration. When the power is constant inside a period, energy = power × time. The line therefore converts 8 Wh, 64 Wh and 72 Wh to heat over the three periods. Add the energies to get 144 Wh, or 0.144 kWh. The number 144 is not in watts. If you add powers from intervals of unequal length, you do not obtain the total energy. You need each interval’s duration as a multiplier. This distinction is transferable to heating, transport, battery use and any SEC question about a changing rate. The full working should show the three interval contributions before the total.

Understand the source voltage ledger

For the deliberately simplified circuit, source voltage is 24 V in every interval. The resistive line’s voltage drop is IR. At currents 2 A, 4 A and 6 A, the line drops are 4 V, 8 V and 12 V. The model therefore leaves 20 V, 16 V and 12 V across the receiver respectively. The receiver potential difference is not fixed. It would be an error to multiply 24 V by the receiver current and label the result as receiver power: 24 V belongs to the source. The formula P = VI requires a voltage and current measured across or through the same chosen component or boundary. Naming the boundary is the simplest defence against this mistake.

Construct source, cable and receiver power

Source electrical power is 24I. Thus the three source powers are 48 W, 96 W and 144 W. Cable heating powers remain 8 W, 32 W and 72 W. The receiver powers are source minus line loss: 40 W, 64 W and 72 W. You can check each row independently: 48 = 8 + 40, 96 = 32 + 64, and 144 = 72 + 72. This is an energy-accounting check, not an assertion that every real system contains only those two destinations. In this fictional model they are the only terms. An unexplained difference in a ledger means an error, an omitted boundary, or an assumption that needs to be stated.

Calculate the full-day energy ledger

Multiply each of the three sets of powers by the corresponding 1 h, 2 h and 1 h. Source energy is 48 + 192 + 144 = 384 Wh. Cable heating is 8 + 64 + 72 = 144 Wh. Receiver energy is 40 + 128 + 72 = 240 Wh. The central cross-check is 384 Wh = 144 Wh + 240 Wh. Converting all three into kilowatt-hours gives 0.384 kWh, 0.144 kWh and 0.240 kWh. Never mix watt-hours on one side with watts on the other. The examiner may reward a logical method even when the story is new, because the method respects conservation and correct units.

Average and peak power answer different questions

The maximum line-heating rate is 72 W during the final hour. The average line-heating rate over the four hours is 144 Wh divided by 4 h, or 36 W. Both statements are true, but neither may replace the other. The peak tells you the largest instantaneous or interval power in the simplified profile; the average tells you how much energy would be transferred per hour if spread evenly across the whole period. A statement that “the line uses 36 W all day” would be wrong as a description of the changing profile. A clear response reports the average and the highest interval separately, including the time basis for each.

Why the average current is not enough

The duration-weighted mean current is (2 × 1 + 4 × 2 + 6 × 1) / 4 = 4 A. If you put this into I²R, you get 32 W. That is not the correct average line-heating power of 36 W. Squaring the mean current discards the extra effect of current variation. The correct method is to square each interval current first, weight the resulting power by its duration, add the energies and divide by total time. This is not an advanced extra formula; it is careful application of the simple formula to each stated period. It provides a strong check on whether a candidate is reasoning or only memorising a procedure.

Why the total-energy fraction is 37.5%

Cable energy loss is 144 Wh out of the 384 Wh supplied by the source. The line-loss fraction for this chosen boundary is therefore 144 / 384 = 0.375, or 37.5%. Receiver energy is 240 / 384 = 62.5% of the source energy in the model. Those two percentages add to 100%. Do not call 62.5% an independently measured transformer or grid efficiency. There is no transformer in this model, and these quantities come from stated assumptions. The same percentage would change if we added further system components or changed the source conditions. A percentage gains meaning only after its numerator, denominator and boundary are named.

Separate power capacity from energy use

A source power of 144 W in hour 4 does not mean the four-hour source energy is 144 Wh. Capacity or peak-rate questions focus on the largest power needed at a moment; energy questions sum power across time. In practical language a family may ask about a total bill while a planner asks about a peak. These are different quantities. For the classroom model, the sum of 384 Wh comes from the whole day, not from choosing the maximum row. In a Physics answer, do not substitute a peak for an integral of energy over time. Even when calculus is unnecessary, the interval method makes that reasoning visible.

Test a 5 A fourth scenario

Suppose a fourth hypothetical operating period used 5 A, still with a 24 V source and 2 Ω line. The line loss would be 5² × 2 = 50 W. The line drop would be 10 V, leaving 14 V across the receiver. The source power would be 120 W; receiver power would be 70 W. The check 120 = 50 + 70 succeeds. This does not tell us which receiver setting would produce 5 A without further information; we have deliberately supplied the current. Students should distinguish a stipulated scenario from a prediction about how an actual load behaves. In exams, conditional calculations are valid when the condition is explicit.

Recognise an impossible proposed row

Consider a draft table claiming that current is 13 A while the source is 24 V and line resistance is 2 Ω. Ohm’s law gives a line drop of 26 V, exceeding the source voltage in the simple passive-receiver model. The claimed row would leave a negative receiver voltage and negative receiver power under that model. That alerts us that the assumptions or the row are incompatible. A strong student does not calculate a cheerful efficiency and move on. They flag the conflict, explain why the model cannot support the claimed positive passive load, and ask which statement must be corrected. This is a boundary check, not an invitation to alter an electrical circuit.

Compare fixed source and fixed receiver carefully

The preceding volume’s comparison held delivered power fixed. In our new profile, source voltage is fixed instead and delivered power changes with current. Students must not copy the conclusion “higher voltage means a lower current for the same delivered power” into a problem where delivered power is not held constant. The formula may still be available, but the surrounding logic is different. Write “in the fixed-load comparison” when referring to earlier work and “in the fixed-source model” here. Exams can change just one condition to test understanding. The accurate answer often begins by noticing that the question is not the same as the previous worked example.

What the model does not include

The model assumes constant line resistance and idealised steady intervals. It does not model temperature-dependent resistance, a.c. power factors, time-dependent start-up transients, the electrical design of actual appliances, or any protective equipment. These omissions do not make the arithmetic useless; they define its limits. A public claim about a real building or grid would require different data and competent engineering review. An examination learner can use this case to practise P = VI, V = IR, P = I²R and E = Pt without pretending to have designed a safe transmission system. Safe classroom work takes place under a teacher’s approved supervision, not by following internet wiring instructions.

How to write the Physics explanation

A high-quality two-sentence explanation might say: “At constant model resistance, line-heating power is proportional to the square of the current. We therefore calculate each period’s loss before multiplying by its duration, because the four-hour energy total depends on both changing current and time.” This response links cause, equation, method and unit without clutter. If asked for a calculation, show each row and a final unit. If asked for an evaluation, add the limitation that current values were provided rather than predicted from a complete receiver model. Clear explanations neither overstate the claim nor hide behind vague language.

What a graph can show without a picture

Describe a graph with time on the horizontal axis and line-heating power on the vertical axis. The 0–1 h plateau is at 8 W; the 1–3 h plateau is at 32 W; the 3–4 h plateau is at 72 W. The area of each rectangular interval represents energy in watt-hours, provided the units and scales are consistent. The total area is 144 Wh. A second graph of current against time would have plateaux at 2 A, 4 A and 6 A, not the same vertical values. The purpose of the graph is to reveal which quantity changes; a correctly labelled table can perform the same analysis when images are not available.

Do not confuse an observation with an assumption

The 2 A, 4 A and 6 A values are inputs to an invented problem. They are not instrument readings from a school experiment. Likewise the fixed resistance of 2 Ω is a declared assumption. The 144 Wh result is derived from them. If a report calls all three categories “measurements”, it misrepresents the evidence. Useful science writing separates what was stipulated, what was observed and what was inferred. That distinction matters when students explain a practical experiment, interpret graphs or evaluate an unfamiliar scientific claim. It also trains honesty beyond the examination hall.

A practical skill without a dangerous practical build

The official G3 Physics syllabus includes experimental planning, observation, data presentation and evaluation, including questions on electrical quantities. This guide can prepare those thinking habits with printed data, teacher-approved simulations or safe existing school demonstrations. It does not provide a mains experiment. To evaluate an investigation description, look for the variable changed, the quantities measured, what was kept fixed, precision of readings, repeats where relevant, and whether the conclusion exceeds the evidence. If a question says no measuring equipment was used, do not invent readings. We can learn experimental reasoning from a hypothetical table while remaining exact about its status.

A twelve-minute exam-style method

Read the question for two minutes and underline fixed source, variable current, total duration and line resistance. Allocate four minutes to calculating per-interval source, line and receiver power. Spend three minutes multiplying by time and adding the energy ledger. Use two minutes to audit units, denominator choices and any claim about efficiency. Reserve a final minute for a clear sentence with the assumption attached. This is not an official time allocation; it is an independent practice routine. Its value is the order of thinking. A confident student will adapt the routine rather than use it mechanically on every problem.

Why this fits the official 2027 G3 SEC

SEAB lists Physics as G3 K323 in 2027, replacing the old reference code 6091. The syllabus includes current electricity, electrical power, circuits and electromagnetic induction, as well as the treatment of cable loss in transmission. G3 Physics has Paper 1 multiple choice, Paper 2 structured/free response, and Paper 3 practical. The three paper weightings are 30%, 50% and 20%. The fictional variable-load case is an original teaching extension using syllabus principles, not a reproduced examination task or a claim about what will appear in a paper. For official assessment rules, candidates and parents should read the linked SEAB syllabus for their exact examination year.

Four weeks from foundation to independent transfer

In week one, revise current, potential difference, resistance and electrical power, writing the unit next to every numerical result. In week two, solve separate steady-interval questions using I²R and E = Pt, then check a complete source–line–receiver ledger. In week three, vary one assumption at a time: fix delivered power instead of source voltage, or change the intervals without changing their order. Explain in words why the answer changes. In week four, answer an unfamiliar variable-load question with no worked example beside you. Mark every numerical, unit and explanation error, then retry a changed question after a delay. Independence is the aim, not repeating one memorised table.

Parents: the question that reveals real understanding

Ask the learner: “Why is the average current 4 A while average line heating is 36 W rather than 32 W?” An effective answer will mention the squaring operation before averaging over time, then reconstruct the interval energies. If the learner can only state that 36 W is correct, the principle is not yet secure. Another useful question is whether 62.5% is a whole-grid efficiency; the answer must identify the model boundary. These two questions diagnose calculation and evidence language together without requiring a parent to be a Physics specialist. The learner should be able to point to the working that supports each sentence.

Independent practice and worked answers

Attempt each question before opening the answer. The most useful habit is to state the fixed conditions first, identify the correct quantity and unit, and only then calculate. The answer notes explain both the method and the common shortcut that would fail.

Practice 1: Rebuild the first interval

Question: At 2 A and 2 Ω, calculate line power and energy over one hour.

Worked answer: P = I²R = 4 × 2 = 8 W; E = 8 W × 1 h = 8 Wh. The unit changes because power has been multiplied by duration. State that this is a modelled line loss, not receiver consumption.

Practice 2: Rebuild the middle interval

Question: The current is 4 A for two hours. How much energy is converted into line heat?

Worked answer: The rate is 4² × 2 = 32 W. Multiply by two hours to obtain 64 Wh. Writing 32 Wh would silently treat two hours as one. Writing 64 W would confuse energy and power.

Practice 3: Rebuild the final interval

Question: For 6 A in the last hour, compare the line’s rate with the first interval.

Worked answer: At 6 A, P = 6² × 2 = 72 W. Compared with 8 W at 2 A, the rate is nine times greater because current is three times greater. The duration happens to be equal in these two rows.

Practice 4: Check the receiver voltage

Question: Find the receiver voltage at 4 A.

Worked answer: The line drops IR = 4 × 2 = 8 V. Receiver voltage is 24 − 8 = 16 V in the simplified series model. Always label the subtraction; a number without a boundary is ambiguous.

Practice 5: Check a source power row

Question: Determine source, cable and receiver power at 4 A.

Worked answer: Source = 24 × 4 = 96 W, cable = 32 W, receiver = 16 × 4 = 64 W. The 96 W source power is partitioned into 32 W and 64 W. Do not treat 96 W as receiver power.

Practice 6: Check the total source energy

Question: Use 48 W for one hour, 96 W for two and 144 W for one. Find total energy.

Worked answer: 48 + 192 + 144 = 384 Wh. The sum of watt figures would not be a valid total energy unless all intervals were identical one-hour intervals.

Practice 7: Check the receiver energy

Question: Use receiver powers 40 W, 64 W and 72 W across the three periods.

Worked answer: 40 × 1 + 64 × 2 + 72 × 1 = 240 Wh. Adding cable loss 144 Wh returns the source energy 384 Wh. That equality is an important independent check.

Practice 8: Find a loss share

Question: What portion of source energy becomes line heat?

Worked answer: 144 / 384 × 100% = 37.5%. This is a source-energy share over the stated four-hour model. It is not a 37.5% reduction relative to another design.

Practice 9: Find an average

Question: What is the average line power over the full four hours?

Worked answer: 144 Wh / 4 h = 36 W. State the averaging period. Average power and highest interval power answer separate questions; the maximum is 72 W.

Practice 10: Catch the false shortcut

Question: A learner squares the 4 A average current to get 32 W. Why is it wrong?

Worked answer: The square must be applied to each interval current before time-weighted averaging. Squaring the mean loses the extra loss due to variation. Demonstrate this by comparing 144 Wh with the 128 Wh shortcut.

Practice 11: Test a claim

Question: “The line always loses 37.5% of source power.” Is that supported?

Worked answer: No. 37.5% is a total energy fraction across the four-hour profile. The instantaneous line shares are 8/48, 32/96 and 72/144, which are not equal.

Practice 12: Identify a model failure

Question: Can the given passive receiver work at 13 A with 24 V and 2 Ω line resistance?

Worked answer: The calculated line drop would be 26 V, greater than the 24 V source. It contradicts the simple positive passive-receiver assumption. The proper response is to flag the incompatible row.

Practice 13: Reverse the thinking

Question: If the 2 A period lasted three hours rather than one, which part of the work changes?

Worked answer: The power for the 2 A row remains 8 W because current and resistance did not change. Its energy becomes 24 Wh. Totals and averages must be recomputed with the revised total duration.

Practice 14: Separate facts from claims

Question: Classify “The model’s line resistance is 2 Ω” and “The line loses 144 Wh”.

Worked answer: The first is a given assumption. The second is a derived result conditional on the stated currents and durations. Neither is evidence of actual school-cable performance.

Practice 15: Explain the limiting sentence

Question: Write one clear final conclusion with a boundary.

Worked answer: For the invented 24 V model with a constant 2 Ω line and the specified four-hour demand profile, 144 Wh is transferred to line heating out of 384 Wh supplied. This does not establish real building performance.

Practice 16: Prepare for transfer

Question: Could you use the same totals if the source voltage changed with time?

Worked answer: Not without recalculating each interval. The source power would depend on the new voltage and current, and the receiver ledger would change accordingly. The method survives, but the fixed source assumption does not.

Frequently asked questions

Is this a real electrical installation guide?

No. The circuit and data are invented for learning. Real electrical systems need approved equipment, qualified professionals and different safety analysis.

Is a watt-hour different from a watt?

Yes. A watt is power, a rate of energy transfer. A watt-hour is energy transferred during time. One kilowatt-hour equals 1000 watt-hours.

Why not use the mean current in I²R?

Because the heating relationship is nonlinear in current. Calculate each interval’s heating power, weight it by duration and then average.

Does higher current always mean more useful power?

Not as a universal claim. In this stated source-and-line model receiver power depends on both current and the receiver voltage, which is not fixed.

What is the largest line-loss rate here?

72 W during the one-hour 6 A period. It is a peak interval rate, not the whole-period average.

What is the total line heating?

144 Wh, or 0.144 kWh, using the three stated intervals and constant modelled line resistance.

What is the difference from Vol 0729?

Vol 0729 compares fixed delivered power under a different operating assumption. This volume tests variable current at fixed source voltage and derives a time ledger.

What is tested in G3 Physics?

SEAB K323 includes electricity, energy transfers and scientific reasoning; its assessment includes MCQ, structured/free response and a practical component. Check the official syllabus each year.

Will a long explanation always score more?

No. Use the fewest sentences that connect the relevant condition, principle, calculation and conclusion. Extra unsupported claims weaken the response.

Can I use the numbers for a school electricity bill?

No. They describe an invented low-voltage teaching case with no measured tariff or real building data.

How should I correct a unit mistake?

Write the quantity name, its defining equation and its final unit, then re-solve a different example so the correction transfers.

Where can a beginner revise earlier ideas?

Use the linked preceding volumes and the site’s PSLE English–Mathematics–Science learner route for earlier quantity, evidence and reading foundations.

Official sources and previous routes

The official syllabus and codes should take priority over any old exam label or third-party study guide. This page describes an original teaching case, not an official specimen question.

Small Group Tutorials

How to Perform in the new G3 SEC Examinations | Learner’s Guide Vol 0733 | Physics Variable-Load Cable Evidence — Compare Source, Line and Receiver Across Changing Demand

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