A circuit can have one resistance at one moment and another a moment later
A student measures an ordinary resistor, a glowing filament lamp and a diode. All three carry electric current under appropriate conditions, yet their current–potential-difference graphs do not tell the same story. For the first device, the points can make a straight line through the origin. For the second, the graph bends because the filament becomes hotter. For the third, the direction of the applied potential difference matters. Treating all three as interchangeable ‘resistors’ hides the science the examination is asking you to notice.
This guide develops one precise skill: read an I–V characteristic graph as evidence about a component. Begin with the axes and units. Describe what the actual curve does. Then explain the relevant mechanism, without inventing measurements or declaring a pattern more universal than the evidence permits. Students should already be able to read basic circuits; the new challenge is recognising when a constant-resistance model is inappropriate.
The previous Vol 0319 on static electricity explained charge, transfer and induction. Electric current is a different situation: it describes the rate of flow of charge through a point in a circuit. It is not a synonym for a stationary accumulation of charge. Keeping this distinction sharp is the first step.
Check the actual 2027 SEC subject boundary
The SEAB 2027 G3 syllabus directory lists Physics as K323. The official K323 Physics syllabus includes current, potential difference, resistance and sketching or interpreting current–potential-difference characteristics of a metallic conductor at constant temperature, a filament lamp and a semiconductor diode. The topic therefore has a definite syllabus home; it is not an invented question prediction.
For K323, Paper 1 is a 1-hour multiple-choice paper carrying 40 marks and 30% of the subject grade; Paper 2 is a 1-hour-45-minute structured and free-response paper carrying 80 marks and 50%; Paper 3 practical is 1 hour 50 minutes, 40 marks and 20%. Use the official syllabus if the assessment rules change. The related combined-science routes are different syllabuses: do not assume every detail of this standalone Physics chapter is specified for K326 or K327 simply because both include electricity.
These explanations and questions are original study exercises, not copied SEAB questions. The exact graph can vary. The safe examination habit is to read the axes and the given information rather than predict a familiar drawing.
Understand the quantities before studying the shape
Electric current I is charge passing a point per unit time: I = Q/t, where I is in amperes, Q in coulombs and t in seconds. A potential difference V between two points is energy transferred per unit charge, measured in volts. Resistance R at a particular operating point is defined by R = V/I when the current is non-zero. These definitions do not require a straight graph.
A steady current of 0.20 A means 0.20 C passes a section of the circuit each second. A potential difference of 6 V across a component means 6 J of energy is transferred for every coulomb passing through it. The two quantities describe different aspects of the process. One is a flow rate; the other concerns energy per charge.
Electrical power transferred to a component is P = VI. That fact matters when explaining a filament lamp, because greater electrical power can make the filament hotter. A temperature change may change its resistance. You do not need to memorise a detailed microscopic story beyond the syllabus, but you must not quietly impose the constant-temperature assumption on a glowing lamp.
Read the axes exactly as drawn
An I–V graph usually places current I on the vertical axis and potential difference V on the horizontal axis. On that arrangement, the ratio I/V is the gradient of a straight line through the origin; the resistance is V/I, its reciprocal. In a V–I graph, where voltage is vertical, the gradient can instead be resistance. Read the labels every time.
A graph can show negative voltages and negative currents to represent reversal of the polarity and current direction relative to chosen reference directions. A straight line through the origin continuing into the third quadrant represents an ohmic conductor under the same constant-temperature conditions in both directions. Do not describe its negative branch as ‘negative resistance’: V/I is positive when both quantities have the same sign.
Check the vertical and horizontal scale separately. A point two grid squares from the origin is not necessarily 2 A or 2 V. Units may be mA, not A. If 200 mA is supplied, write 0.200 A before calculating resistance in ohms. Correct graphs begin with ordinary reading discipline.
Why a constant-temperature metal can obey Ohm’s law
The syllabus language matters: current through a metallic conductor is directly proportional to the potential difference across it provided the temperature and other physical conditions remain constant. That is Ohm’s law in this setting. A straight I–V line through the origin indicates that doubling V doubles I, so V/I remains constant.
A straight line tells a relationship, not a universal property of every component. Some students see one straight-line demonstration and write that every conductor keeps the same resistance. Others observe that a copper wire can get warm and treat the straight-line statement as false. The correct answer keeps the condition attached: at constant temperature the model applies; when temperature changes, the resistance may also change.
For example, if a component has 2.0 V across it and current 0.40 A, R = 5.0 Ω. If it remains ohmic at constant temperature, applying 4.0 V predicts 0.80 A. The prediction depends on the assumption that its physical conditions remain the same. Without that assumption, the arithmetic alone cannot guarantee the new current.
What the filament lamp curve says
A filament lamp typically gives a curve that becomes less steep on an I–V graph as the magnitude of potential difference grows. As current increases, the filament heats up. In a metal, the greater temperature increases resistance, so each additional volt produces a smaller increase in current than earlier. The curve is not a straight line.
Do not confuse ‘flatter graph’ with ‘less current’. On the positive side, the current may still increase as voltage rises, while its rate of increase becomes smaller. The filament resistance at a particular non-zero point is V/I. It should generally grow with hotter operating conditions. You cannot obtain one universal constant resistance by measuring the graph at a single point and using it everywhere.
Nor should you declare the filament fully hot the instant a switch closes. Heating and cooling take time. Actual observations can depend on how quickly the readings are taken. An examination graph provides the pattern you need, but good experimental reasoning acknowledges conditions when interpreting it.
What a semiconductor diode does differently
An idealised semiconductor diode allows substantial current in its forward direction once a suitable forward potential difference is applied, while allowing very little reverse current over the ordinary range shown in a school graph. This asymmetric characteristic is the key distinction: reversing the polarity does not normally produce a mirror-image current of the same magnitude.
Students should not write that every diode turns on at one exact universal voltage. A graph may show an approximate knee, but the precise value depends on diode type and conditions; the K323 skill is interpretation, not memorising one magic threshold. A light-emitting diode is a particular type of diode, but not every diode glows. Forward-bias a diode in the expected direction and a suitable circuit may conduct; reverse it and ordinary current may be negligible in the plotted region.
Real diodes have limits. Reverse breakdown at sufficiently large voltage is not a classroom instruction to apply unsafe voltages, and it should not be inferred if not displayed. This chapter only interprets supplied diagrams or teacher-supervised low-voltage observations; do not experiment with mains circuits.
Resistance on a curved graph is not the gradient
A frequent error is to draw a tangent to a filament-lamp I–V curve and call its gradient ‘the resistance’. With current vertical and potential difference horizontal, the gradient is a local change in current divided by a local change in voltage, not V/I. At a point, R = V/I uses the point’s coordinates. These are different calculations on a nonlinear curve.
Suppose a lamp operates at 6.0 V and 0.30 A. Its resistance at that point is 20 Ω. A small increase in voltage may raise current by only 0.02 A; that local rate of change can differ greatly from 0.30/6.0 A/V. There is no contradiction. The point ratio describes resistance under those conditions; the tangent slope describes how quickly the current changes near that point.
Later Mathematics Vol 0222 introduces tangent gradients as a separate mathematical reading skill. Keep the physical meaning of the axes attached before importing any gradient method.
A practical method for gathering honest readings
Use a safe, low-voltage school apparatus under your teacher’s guidance. Connect an ammeter in series with the component and a voltmeter in parallel across it. Change the supplied potential difference gradually within the component and equipment ratings; note the current each time. Record both readings with their units and a consistent convention for polarity.
A variable resistor may help control current, but a school investigation is not a licence to connect an unknown component directly across a high-power source. The order of readings may matter for a filament lamp because its temperature changes. If a reading looks strange, first check connections, ranges, polarity and heating before treating the value as a new law of Physics.
Use enough readings to test shape, especially where a diode characteristic turns sharply. Plot an appropriate scale; do not force a ruler through curved data. The sensible conclusion is about the observed trend within the measured range, not a claim that every voltage everywhere must obey the same curve.
The examination answer should describe, explain and qualify
For a description, state observable graph features: through the origin, straight, curved, steeper or flatter, approximately zero reverse current, or asymmetry. For an explanation, give the reason the syllabus supports: constant temperature and constant resistance for an ohmic conductor; filament heating and increased resistance for the lamp; direction-dependent conduction for a diode.
For a calculation, quote a point and write R = V/I with units. For comparison, name the same axis, voltage interval and component. For a judgement, state the limitation of the data. The command word controls how much interpretation is required; a graph description alone may not answer an ‘explain’ question.
Build answers around two clauses rather than a long paragraph full of repeated statements: ‘The lamp’s I–V curve becomes less steep as voltage rises because its filament heats up and its resistance increases.’ Then, if the question asks about a particular point, calculate that point’s resistance separately.
The bridge from primary school and earlier electricity chapters
At PSLE level, students learn to identify circuit components, use evidence from a switch or lamp and explain what a change in a circuit means. The PSLE Science revision route emphasises understanding before memorising phrasing. Secondary Physics adds measurable quantities, their units and models with stated conditions.
If you cannot yet tell series from parallel connections, return to Vol 0159 on series circuits and Vol 0171 on meters. A component’s graph is only useful when the measured potential difference is across that component and the measured current is through it.
That is the learning sequence: correct circuit, correct quantity, correct graph, correct physical explanation. Trying to memorise three shapes before understanding current and voltage reverses the order and makes each new drawing seem like an exception.
Forty original examination checks with complete reasoning
The following short cases move from recognizing an axis to diagnosing an experimental conclusion. Read the prompt, make a decision before reading the explanation, and then change one condition to see whether your answer still holds. A correct number without the right physical meaning is not a complete answer.
1. A wire gives readings (1 V, 0.2 A), (2 V, 0.4 A), (3 V, 0.6 A).
These three points fit a line through the origin. The calculated resistance is 1/0.2 = 5 Ω and remains 5 Ω at the other two readings. State that they support ohmic behaviour over the measured range, assuming constant temperature. The data do not prove every possible operating condition has constant resistance.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
2. The vertical graph axis says I/A; the horizontal says V/V. What does slope mean?
For a straight line, slope is change in current divided by change in potential difference, in A/V. It is conductance in the straight-line case, not resistance in ohms. To find the resistance, take V/I at a suitable non-zero point. Students lose marks when they recall a generic ‘gradient equals resistance’ rule without reading axes.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
3. A graph instead places V vertically and I horizontally.
The gradient of a straight through-origin V–I graph is V/I and can represent resistance in ohms. The physical component may be unchanged; only the graph convention has reversed. Write the units of the gradient as a check. The labels, not the name ‘characteristic graph’, determine the calculation.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
4. A current of 250 mA passes at 5.0 V.
Convert 250 mA to 0.250 A. The operating resistance is 5.0/0.250 = 20 Ω. Dividing 5 by 250 would give the wrong numerical value if then labelled in ohms. This is a unit-conversion problem before it becomes a physics calculation.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
5. A metallic conductor’s voltage doubles at constant temperature.
For an ohmic metallic conductor under unchanged physical conditions, current also doubles. The ratio V/I remains constant. Do not say twice the voltage creates twice the electrons; the law relates measurable current and potential difference, not production of charged particles.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
6. The same wire is allowed to become hotter.
A metallic wire’s resistance generally increases as its temperature rises. That breaks the constant-temperature condition needed for a straight ohmic relationship across changing voltages. The student must distinguish a wire’s material from the experimental conditions in which it behaves as a fixed resistance.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
7. A filament lamp carries more current at 6 V than at 3 V.
This does not make it ohmic. The relevant test is whether current increases in direct proportion to voltage, keeping V/I constant. On a lamp, rising temperature usually raises resistance. Compare the ratios at actual readings rather than deciding from the fact that both quantities increased.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
8. A lamp gives 2 V, 0.20 A and 6 V, 0.30 A.
At the first operating point R = 10 Ω; at the second R = 20 Ω. Since resistance has changed, these data do not fit one constant-resistance straight line. The filament’s rising temperature provides the usual physical explanation. Check that each resistance uses voltage and current measured together.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
9. A student calls a gently sloping positive lamp curve ‘decreasing current’.
The curve may still rise as voltage increases. Becoming less steep means the additional current per additional volt becomes smaller, not necessarily negative. Read whether the curve actually falls; if it does not, ‘decreasing current’ is a false description of the graph.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
10. A diode shows almost no current at negative voltages.
The displayed range suggests negligible reverse current under those conditions. Do not conclude ‘no charge exists’ or ‘the diode cannot ever conduct in reverse under any possible voltage’. The task is to interpret the plotted observations, not to claim universal absolute behaviour.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
11. Reverse polarity is applied to a diode.
Reversing polarity changes the direction of attempted conduction relative to the component’s forward direction. In the simplified school characteristic, its reverse current is very small. A diode is not equivalent to a fixed resistor with a symmetric positive and negative characteristic.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
12. A line crosses the origin but bends farther from it.
Passing through the origin is not enough to establish Ohm’s law. Direct proportionality requires a straight line through the origin over the relevant range and unchanged conditions. A curved I–V characteristic gives different V/I ratios at different operating points, even if it includes (0,0).
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
13. A diode characteristic has a steep turn in the positive region.
A steep positive I–V rise means that a relatively small extra forward potential difference is associated with a large extra current in that region. It does not mean infinite current or an identical turn-on voltage for every semiconductor. Read the supplied graph scales before quoting any value.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
14. The question says a semiconductor diode and the student writes LED.
A light-emitting diode is one type of semiconductor diode, but the two terms are not identical in every context. Unless the question supplies light output or states LED, avoid adding claims about emitted colour, brightness or indicator design. Electrical rectification and light emission are not interchangeable observations.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
15. A component has 4 V and 0 A.
The ratio R = V/I cannot be calculated by division by zero. In a diode’s non-conducting part of an idealised characteristic, describe negligible current or very high effective resistance instead of presenting a finite numerical resistance. The formula has a domain condition that arithmetic must respect.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
16. Current is negative and voltage is negative.
With the usual consistent sign convention, the calculated V/I can be positive because a negative divided by a negative is positive. The sign tells direction relative to the reference, not automatically a physically ‘negative resistor’. The graph’s quadrant and axis definitions must be considered together.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
17. A student draws a straight best-fit line across lamp points.
A forced straight line removes the nonlinearity the investigation aims to reveal. Plot the readings and draw a suitable smooth curve where justified. In an examination drawing task, follow the given data and instruction; do not impose an ohmic relationship merely because a ruler is convenient.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
18. An ammeter is connected across the lamp.
An ammeter is normally placed in series to measure the current through the lamp. Connecting it in parallel is the wrong measurement topology and can be unsafe in a real circuit. The voltmeter should be connected across the relevant component. Identify the meter placement before trusting a table of readings.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
19. A voltmeter is placed across both lamp and another resistor.
Its reading is the potential difference across the combination, not automatically across the lamp alone. Using that reading with the lamp current to calculate a lamp-only resistance can be wrong. The student must specify exactly what the instrument measures before interpreting the I–V plot.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
20. A 4 V supply is reported but only 3 V is across the test component.
Use the 3 V measured across that component for its characteristic, not the full 4 V supply. Other circuit elements may share the potential difference. A calculation can be arithmetically tidy while physically mistaken if the wrong pair of measurements is used.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
21. The filament is allowed to cool between readings.
Temperature differences can affect the current observed at the same potential difference. A fair investigation must control or at least describe the relevant procedure. Do not reject all readings as useless; instead explain which condition was altered and why the resulting characteristic may differ.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
22. The lamp is measured immediately and again after warming.
It may draw a different current at the same applied voltage because the filament resistance changes with temperature. To describe it, separate the independent voltage setting from the time and temperature of observation. Never turn a time-dependent heating effect into a claim that current itself always falls.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
23. The conductor shows 0.12 A at 3 V. Predict current at 8 V if ohmic.
R = 3/0.12 = 25 Ω. At 8 V the predicted I = V/R = 8/25 = 0.32 A. The phrase ‘if ohmic’ is essential: the prediction assumes the conductor’s resistance stays 25 Ω. For a filament lamp the same proportional scaling would be unjustified.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
24. A graph uses 0.1 A per vertical square and 2 V per horizontal square.
The graphical slope per square is not the numerical slope in A/V. One diagonal square rise per horizontal square corresponds to 0.1/2 = 0.05 A/V. The matching resistance is 20 Ω for a through-origin straight line. Axis scales determine the physical gradient.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
25. A learner says a current of 2 A means 2 J per second.
Amperes measure coulombs per second, not joules per second. Current I = Q/t. Joules per second measure power, for which P = VI gives a relevant electrical expression. Identify whether the answer describes charge flow or energy transfer; the units expose the mix-up.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
26. A potential difference of 9 V is applied across a component.
Nine volts means 9 joules of energy transferred per coulomb between the relevant points. It does not by itself tell us current or resistance. To infer either, we need an I–V characteristic, another measured quantity or an assumption about the component.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
27. Two straight lines through the origin have different steepness.
On current-vertical I–V axes, the steeper line has greater current for a given voltage, and therefore smaller resistance. On voltage-vertical V–I axes, the steeper line has greater resistance. An exam answer that omits the axis convention can reverse the physics.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
28. One diode carries 0.12 A at 1.5 V and almost none at −1.5 V.
The forward and reverse readings demonstrate asymmetry at those operating points. You may calculate forward operating resistance as 1.5/0.12 = 12.5 Ω, but this is not evidence that the entire forward curve is a line with constant resistance.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
29. A question asks why the filament curve flattens.
An appropriate chain is: larger current increases electrical heating; the filament becomes hotter; a hotter metal has greater resistance; therefore current increases less rapidly as applied potential difference rises. A vague answer such as ‘energy is lost’ neither states the graph pattern nor identifies the resistance mechanism.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
30. A student plots points at −4 V and +4 V for a fixed resistor.
For an ideal ohmic fixed resistor kept at constant temperature, equal and opposite applied voltages produce equal and opposite currents according to the chosen sign convention. The I–V line passes through the origin with symmetry about it. Do not apply that conclusion to a diode.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
31. The plotted lamp curve has only a few scattered readings.
Do not claim an exact function or threshold from sparse data. Describe the apparent tendency and any uncertainty. A good practical conclusion can be ‘the data suggest resistance increases as voltage rises’ while acknowledging that additional readings would make the shape more reliable.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
32. A point appears off the smooth curve.
Check for transcription, misread scales, unstable heating, loose contacts or instrument error before discarding it. An anomaly does not automatically prove faulty physics, nor may it be deleted to improve a desired line. Record the issue and, if appropriate, repeat the measurement under the same conditions.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
33. A teacher asks for the ratio V/I from two different lamp points.
Use each point’s own V and I. A lamp may give 2 V / 0.20 A = 10 Ω and 6 V / 0.30 A = 20 Ω. The comparison directly tests whether resistance is constant. Do not average the two voltages and divide by an unrelated average current to describe each operating state.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
34. The graph shows negative current for positive voltage.
Before declaring it impossible, check which direction the current arrow defines as positive and whether the axes or polarity conventions are given. In a usual passive resistor measurement the signs align; unconventional labels can change the recorded sign. Interpret the defined experiment, not an assumed picture.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
35. A student describes a diode as a perfect one-way switch.
That can be a starting simplification, but the characteristic shows gradual forward response and finite practical limits. A better answer describes very small reverse current in the indicated range and much greater forward current at suitable positive voltages. Avoid absolute claims not warranted by the graph.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
36. A graph is labelled ‘typical characteristic’ without numerical scales.
You may compare qualitative shapes, forward/reverse behaviour and whether resistance stays constant; you cannot compute a numerical operating resistance without numbers. Do not invent a voltage threshold, slope or exact current just because another textbook’s example had one.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
37. A student is asked to contrast all three components in one sentence.
A concise complete comparison is: at constant temperature an ohmic conductor gives a straight I–V line through the origin; a filament lamp gives a curve because its temperature and resistance change; a diode conducts strongly in one direction compared with the opposite direction. Then match each explanation to the command word.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
38. The question changes only the graph orientation.
Re-read the quantities. If the physical device and readings stay unchanged, the same point has coordinates exchanged, not different physics. Resistance still means potential difference divided by current. Translate the graph before deciding whether ‘steeper’ means larger or smaller resistance.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
39. A model answer says ‘current is the speed of electrons’.
Current is rate of charge flow, I = Q/t. Charge carriers in a wire have a drift speed, but current also depends on how much charge passes a cross-section each second. Avoid replacing a defined electrical quantity with an everyday word that changes its meaning.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
40. A student wants one final check before handing in.
Read the labels; state the observed shape; quote a point if calculating; use V/I rather than a guessed gradient; identify temperature or direction where relevant; include units; avoid claims outside the plotted range. This short sequence catches more mistakes than drawing the characteristic from memory again.
A useful self-check is to say which quantity, condition or graph feature made that conclusion possible. If your reason would sound equally plausible for all three components, it probably needs more precision. Try explaining it aloud without looking at the answer, then compare the wording with the plotted evidence.
Turn a weak answer into a defensible one
Consider the question: ‘Why does the filament-lamp graph become less steep?’ A weak answer is ‘because the electricity gets weaker’. This is not a defined relationship. A better answer says the current still increases, but by smaller amounts for equal increases in potential difference because the filament becomes hotter and its resistance rises. The improved answer identifies the observable graph feature and a causal mechanism without inventing a new quantity.
Consider the diode: ‘The current is zero backwards because diodes only work forwards.’ The improved answer describes the small reverse current in the ordinary plotted range, the much larger current for suitable forward bias and the component’s asymmetric characteristic. There is no need to discuss exotic failure conditions unless the question supplies them.
Finally, consider a fixed conductor. ‘The steeper line has the greater resistance’ is wrong if the graph is I on V axes. Fix it by reading the axes and explaining that a greater I/V gradient means more current at the same voltage, so V/I is smaller. The correction is quick only when the student has linked the mathematics to the physical definitions.
An honest final audit and the next bridge
Use an unfamiliar graph for the final test. Name the two axes, decide whether the plotted pattern is straight or curved, and tell a reader what one point means. Predict only when the conditions support prediction. Distinguish the graph’s shape from the explanation for that shape. If you need to use a ratio, write the units before pressing the calculator.
The G3 Science learning owner covers the broader route; the post-PSLE pathway explains how to build the habits that make Secondary questions manageable. The Secondary 1 Mathematics foundation reference shows why labels, representation and checking matter before difficult algebra.
The next mathematics chapter will look at the separate task of building a straight-line model from limited readings and then checking it against new data. That is not permission to draw a line through a lamp or diode curve; it is a chance to learn exactly when a simple model can and cannot be trusted.
