Revision Notes
OCR GCSE Physics A: Resistance, Ohm's law and I-V characteristics (J249) – Revision Notes
Condensed revision notes on V = IR, constant and changing resistance, circuit design and I-V graphs for OCR GCSE (9-1) Gateway Science Physics A.
- Subject
- Physics
- Level
- GCSE
- Topic
- Resistance, Ohm's law and I-V characteristics
- Author
- Marlbridge Academic Team
- Updated
Aligned to OCR GCSE Physics (J249), Version 5.0 (July 2026), for first assessment in 2018. Official specification .
Syllabus page (what it covers and how it is assessed): OCR GCSE Physics.
Syllabus points this page covers
J249
- P3.2 Simple circuits
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These revision notes are for OCR GCSE (9-1) Gateway Science Physics A (J249), Version 5.0 (July 2026), and cover statements P3.2c to P3.2h in topic P3, Electricity. All of this content can be tested in both Foundation and Higher Tier papers. For full explanations and extra worked examples, read the study guide, then test yourself with the practice questions. See also the OCR GCSE Physics hub and the printable checklist. Unsure where to begin? Take the free 10-minute diagnostics. For paper technique, see exam preparation.
Quantities and units (P3.2c)
| Quantity | Symbol | Unit | Measured with |
|---|---|---|---|
| Current | I | ampere (A) | Ammeter, in series |
| Potential difference | V | volt (V) | Voltmeter, in parallel across the component |
| Resistance | R | ohm (Ω) | Calculated from V ÷ I |
- Potential difference: energy transferred per unit charge between two points. 1 V = 1 J per coulomb.
- Resistance: how hard it is for charge to flow through a component.
- More p.d. across a component gives more current. More resistance gives less current for the same p.d.
Prefixes to recall: 1 mA = 0.001 A, 1 kΩ = 1000 Ω, 1 MΩ = 1 000 000 Ω, 1 µA = 0.000 001 A.
The equation (P3.2d)
| Find | Equation | Unit |
|---|---|---|
| p.d. | V = I × R | V |
| current | I = V ÷ R | A |
| resistance | R = V ÷ I | Ω |
Method in four steps
- Write down the quantity you want and the two you know, with units.
- Convert mA, kΩ and so on to A and Ω.
- Write the equation, rearranged for the unknown, then substitute.
- Finish with the unit, and round sensibly.
Small worked reminder. A component has 0.15 A through it and a resistance of 40 Ω. V = I × R = 0.15 × 40 = 6.0 V.
Second reminder, with a prefix. A 2.2 kΩ resistor has 11 V across it. I = 11 ÷ 2200 = 0.0050 A = 5.0 mA.
Constant or changing resistance (P3.2e)
| Constant R (ohmic) | R changes |
|---|---|
| Fixed resistor | Filament lamp |
| Metal wire at constant temperature | Diode |
| NTC thermistor | |
| LDR |
- V = IR always holds at a given moment. What changes is the value of R.
- Wire: longer wire means larger R (R is proportional to length at constant temperature). Thicker wire means smaller R.
Must-know distinction. The key idea is that temperature changes resistance. A metal wire stays ohmic only if its temperature stays constant. That is why wire experiments use small currents and switch off between readings.
Circuit design (P3.2f)
Standard circuit
- Component, ammeter and variable resistor all in series with the supply.
- Voltmeter in parallel across the component only.
- Alter the variable resistor (or use different numbers of cells) to get a range of readings. Reverse the connections for negative values.
Suggested practical work from the specification: record the p.d. across and current through different components and calculate resistances, investigate resistance in a wire, and investigate the effect of length on resistance in a wire.
| Component | What to vary | What to keep constant or watch |
|---|---|---|
| Wire | Length (move a clip along it) | Same wire, small current, wire not hot |
| Filament lamp | P.d. | Read current at each p.d.; the lamp glows brighter as current rises |
| Diode | P.d., in both directions | Protective resistor in series |
| NTC thermistor | Temperature (water bath) | Same p.d. each time |
| LDR | Light intensity | Other light kept constant |
I-V graphs (P3.2g and P3.2h)
Current on the vertical axis, p.d. on the horizontal axis.
| Element | Graph | R |
|---|---|---|
| Resistor, wire at constant temperature | Straight line through the origin | Constant |
| Filament lamp | S-shaped curve through the origin, flatter at high p.d. | Rises as filament heats |
| Diode | Zero or tiny current in reverse and at low forward p.d., then steep rise | Huge in reverse, small when conducting |
| NTC thermistor | Curve that steepens if the current warms it | Falls as temperature rises |
| LDR | Steeper line in bright light | Falls as light intensity rises |
Rules for graphs
- Straight line through origin means linear. Anything else means non-linear.
- Straight line, current vertical: gradient = 1 ÷ R. Steeper means smaller R.
- Curve: find R at a point with V ÷ I. The gradient of the curve is not R.
- Draw a smooth line of best fit through plotted points.
Why a filament lamp’s resistance rises. More current heats the filament. Its ions vibrate more, so free electrons collide with them more often and charge flows less easily.
Reading a table or graph in four steps
- Pick a pair of readings, V and I, from the same point.
- Calculate R = V ÷ I for that pair.
- Repeat for a second pair. If R is the same, the element is linear over that range. If R changes, it is non-linear.
- Say which way R changes and link it to the cause: temperature for a lamp or thermistor, light for an LDR, direction of current for a diode.
Worked reminder. A component carries 0.10 A at 4.0 V and 0.16 A at 8.0 V. R = 4.0 ÷ 0.10 = 40 Ω, then R = 8.0 ÷ 0.16 = 50 Ω. The resistance has risen, so it is non-linear, and a filament lamp is a likely match because more current heats the filament.
Second worked reminder. A resistor gives 0.05 A at 2.0 V and 0.15 A at 6.0 V. R = 2.0 ÷ 0.05 = 40 Ω and R = 6.0 ÷ 0.15 = 40 Ω. The ratio is constant, so it is linear.
Command words in this topic
- State or recall: one fact, no reasons needed, such as the unit of resistance.
- Calculate: write the equation, substitute, give the answer with a unit.
- Describe: say what happens, such as the shape of a graph or how R changes.
- Explain: give a reason as well, using the idea of heating, vibrating ions and collisions, or the effect of light or temperature on resistance.
- Suggest or evaluate: use the data given, quote figures, and say what the evidence can and cannot show.
Applications
- Thermistor: temperature sensing, such as a thermostat or fire alarm.
- LDR: light sensing, such as automatic night lights.
- Diode: current in one direction only.
Quick self-test
- A component passes 0.60 A when the p.d. across it is 12 V. Find R.
- A 120 Ω resistor has 3.0 V across it. Find the current in mA.
- A student measures 5.0 V across a 250 Ω resistor. Find the current in A.
- State the unit and symbol for p.d.
- A 2.2 kΩ resistor is connected across 11 V. Find the current in mA.
- A wire of length 0.90 m has resistance 5.4 Ω. What is the resistance of 0.30 m of the same wire at constant temperature?
- Describe the shape of the I-V graph for a fixed resistor.
- Explain why the resistance of a filament lamp increases as the current increases.
- State how the resistance of an NTC thermistor changes as temperature rises.
- What happens to an LDR’s resistance when it is moved into brighter light?
- Which component’s I-V graph shows almost no current in one direction?
- A voltmeter is placed in series with a resistor by mistake. What is wrong with the connection?
Answers
- R = V ÷ I = 12 ÷ 0.60 = 20 Ω
- I = V ÷ R = 3.0 ÷ 120 = 0.025 A = 25 mA
- I = 5.0 ÷ 250 = 0.020 A
- Volt, V (1 V = 1 J per coulomb).
- I = 11 ÷ 2200 = 0.0050 A = 5.0 mA
- 0.30 m is one third of 0.90 m, so R = 5.4 ÷ 3 = 1.8 Ω
- A straight line through the origin.
- The filament gets hotter, so its ions vibrate more, so electrons collide with them more often, so it is harder for charge to flow.
- It decreases.
- It decreases.
- A diode (in the reverse direction).
- A voltmeter must be connected in parallel across the component. In series, its very high resistance would let almost no current through the circuit.
Where marks are usually lost
- Not converting mA or kΩ before substituting.
- Leaving out the unit, or writing the wrong one such as A for resistance.
- Writing “the resistance goes up because it is hot” without naming the cause: ions vibrate more and electrons collide more often.
- Mixing up the thermistor (temperature) and the LDR (light).
- Using the gradient of a curve instead of V ÷ I for resistance.
- Drawing the diode curve symmetrically, or the lamp curve as a straight line.
- Putting the voltmeter in series, or leaving out the variable resistor from a circuit design.
- Describing an experiment without saying what is kept constant, what is measured, and how readings are repeated.
- Dot-to-dot lines on a graph instead of a smooth best-fit line.
Official syllabus
OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), for first assessment in 2018, published by OCR.
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Study guides
OCR GCSE Physics A: Resistance, Ohm's law and I-V characteristics (J249)
Study guide to resistance, V = IR, resistors, lamps, diodes, thermistors, LDRs and I-V graphs for OCR GCSE (9-1) Gateway Science Physics A.
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Practice questions
OCR GCSE Physics A: Resistance, Ohm's law and I-V characteristics (J249) – Practice Questions
Original practice questions with worked answers on V = IR, wire resistance, lamps, diodes, thermistors, LDRs and I-V graphs for OCR GCSE Physics A.
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OCR GCSE Physics A: Series and parallel circuit calculations and electrical power (J249)
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