OCR GCSE Physics A: Resistance, Ohm's law and I-V characteristics (J249) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions are for OCR GCSE (9-1) Gateway Science Physics A (J249), Version 5.0 (July 2026), statements P3.2c to P3.2h, on resistance, V = IR and I-V characteristics. All are suitable for Foundation and Higher Tier candidates. Read the study guide first if you need to, and use the revision notes to check your recall. See also the OCR GCSE Physics hub and the printable checklist. Unsure where to begin? Take the free 10-minute diagnostics.

Questions

1. State the unit, and its symbol, for each of: current, potential difference and resistance. [3]

2. A resistor of resistance 330 Ω has a p.d. of 4.5 V across it. Calculate the current through it in milliamperes. [3]

3. An ammeter reads 0.28 A and a voltmeter across the same component reads 7.0 V. The component has constant resistance.

(a) Calculate its resistance. [2] (b) Calculate the p.d. needed to make a current of 0.12 A flow through it. [2]

4. (a) State what is meant by potential difference. [1] (b) What does a p.d. of 9 V across a component tell you about the energy transferred? [1] (c) A 6.0 V supply is connected across a 20 Ω resistor. The resistor is then replaced by a 40 Ω resistor. Calculate the current in each case and describe the change. [3]

5. Describe how you would investigate the effect of the length of a wire on its resistance. Include the circuit and how you would obtain reliable results. [5]

6. In an investigation, the current through each length of one wire was kept at 0.60 A. The p.d. across the wire was recorded.

Length (m)	0.30	0.60	0.90
P.d. (V)	1.44	2.88	4.32

(a) Calculate the resistance for each length. [3] (b) State the relationship between resistance and length shown by the data. [1] (c) Predict the resistance of 1.20 m of the same wire at the same temperature. [1]

7. The table shows readings for a filament lamp.

P.d. (V)	2.0	4.0	6.0	8.0
Current (A)	0.30	0.50	0.64	0.75

(a) Calculate the resistance at 2.0 V and at 8.0 V. [2] (b) Describe how the resistance changes and explain why. [3]

8. Describe the I-V graph of a diode and state what it shows about how the diode works. [4]

9. (a) An NTC thermistor is connected across a 3.0 V supply. Its resistance is 1800 Ω at 25 °C and 300 Ω at 75 °C. Calculate the current in mA at each temperature. [3] (b) An LDR is in the same type of circuit. State how its resistance changes when the light becomes brighter, and what happens to the ammeter reading. [2]

10. Two components, X and Y, are tested. The table gives their readings for both directions of current.

P.d. (V)	-6.0	-3.0	3.0	6.0
Current through X (A)	-0.12	-0.06	0.06	0.12
Current through Y (A)	-0.15	-0.10	0.10	0.15

(a) State which component is linear and give a reason from the data. [2] (b) Calculate the resistance of X. [1] (c) Calculate the resistance of Y at 3.0 V and at 6.0 V. [2] (d) State what the results show about Y and name a component that could be Y. [1]

11. A student investigates an NTC thermistor.

(a) Describe a circuit and method for finding the thermistor’s resistance at different temperatures. [3] (b) The student keeps the p.d. at 4.0 V. The current is 2.0 mA at 10 °C and 20.0 mA at 70 °C. Calculate the resistance at each temperature. [2] (c) State the pattern shown. [1] (d) The student says: “This proves the thermistor does not obey V = IR.” Evaluate this claim. [2]

Answers

1. Current: ampere, A [1]. Potential difference: volt, V [1]. Resistance: ohm, Ω [1]. [3] Examiner insight: Give both the name and the symbol, and use the Greek letter omega for the ohm. Writing only a symbol or only a name risks losing the credit for that quantity.

2. I = V ÷ R [1]. I = 4.5 ÷ 330 [1]. I = 0.0136 A = 13.6 mA [1]. [3] Examiner insight: Write the equation and the substitution as separate lines. A wrong unit, such as leaving the answer as 0.0136 when milliamperes were asked for, can cost the final mark.

3. (a) R = V ÷ I = 7.0 ÷ 0.28 [1] = 25 Ω [1]. (b) V = I × R = 0.12 × 25 [1] = 3.0 V [1]. [4] Examiner insight: Use your answer to part (a) in part (b), and set out each substitution on its own line so your method is easy to follow.

4. (a) The energy transferred per unit charge (per coulomb) as charge moves between two points [1]. (b) Each coulomb of charge transfers 9 J of energy [1]. (c) 20 Ω: I = 6.0 ÷ 20 = 0.30 A [1]. 40 Ω: I = 6.0 ÷ 40 = 0.15 A [1]. Doubling the resistance halves the current (same p.d.) [1]. [5] Examiner insight: Definitions of p.d. should mention energy and charge. “Voltage pushes the current” is too vague to describe it.

5. Connect the wire in series with a power supply, an ammeter and a switch [1]. Connect a voltmeter across the length of wire being tested [1]. Use a metre rule and a crocodile clip to change the length tested, and record the length [1]. Calculate R = V ÷ I for each length, repeat and take a mean [1]. Use a small current and switch off between readings so the wire stays at constant temperature [1]. [5] Examiner insight: Method questions are easiest to answer in a logical order: circuit, what you change, what you measure, calculation, control. Naming a control variable and a reason is a typical way to earn the last point.

6. (a) R = V ÷ I: 1.44 ÷ 0.60 = 2.4 Ω [1]; 2.88 ÷ 0.60 = 4.8 Ω [1]; 4.32 ÷ 0.60 = 7.2 Ω [1]. (b) Resistance is directly proportional to length [1]. (c) 4 × 2.4 = 9.6 Ω [1]. [5] Examiner insight: “Directly proportional” is more complete than “as length goes up, resistance goes up”, because it describes the pattern in the numbers. Show the division for each length.

7. (a) 2.0 ÷ 0.30 = 6.7 Ω [1]; 8.0 ÷ 0.75 = 10.7 Ω [1]. (b) Resistance increases as the p.d. and current increase [1]. The filament gets hotter [1]. Its ions vibrate more, so electrons collide with them more often [1]. [5] Examiner insight: Explanations earn more when they name the cause (temperature rise) and the effect on charge flow (more collisions). Stating only “it heats up” is incomplete.

8. In the forward direction there is almost no current until a small p.d. is reached [1]. Above that p.d. the current rises steeply [1]. In the reverse direction the current is almost zero [1]. A diode lets current flow in one direction only [1]. [4] Examiner insight: A sketch or description must show both directions. A graph that is symmetrical about the origin is a typical error that loses the reverse-direction point.

9. (a) I = V ÷ R. At 25 °C: 3.0 ÷ 1800 [1] = 0.00167 A = 1.67 mA [1]. At 75 °C: 3.0 ÷ 300 = 0.010 A = 10 mA [1]. (b) Resistance decreases [1]. The ammeter reading increases [1]. [5] Examiner insight: State the direction of the change in both the resistance and the current for part (b). Answering only one of them may leave a mark unclaimed.

10. (a) X is linear [1]. Its current is proportional to the p.d. (it doubles when p.d. doubles), and the ratio V ÷ I is the same each time [1]. (b) R = 6.0 ÷ 0.12 = 50 Ω [1]. (c) At 3.0 V: 3.0 ÷ 0.10 = 30 Ω; at 6.0 V: 6.0 ÷ 0.15 = 40 Ω [2]. (d) Y is non-linear: its resistance increases as p.d. increases, like a filament lamp [1]. [6] Examiner insight: Quote figures from the table to support the conclusion. “It is a straight line” cannot be seen from four readings, whereas the constant ratio V ÷ I is evidence.

11. (a) Connect the thermistor in series with a power supply and ammeter, with a voltmeter across the thermistor [1]. Place the thermistor in a water bath and read the temperature with a thermometer [1]. At each temperature read V and I, calculate R = V ÷ I, and repeat as the temperature changes [1]. (b) 10 °C: R = 4.0 ÷ 0.0020 = 2000 Ω [1]. 70 °C: R = 4.0 ÷ 0.0200 = 200 Ω [1]. (c) Resistance decreases as the temperature increases [1]. (d) The claim is not supported [1]. The p.d. was constant, and the resistance changed because the temperature changed. V = IR still holds at each temperature, and to test linearity you need an I-V graph at constant temperature [1]. [8] Examiner insight: In an evaluation, say what the data can show and what it cannot. Pointing out that temperature, not p.d., was varied is the key idea.

Where marks are usually lost

• Leaving mA or kΩ unconverted before substituting.
• Missing or incorrect units, including A written for resistance.
• Writing the answer with no working when the question says “calculate”.
• Confusing the thermistor (temperature) with the LDR (light).
• Using the gradient of a curve as the resistance.
• Drawing a symmetrical diode graph.
• Not naming the cause when explaining a lamp’s changing resistance: temperature rises, ions vibrate more, more collisions.
• Method answers with no mention of what is kept constant or how readings are repeated.
• Claiming that V = IR does not apply to non-linear components. It applies at every point.

Next steps

• Review the revision notes and the study guide.
• Visit the course hub and the checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

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.