OCR A Level Physics A: Resistance, resistivity and electrical power (H556) – 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 cover sections 4.2.3 (Resistance), 4.2.4 (Resistivity) and 4.2.5 (Power) of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. Where a resistivity is needed it is given in the question. Revise first with the study guide or the revision notes. Course pages: OCR A Level Physics hub and checklist.

Questions

1. (a) Define the resistance of a component. [1] (b) A p.d. of 9.0 V produces a current of 15 mA in a resistor. Calculate its resistance. [2]

2. (a) State Ohm’s law. [2] (b) A filament lamp passes 0.20 A when the p.d. is 2.0 V and 0.50 A when the p.d. is 10.0 V. Calculate the resistance at each p.d. and state whether the lamp is ohmic. [3]

3. Describe the I-V graph of a filament lamp for p.d. values of both signs, and explain its shape. [4]

4. The table gives data for an LED.

p.d. / V	0	1.0	1.6	1.8	2.0
current / mA	0	0	0.1	5	20

(a) Estimate the p.d. at which the LED starts to conduct appreciably. [1] (b) Calculate the resistance of the LED at 2.0 V. [2] (c) The LED is to run at 20 mA from a 9.0 V supply with a series resistor. Calculate the resistor’s value. [2]

5. Describe how you would obtain the I-V characteristic of a semiconductor diode in the laboratory. Include the circuit and any precautions. [5]

6. (a) State and explain how the resistance of a light-dependent resistor changes as the light intensity increases. [2] (b) An NTC thermistor has resistance 4.7 kΩ at 20 °C and 1.2 kΩ at 50 °C. A 6.0 V cell pack is placed across it at each temperature. Calculate the current each time. [2]

7. A constantan wire has length 2.40 m and diameter 0.20 mm. Its resistance is 37.4 Ω. (a) Calculate the resistivity of constantan. [3] (b) A second wire of the same material and length has diameter 0.40 mm. Calculate its resistance. [2]

8. A student determines the resistivity of a metal wire. She measures the diameter with a micrometer and the resistance of different lengths of wire. The graph of resistance against length is a straight line with gradient 10.2 Ω m⁻¹. The mean diameter is 0.25 mm. (a) Explain why she measures the diameter at several places and in perpendicular directions. [2] (b) Calculate the resistivity. [3] (c) The diameter has an uncertainty of ±0.01 mm. Calculate the percentage uncertainty in the area. [1] (d) Explain why she uses a small current. [1]

9. Compare and explain how the resistivity of a metal and of a semiconductor change when the temperature rises. [4]

10. A heated rear window is rated 90 W at 12 V. (a) Calculate the current. [2] (b) Calculate its resistance. [1] (c) Calculate the energy transferred in 10 minutes. [2]

11. An electric shower is rated 7.5 kW. It runs for 6.0 minutes. (a) Calculate the energy transferred in kW h. [2] (b) Electricity costs 28 p per kW h. Calculate the cost. [1] (c) Convert the energy to joules. [1] (d) The water gains 2.43 MJ of thermal energy. Calculate the efficiency. [2]

12. A heating element is made from nichrome wire of diameter 0.40 mm. It is to dissipate 1.2 kW from a 230 V supply. Take the resistivity of nichrome as 1.1 × 10⁻⁶ Ω m. (a) Calculate the current. [1] (b) Calculate the resistance needed. [2] (c) Calculate the cross-sectional area of the wire. [1] (d) Calculate the length of wire needed. [2] (e) The element runs for 40 minutes at 30 p per kW h. Calculate the cost. [2]

Answers

1. (a) Resistance is the p.d. across a component divided by the current through it (R = V/I). [1] (b) R = V/I = 9.0 / 0.015 [1] = 600 Ω [1] Examiner insight: A definition scores best when it says “p.d. across” and “current through” the same component. Convert milliamperes to amperes before dividing.

2. (a) The current through a conductor is directly proportional to the p.d. across it [1], provided the temperature (and other physical conditions) stay constant. [1] (b) R = 2.0 / 0.20 = 10 Ω [1]; R = 10.0 / 0.50 = 20 Ω [1]. The resistance changes, so I is not proportional to V and the lamp is non-ohmic. [1] Examiner insight: Leaving out the constant-temperature condition loses the second point. A conclusion needs to refer to the calculated values.

3. The graph passes through the origin and is symmetrical, so the curve for negative p.d. is a rotation of the positive one. [1] The gradient decreases (the line flattens) as the magnitude of p.d. increases. [1] The current heats the filament, so the metal ions vibrate with larger amplitude. [1] There are more collisions with conduction electrons, so resistance (V/I) rises. [1] Examiner insight: A good answer links the shape to temperature and collisions. “The resistance increases” with no cause is only a restatement of the graph.

4. (a) About 1.6 V to 1.8 V (current becomes appreciable there). [1] (b) R = V/I = 2.0 / 0.020 [1] = 100 Ω [1] (c) p.d. across the resistor = 9.0 - 2.0 = 7.0 V [1]; R = 7.0 / 0.020 = 350 Ω [1] Examiner insight: In a series-resistor question, subtract the LED’s p.d. from the supply before using V/I. Using the full supply p.d. gives 450 Ω.

5. Circuit: the supply connected across the whole track of a variable resistor so the p.d. can start from zero (potential divider). [1] The diode in series with an ammeter, the voltmeter connected across the diode only. [1] A protective resistor in series so the forward current stays below the rated maximum. [1] Increase the p.d. in equal steps, recording V and I at each, and take extra readings near the threshold. [1] Reverse the diode (or the supply) and repeat to get the reverse current; plot I against V. [1] Examiner insight: A reason for each feature is rewarded more than a list of apparatus. Mention that the diode is damaged by a large forward current if you name the protective resistor.

6. (a) The resistance decreases. [1] More light releases more charge carriers in the semiconductor. [1] (b) At 20 °C: I = 6.0 / 4700 = 1.3 mA. [1] At 50 °C: I = 6.0 / 1200 = 5.0 mA. [1] Examiner insight: Convert kΩ to Ω before dividing. Give the explanation as more charge carriers, not faster electrons.

7. (a) A = π(0.10 × 10⁻³)² = 3.14 × 10⁻⁸ m² [1]; ρ = RA/L = 37.4 × 3.14 × 10⁻⁸ / 2.40 [1] = 4.9 × 10⁻⁷ Ω m [1] (b) Doubling the diameter makes A four times larger, so R is a quarter: [1] 37.4 / 4 = 9.35 Ω. [1] Examiner insight: Diameter must be halved to a radius, and mm to m, before squaring. Scaling arguments like part (b) are quicker and show understanding.

8. (a) The wire may not have a uniform diameter or a circular cross-section. [1] Repeating at several places and orientations and taking a mean reduces the effect of this. [1] (b) A = π(0.125 × 10⁻³)² = 4.91 × 10⁻⁸ m² [1]; ρ = gradient × A = 10.2 × 4.91 × 10⁻⁸ [1] = 5.0 × 10⁻⁷ Ω m [1] (c) % uncertainty in d = 0.01 / 0.25 × 100 = 4%, so in A it is 8%. [1] (d) A large current heats the wire, which changes its resistivity. [1] Examiner insight: The gradient of an R-L graph is ρ/A, so multiply by A and not by L. The doubling for a squared quantity is a common check-point.

9. Metal: resistivity increases. [1] The ions vibrate more, so conduction electrons collide with them more often (the number density of electrons hardly changes). [1] Semiconductor: resistivity decreases. [1] More charge carriers are released as thermal energy rises, which outweighs the extra collisions. [1] Examiner insight: Each material needs both a direction and a reason. Writing “electrons move faster” for the semiconductor does not explain the fall in resistivity.

10. (a) I = P/V = 90 / 12 [1] = 7.5 A [1] (b) R = V/I = 12 / 7.5 = 1.6 Ω (or V²/P). [1] (c) t = 600 s; W = Pt = 90 × 600 [1] = 5.4 × 10⁴ J (54 kJ) [1] Examiner insight: Convert minutes to seconds when the answer is in joules. Check part (c) with W = VIt = 12 × 7.5 × 600.

11. (a) t = 0.10 h; E = 7.5 × 0.10 [1] = 0.75 kW h [1] (b) Cost = 0.75 × 28 = 21 p [1] (c) 0.75 × 3.6 × 10⁶ = 2.7 × 10⁶ J [1] (d) efficiency = 2.43 / 2.70 × 100% [1] = 90% [1] Examiner insight: Use input energy in the denominator. Work in kW and hours throughout part (a) and only convert to joules at the end.

12. (a) I = P/V = 1200 / 230 = 5.2 A [1] (b) R = V²/P = 230² / 1200 [1] = 44 Ω [1] (c) A = π(0.20 × 10⁻³)² = 1.26 × 10⁻⁷ m² [1] (d) L = RA/ρ = 44.1 × 1.26 × 10⁻⁷ / 1.1 × 10⁻⁶ [1] = 5.0 m [1] (e) Energy = 1.2 × (40/60) = 0.80 kW h [1]; cost = 0.80 × 30 = 24 p [1] Examiner insight: Keep unrounded values until the end: using 44 Ω rather than 44.08 Ω shifts the length slightly. Rearrange R = ρL/A before substituting.

Where marks are usually lost

• Stating Ohm’s law without the constant-temperature condition.
• Calling a lamp “ohmic” or “not ohmic” without calculated values of V/I.
• Taking the slope of a curved graph as resistance.
• Using diameter as radius, or not converting mm and mm² to metres.
• Subtracting nothing for the LED’s p.d. in a series-resistor question.
• Explaining thermistors and LDRs with “electrons move faster” instead of “more charge carriers”.
• Mixing kW with seconds or W with hours, so kW h and joules get confused.
• Using output energy rather than input energy when finding the cost or efficiency.
• Omitting units or giving too many significant figures.

Next steps

• Re-read the revision notes and the study guide.
• Go to the OCR A Level Physics hub and the printable checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

This page is aligned to the OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), first assessment 2017, published by OCR, sections 4.2.3 to 4.2.5.