AQA GCSE Physics 8463: Electricity – 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 section 4.2 Electricity (4.2.1.1 to 4.2.5.2) of the AQA GCSE Physics (8463) specification, for teaching from September 2016 and exams from 2018 onwards. The topic is assessed on Paper 1, set at Foundation and Higher Tier. No content in section 4.2 is marked (HT only), so every question suits both tiers. Questions 3 and 4 are based on required practicals 4 and 3.

Learn the content first in the Electricity study guide and the revision notes. The course hub is AQA GCSE Physics, and the printable checklist lists every point.

Questions

1.

(a) State what is meant by electric current. [1] (b) A current of 0.15 A flows through a lamp for 4.0 minutes. Calculate the charge that flows. [2]

2. Name the component described in each case.

(a) Its resistance decreases as light intensity increases. [1] (b) Current flows through it in one direction only. [1] (c) Its resistance decreases as temperature increases. [1]

3. A student investigates the I–V characteristic of a filament lamp.

(a) Describe how she should set up the circuit and take readings. [3] (b) Describe the shape of the graph of current against potential difference and explain it. [3]

4. A student measures the resistance of a wire at different lengths. The pd across the wire is kept at 1.50 V.

Length (cm)	20	40	60	80
Current (A)	0.75	0.38	0.25	0.19
Resistance (Ω)	2.0	?	6.0	7.9

(a) Calculate the resistance at 40 cm. Give your answer to 2 significant figures. [2] (b) Describe the relationship between length and resistance, using the data. [2] (c) Explain why the student switches off the circuit between readings. [1]

5. A 30 Ω resistor and a resistor R are in series with a 12 V supply. The current is 0.25 A.

(a) Calculate the total resistance of the circuit. [2] (b) Calculate the resistance of R. [1] (c) Calculate the pd across the 30 Ω resistor. [2]

6. Lamps A and B are connected in parallel to a 6.0 V battery. The current from the battery is 1.2 A and the current through lamp A is 0.50 A.

(a) State the pd across lamp B. [1] (b) Calculate the current through lamp B. [1] (c) A third lamp is added in parallel. Explain why the total resistance of the circuit decreases. [2]

7. A toaster is connected to the mains with three-core cable. Explain the job of each wire, why the live wire can be dangerous even when the toaster’s switch is open, and why a connection between the live wire and earth is dangerous. [6]

8. An electric heater runs on a 230 V supply and draws a current of 5.0 A.

(a) Calculate the power of the heater. [2] (b) Calculate the energy transferred in 20 minutes. [3] (c) Calculate the charge that flows in 20 minutes, and use it to confirm your answer to (b). [2]

9. A power station sends 200 kW through transmission cables with a total resistance of 0.50 Ω.

(a) The cables are at 20 000 V. Calculate the current in the cables. [2] (b) Calculate the power wasted as heat in the cables. [2] (c) Explain why the National Grid uses step-up and step-down transformers. [3]

10. A student rubs an acetate strip with a cloth. The strip becomes positively charged.

(a) Explain, in terms of electrons, how the strip and the cloth become charged. [2] (b) The strip is brought near a second positively charged strip hanging from a thread. Describe what happens and state what type of force acts. [2] (c) Describe the electric field pattern around an isolated positively charged sphere. [2] (d) Explain how a spark can jump from a charged object to a nearby earthed metal pipe. [2]

11. A thermistor and a 4.0 kΩ fixed resistor are in series with a 12 V battery. At 20 °C the thermistor’s resistance is 2.0 kΩ. At 5 °C it is 8.0 kΩ.

(a) Calculate the current at 20 °C. [3] (b) Calculate the pd across the thermistor at 20 °C. [2] (c) Calculate the pd across the thermistor at 5 °C. [3] (d) Explain how this circuit could be used to switch on a heater when the room gets cold. [2] (e) Calculate the power transferred in the fixed resistor at 20 °C. [2]

Answers

1. (a) The rate of flow of electrical charge [1] (b) t = 4.0 × 60 = 240 s; Q = 0.15 × 240 [1]; Q = 36 C [1] Examiner insight: Leaving t as 4.0 gives 0.60 C, which usually scores 0 for the calculation because the time conversion is the method step.

2. (a) LDR (light-dependent resistor) [1] (b) Diode [1] (c) Thermistor [1] Examiner insight: Each mark needs the exact component; “resistor” or “variable resistor” gets no credit.

3. (a) Ammeter in series with the lamp [1]; voltmeter in parallel across the lamp [1]; change the pd with a variable resistor (or variable supply) and record I and V, reversing the connections for negative values [1] (b) A curve through the origin that gets less steep as pd increases, in both directions [1]; as the current increases the filament gets hotter [1]; so its resistance increases [1] Examiner insight: “Resistance increases” alone is one mark; the explanation mark needs the link to filament temperature.

4. (a) R = 1.50 ÷ 0.38 [1] = 3.9 Ω [1] (b) Resistance increases as length increases [1]; doubling the length (20 cm to 40 cm, or 40 cm to 80 cm) roughly doubles the resistance, so R is directly proportional to length [1] (c) So the wire does not heat up, because resistance changes with temperature [1] Examiner insight: “Directly proportional” needs data quoted as evidence; “as one goes up the other goes up” only earns the first mark in (b).

5. (a) R_total = 12 ÷ 0.25 [1] = 48 Ω [1] (b) R = 48 − 30 = 18 Ω [1] (c) V = 0.25 × 30 [1] = 7.5 V [1] Examiner insight: An error in (a) is carried forward into (b), so a wrong total can still earn the mark in (b) if the subtraction is right.

6. (a) 6.0 V [1] (b) 1.2 − 0.50 = 0.70 A [1] (c) The new lamp gives another path for the charge [1]; so the total current increases for the same pd, meaning the total resistance is lower [1] Examiner insight: “More lamps means less resistance” repeats the question; both marks need the idea of an extra path and more current for the same pd.

7. Six points needed:

• Live wire (brown) carries the alternating pd from the supply [1]
• Neutral wire (blue) completes the circuit and is at or close to 0 V [1]
• Earth wire (green and yellow) is a safety wire that stops the casing becoming live; it carries current only if there is a fault [1]
• With the switch open, the live wire is still at about 230 V [1]
• Touching it puts a large pd across your body (at 0 V), so a current flows through you: an electric shock [1]
• A live-to-earth connection is a low-resistance path, so a very large current flows, risking fire or shock [1]

Examiner insight: On a 6-mark explain question, listing colours alone caps the answer low; each mark needs a function or a reason, not just a name.

8. (a) P = 230 × 5.0 [1] = 1150 W [1] (b) t = 20 × 60 = 1200 s [1]; E = 1150 × 1200 [1] = 1 380 000 J (1.38 MJ) [1] (c) Q = 5.0 × 1200 = 6000 C [1]; E = QV = 6000 × 230 = 1 380 000 J, which agrees [1] Examiner insight: Error carried forward applies: a wrong power in (a) used correctly in (b) still earns the method marks.

9. (a) I = P ÷ V = 200 000 ÷ 20 000 [1] = 10 A [1] (b) P = I²R = 10² × 0.50 [1] = 50 W [1] (c) Step-up transformers increase the pd, so the current in the cables is smaller for the same power [1]; a smaller current means less heating of the cables, so less energy is wasted [1]; step-down transformers reduce the pd to a safe value for homes [1] Examiner insight: Writing kW as W (200 instead of 200 000) is the usual slip; a power-of-ten error loses the accuracy mark.

10. (a) Electrons are transferred from the strip to the cloth [1]; the strip loses electrons so it is positive, and the cloth gains an equal negative charge [1] (b) The hanging strip moves away (is repelled) [1]; this is a non-contact force [1] (c) Straight lines radiating out from the surface [1]; arrows pointing away from the sphere [1] (d) Charge builds up, so the pd between the object and the pipe becomes large [1]; electrons jump across the gap through the air, which is the spark [1] Examiner insight: Any answer that says positive charges or protons move loses the mark in (a).

11. (a) R_total = 4000 + 2000 = 6000 Ω [1]; I = 12 ÷ 6000 [1] = 0.0020 A (2.0 mA) [1] (b) V = 0.0020 × 2000 [1] = 4.0 V [1] (c) R_total = 4000 + 8000 = 12 000 Ω [1]; I = 12 ÷ 12 000 = 0.0010 A [1]; V = 0.0010 × 8000 = 8.0 V [1] (d) As the room cools the thermistor’s resistance increases, so it takes a bigger share of the pd [1]; the rising pd across the thermistor can operate a switch that turns on the heater [1] (e) P = I²R = 0.0020² × 4000 [1] = 0.016 W [1] Examiner insight: Using 2.0 kΩ as 2.0 Ω gives currents 1000 times too big; kΩ must be converted before substituting.

Where marks are usually lost

• Not converting minutes to seconds, or kW, kΩ and mA to base units.
• Giving a bare numerical answer with no unit on a calculation.
• Describing I–V graphs without saying whether the line passes through the origin.
• Explaining parallel resistance by repeating the question instead of giving the extra-path idea.
• Saying current is “used up” by the first lamp in a series circuit.
• Writing that the earth wire carries current all the time.
• Saying protons or positive charges move when objects are rubbed.
• In National Grid answers, missing the link between lower current and less heating in the cables.

Next steps

• Recap with the Electricity revision notes
• Relearn weak areas in the Electricity study guide
• Course hub: AQA GCSE Physics
• Printable AQA GCSE Physics checklist
• Try all free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA GCSE Physics (8463) specification, for teaching from September 2016 onwards, for exams in 2018 onwards (Version 1.1, 30 September 2019), published by AQA. These questions cover section 4.2 Electricity.