AQA GCSE Combined Science: Trilogy Physics Paper 1 (Foundation), June 2025: Question 2
16 marks · Low Demand difficulty · Short Answer
Construct a circuit to measure current in a filament lamp, perform charge and resistance calculations, interpret an I-V graph, and explain the advantages of LED bulbs over filament lamps using power data.
Practise this questionQuestion
Question text
02 Figure 2 shows circuit symbols for some electrical components.
Figure 2
A student connects a circuit to measure the current in a filament lamp.
The student uses:
• an ammeter
• a cell
• a filament lamp.
02.1 Draw a circuit diagram for a circuit the student could use to measure the current in a
filament lamp.
You should use three of the circuit symbols from Figure 2.
[3 marks]
02.2 Which component is the source of energy for the circuit?
[1 mark]
Tick ( ) one box.
Ammeter
Cell
Filament lamp
02.3 There is a current of 1.5 A in the filament lamp for a time of 30 s.
Calculate the charge flow in the filament lamp.
Use the equation:
*07* charge flow = current × time
[2 marks]
9 Charge flow = C
02.4 The current in the filament lamp is 1.5 A when the potential difference across the
filament lamp is 12 V.
Calculate the resistance of the filament lamp.
Use the equation:
potential difference
resistance =
current
Choose the unit from the box.
[3 marks]
Ω °C W
Resistance = Unit
Figure 3 shows how the current varies with potential difference for the filament lamp.
Figure 3
02.5 Describe the relationship between potential difference and current for the
filament lamp.
[2 marks]
02.6 As the current in the filament lamp increases, the temperature of the filament
lamp increases.
What happens to the resistance of the filament lamp as the temperature increases?
[1 mark]
02.7 In most homes, filament lamps have been replaced with LED bulbs.
Table 1 shows information about a filament lamp and an LED bulb.
Table 1
Input power in watts Brightness in arbitrary units
Filament lamp 40 800
LED bulb 8 800
Explain the advantages of using LED bulbs instead of filament lamps.
You should include a calculation in your answer.
[4 marks]
Extra space
Mark scheme
Show the mark scheme
Question 2
AO /
Question Answers Extra information Mark
Spec. Ref.
02.1 symbols for filament lamp, allow 1 mark for 2 correct circuit 2 AO1
ammeter and cell used symbols 6.2.1.1
a simple complete series circuit MP3 dependent on scoring at 1
least 1 other mark
AO /
Spec. Ref.
02.2 cell 1 AO1
6.2.1.2
AO /
Spec. Ref.
02.3 Q = 1.5 × 30 1 AO2
6.2.1.2
Q = 45 (C) 1
AO /
Spec. Ref.
02.4 12 1 AO2
R =
1.5
R = 8 allow 8.0 1 AO2
Ω allow ohms 1 AO1
6.2.1.3
AO /
Spec. Ref.
02.5 as potential difference 1 AO3
increases, the current increases 6.2.1.4
non-linear relationship– allow at a decreasing rate – 8464/P1/1F –
AO /
Spec. Ref.
02.6 (resistance) increases 1 AO1 9
6.2.1.4
AO /
Question Answers Mark
Spec. Ref.
02.7 Level 2: Relevant points (reasons / causes) are identified, given in 3–4 AO3
detail and logically linked to form a clear account.
Level 1: Points are identified and stated simply, but their relevance 1–2 AO2
is not clear and there is no attempt at logical linking.
6.2.4.2
No relevant content 0
Indicative content
• both types of bulb have equal brightness
• the same number of bulbs are needed
• LED bulbs do not need to be replaced as often
power comparison
• the input power of filament lamp is greater
• the input power of the filament lamp is 32 W greater
• the input power of filament lamp is 5 times greater
• the brightness per watt for the LED bulb is 100 arbitrary units
• the brightness per watt for the filament lamp is 20 arbitrary units
consequence
• LED bulb transfers less energy
• (number of units of) electricity used is lower for LED bulb
• LED bulb wastes less energy
• LED bulb running costs are lower
• the LED bulb is more efficient
• the LED bulb is 5 times more efficient
for Level 2 answers must include a valid calculation
Total Question 2 16
How to answer it
Electricity: Circuits, Resistance & Bulb Efficiency
What this question tests
This 16-mark foundational electricity question assesses circuit symbols and simple series wiring, calculating charge flow ( Q = I × t ) and resistance ( R = V / I ), interpreting non-linear current–potential difference graphs for a filament bulb, explaining temperature-dependent resistance, and evaluating energy efficiency between filament and LED bulbs using quantitative comparisons.
Drawing a Circuit to Measure Current
A simple series circuit with a cell, ammeter, and filament lamp
✅ What your diagram must show
- Cell symbol: Long thin line paired with a shorter thicker line ( + | ' - ).
- Ammeter symbol: Circle with capital letter A inside.
- Filament lamp symbol: Circle with an X inside.
- Complete series loop: Continuous wire connecting all three components with no gaps and no extra branches.
🧠 Diagram Description & Marking
Draw a single closed rectangular loop containing:
- The cell on one side.
- The ammeter in series along the line.
- The lamp in series.
• 2 marks: All 3 correct symbols used (1 mark if 2 are correct).
• 1 mark: Connected in a simple, unbroken series circuit.
❌ Common Errors
- Connecting the ammeter in parallel across the lamp (ammeters must always be in series).
- Leaving wires open or crossing over components without touching terminals.
- Confusing the filament lamp symbol (circle with an X) with the resistor (plain rectangle) or thermistor.
Circuit Energy Source
Identifying which component supplies electrical energy
✅ Correct Answer
Tick: Cell
💡 Key Knowledge
The cell stores chemical energy and converts it to electrical energy, providing the potential difference needed to push charge around the circuit. The ammeter only measures current, and the lamp dissipates energy as light and heat.
Calculating Charge Flow
Applying the equation: charge flow = current × time
📐 Step-by-Step Calculation
- Identify values: Current I = 1.5 A , Time t = 30 s .
- Use formula: Q = I × t
- Substitute: Q = 1.5 × 30
- Final Answer: 45 C
🧠 Exam Technique
- Units are already in standard form (seconds and amperes), so no unit conversions are needed.
- Always write the substitution line 1.5 × 30 to secure the method mark even if you make a calculation slip.
[1 mark] for final value 45
Calculating Resistance and Identifying Units
Applying: resistance = potential difference / current
📐 Calculation Steps
- Values: V = 12 V , I = 1.5 A
- Substitute into formula:
R = 12 / 1.5 [1 mark] - Calculate value:
R = 8 (or 8.0) [1 mark] - Select unit:
Ω (ohms) [1 mark]
❌ Common Traps
- Picking the wrong unit: °C is temperature, W (watts) is power, Ω (ohms) is resistance.
- Inverting the division ( 1.5 / 12 = 0.125 ). Remember: R = V / I .
Interpreting the I–V Graph for a Filament Lamp
Describing the relationship shown in Figure 3
✅ Full-Mark Response
- As the potential difference increases, the current increases. [1 mark]
- The relationship is non-linear (or current increases at a decreasing rate). [1 mark]
🧠 Exam Technique: Two-Part Description
For curved graphs in physics, describing the general trend (both increase) is only worth 1 mark. To get the 2nd mark, you must state how the rate changes:
- Say "non-linear" or "it curves".
- Or say "increases at a decreasing rate" (the curve flattens off).
❌ Major Misconception
Do not say "current is directly proportional to potential difference". A directly proportional relationship produces a straight line through the origin, but this graph curves!
Temperature and Resistance in a Filament
Recall what happens to resistance when temperature rises
✅ Correct Answer
The resistance increases. [1 mark]
💡 Why does this happen?
As the filament gets hotter, the metal lattice ions vibrate faster and with greater amplitude. This causes more frequent collisions with the flowing electrons (charge carriers), increasing the opposition to current flow (resistance).
Evaluating LED Bulbs vs Filament Lamps
Extended response comparing power, brightness, and efficiency
📐 Valid Calculations (Pick at least one!)
- Power difference:
40 W - 8 W = 32 W greater power for filament lamp. - Power ratio / factor:
40 W / 8 W = 5 (filament uses 5× more power; LED uses 1/5 the power). - Brightness per watt:
LED: 800 / 8 = 100 units/W
Filament: 800 / 40 = 20 units/W
✅ Model Answer Structure
1. Brightness comparison: Both bulbs produce the exact same brightness (800 arbitrary units).
2. Calculation & Power: The LED bulb only requires 8 W compared to 40 W for the filament lamp. That means the filament lamp uses 32 W more power (or 5 times more power) for the same light output.
3. Advantages / Consequences: Because the LED bulb has a much lower power input, it transfers less electrical energy per second, wastes less energy as heat, costs less money to run, and is significantly more efficient.
🧠 Level 2 Marking Criteria (3–4 Marks)
- Must provide a valid calculation (e.g. 32 W difference or 5× ratio). Without a calculation, you are capped at Level 1 (max 2 marks)!
- Link the lower power input to practical benefits: lower electricity bills / running costs and greater efficiency.
- Note that LED bulbs also last longer / need replacing less often.
❌ Why Students Lost Marks
- Simply copying numbers from Table 1 without doing an actual mathematical calculation.
- Failing to link "lower power" to "less energy used" or "cheaper to run".
Topics
Physics · P1: Energy · P2: Electricity
Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Physics Paper 1 (Foundation), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.