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 question

Question

Exam question with seven parts regarding an electrical circuit with a filament lamp. Figure 2 displays six circuit symbols: filament lamp, ammeter, LDR, voltmeter, cell, and fuse. Question 02.1 asks to draw a circuit diagram to measure current in a filament lamp using three symbols. 02.2 asks to tick the source of energy from ammeter, cell, or filament lamp. 02.3 asks to calculate charge flow given current of 1.5 A and time 30 s using charge flow = current × time. 02.4 asks to calculate resistance given 12 V and 1.5 A using resistance = potential difference / current, and select the correct unit from ohms, degrees Celsius, or watts. Figure 3 shows a non-linear I-V graph of current against potential difference curving downwards. 02.5 asks to describe this relationship. 02.6 asks what happens to resistance as temperature increases. Table 1 lists input power and brightness for a filament lamp (40 W, 800 arbitrary units) and an LED bulb (8 W, 800 arbitrary units), followed by a 4-mark question to explain the advantages of LEDs including a calculation.
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 Mark scheme for Question 2 totaling 16 marks. 02.1 awards 2 marks for symbols of filament lamp, ammeter, and cell, and 1 mark for a simple series circuit. 02.2 awards 1 mark for cell. 02.3 awards 1 mark for 1.5 × 30 and 1 mark for 45 C. 02.4 awards 1 mark for 12 / 1.5, 1 mark for 8, and 1 mark for ohms (Ω). 02.5 awards 1 mark for current increasing as potential difference increases, and 1 mark for non-linear / at a decreasing rate. 02.6 awards 1 mark for resistance increases. 02.7 uses a 2-level scheme (up to 4 marks) based on brightness comparison, power calculation (e.g. 5 times greater / 32 W difference), and consequences like lower running costs, less energy wasted, or higher efficiency.

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

Revision Summary

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.

Question 02.1 • 3 Marks

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.
Mark Breakdown:
• 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.
Question 02.2 • 1 Mark

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.

Question 02.3 • 2 Marks

Calculating Charge Flow

Applying the equation: charge flow = current × time

📐 Step-by-Step Calculation

  1. Identify values: Current I = 1.5 A , Time t = 30 s .
  2. Use formula: Q = I × t
  3. Substitute: Q = 1.5 × 30
  4. 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 1.5 × 30
[1 mark] for final value 45
Question 02.4 • 3 Marks

Calculating Resistance and Identifying Units

Applying: resistance = potential difference / current

📐 Calculation Steps

  1. Values: V = 12 V , I = 1.5 A
  2. Substitute into formula:
    R = 12 / 1.5 [1 mark]
  3. Calculate value:
    R = 8 (or 8.0) [1 mark]
  4. 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 .
Question 02.5 • 2 Marks

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!

Question 02.6 • 1 Mark

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).

Question 02.7 • 4 Marks

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.