AQA GCSE Physics Physics Paper 1 (Foundation), June 2023: Question 1
13 marks · Low Demand difficulty · Short Answer
Identify circuit components and errors, determine the effects of resistance changes, calculate current and resistance using given equations, and match components to their I–V characteristic graphs.
Practise this questionQuestion
Question text
01 A student investigated how the current in a filament lamp varies with the potential
difference across the lamp.
Figure 1 shows the circuit used.
Figure 1
01.1 What is component P?
[1 mark]
01.2 Complete the sentences.
Choose answers from the box.
[2 marks]
charge current energy potential difference power
The ammeter in the circuit measures .
The voltmeter in the circuit measures3 .
01.3 How will increasing the resistance of the variable resistor in Figure 1 affect each of
the following quantities?
[3 marks]
Tick ( ) one box in each row.
Stays the
Quantity Decreases Increases
same
Current in the circuit
Potential difference across the lamp
Total resistance of the circuit
01.4 A charge flow of 15 coulombs passed through the filament lamp in a time
of 60 seconds.
Calculate the current in the lamp.
Use the equation:
charge flow
current =
time
[2 marks]
4 Current = A
01.5 When the current in the filament lamp is 0.12 A, the potential difference across the
lamp is 6.0 V.
Calculate the resistance of the filament lamp.
Use the equation:
potential difference
resistance =
current
[2 marks]
*03* 5 Resistance = Ω
01.6 The student repeated the investigation after replacing the lamp with a resistor at
constant temperature and then a diode.
The student plotted a graph for each component.
Draw one line from each component to its graph.
[2 marks]
01.7 Figure 2 shows an ammeter.
The ammeter is not connected to a circuit.
Figure 2
What type of error does the ammeter display?
[1 mark]
Tick ( ) one box.
A positive error
A random error
A zero error
Mark scheme
Show the mark scheme
Question 1
AO /
Question Answers Extra information Mark
Spec. Ref.
01.1 switch 1 AO1
4.2.1.1
RPA4
AO /
Spec. Ref.
01.2 current 1 AO1
4.2.1.4
potential difference allow p.d. 1 RPA4
allow voltage
in this order only
AO /
Spec. Ref.
01.3 AO1
4.2.2
Stay the
Quantity Decrease Increase 4.2.1.3
same
RPA4
Current in the circuit ✓ 1
Potential difference across
✓ 1
the lamp
Total resistance of the
✓ 1
circuit
any extra tick in a row negates the mark for that row
AO /
Spec. Ref.
01.4 15 1 AO2
current = 4.2.1.2
current = 0.25 (A) – HYSICS – 1 –
AO /
Question Answers Extra information Mark 7
Spec. Ref.
01.5 6.0 1 AO2
R = 4.2.1.3
0.12
R = 50 (Ω) 1
AO /
Question Answers Mark
Spec. Ref.
01.6 AO1
2 4.2.1.4
RPA4
2 marks for all 3 correct
1 mark for 1 or 2 correct
additional line from a box on the left negates the mark for that box
AO /
Spec. Ref.
01.7 a zero error 1 AO3
4.2.1.4
RPA4
Total Question 1 13
How to answer it
AQA GCSE Physics: Circuit Basics, Resistance & I–V Graphs
What This Question Tests
This question covers foundational electrical physics from Required Practical 4 (Investigating I–V Characteristics) and Topic 4.2 (Electricity):
- Identifying circuit symbols (open switch).
- Knowing the functions of ammeters and voltmeters.
- Understanding the effect of changing resistance in a series circuit.
- Substituting values into charge ( I = Q / t ) and Ohm's law ( R = V / I ) equations.
- Recognising characteristic current–potential difference graphs for a resistor, lamp, and diode.
- Identifying experimental measurement errors (zero error).
Component Identification
Identifying component P from the circuit diagram
✅ Correct Answer
switch (or open switch)
🧠 Exam Technique
Look closely at how component P is drawn. Two open circles with a tilted line between them indicates a switch. Don't overcomplicate: standard symbol recall only requires the simple name.
Meters and Measured Quantities
Completing sentences using the provided word box
✅ Correct Answer
- The ammeter in the circuit measures current.
- The voltmeter in the circuit measures potential difference (allow: p.d. or voltage).
💡 Key Knowledge
- Ammeter: Connected in series to measure the rate of flow of charge (current in Amperes, A).
- Voltmeter: Connected in parallel across a component to measure energy transferred per unit charge (potential difference in Volts, V).
Increasing Resistance in a Series Circuit
Effect of increasing variable resistor resistance on circuit quantities
✅ Correct Table Responses
| Quantity | Decreases | Stays the same | Increases |
|---|---|---|---|
| Current in the circuit | ✔ | ||
| Potential difference across the lamp | ✔ | ||
| Total resistance of the circuit | ✔ |
💡 Step-by-Step Physics Reasoning
- Total resistance: In a series circuit, resistances add up ( R_total = R_lamp + R_variable ). Increasing the variable resistor directly increases total resistance.
- Circuit current: Since the battery voltage is fixed, a higher total resistance causes circuit current to decrease ( I = V / R_total ).
- P.d. across lamp: With lower current flowing through the lamp, the potential difference across it must decrease ( V = I × R ).
❌ Common Errors
- Assuming potential difference across the lamp stays the same because the battery voltage hasn't changed. (The variable resistor takes a larger share of the total voltage).
- Placing more than one tick per row: if you tick two boxes in a single row, you lose the mark automatically.
Calculation: Current from Charge and Time
Using: current = charge flow / time
📐 Step-by-Step Calculation
Charge flow ( Q ) = 15 coulombs (C)
Time ( t ) = 60 seconds (s)
current = 15 / 60 [1 mark]
current = 0.25 A [1 mark]
🧠 Examiner Tip
Always show your substitution step! Even if you make a silly arithmetic mistake on your calculator, writing 15 / 60 guarantees 1 method mark.
Check units: time was already given in seconds. If it were given in minutes, you would first have had to multiply by 60.
Calculation: Resistance of the Lamp
Using: resistance = potential difference / current
📐 Step-by-Step Calculation
Potential difference ( V ) = 6.0 V
Current ( I ) = 0.12 A
R = 6.0 / 0.12 [1 mark]
R = 50 Ω [1 mark]
❌ Common Pitfall
Dividing upside-down: Students frequently divide current by voltage ( 0.12 / 6.0 = 0.02 ). Always double-check the formula given right in the question prompt: potential difference / current .
Matching I–V Characteristic Graphs
Linking circuit components to their Current–Voltage graphs
✅ Correct Matches
- Diode: Connects to Graph 4 (bottom graph — flat horizontal zero line on left/negative axis, sharp upward bend in positive quadrant).
- Filament lamp: Connects to Graph 1 (top graph — smooth S-shaped curve flattening as potential difference increases).
- Resistor (at constant temperature): Connects to Graph 2 (second graph — straight diagonal line passing directly through origin).
💡 How to Recognise Each Graph
- Resistor (Ohmic Conductor): Resistance is constant, so current is directly proportional to p.d. → Straight line through (0,0).
- Filament Lamp: As current increases, temperature increases → resistance increases → Curve bends/flattens towards voltage axis.
- Diode: Extremely high resistance in reverse direction, only lets current flow one way → Flat at zero on negative side, rises rapidly past threshold voltage (~0.6V).
Experimental Errors
Ammeter reading 0.02 A when NOT connected to a circuit
✅ Correct Answer
Tick: A zero error
💡 Key Science Skill: Types of Error
- Zero error: When an instrument gives a reading when the true value being measured is zero. It is a specific type of systematic error.
- How to fix it: Either reset/tare the meter to zero, or subtract the false reading (0.02 A) from every measurement taken.
❌ Distractor Warning
"A positive error" sounds plausible because +0.02 is a positive number, but it is not a standard scientific classification of error. Stick to syllabus terminology: zero error.
Topics
Physics · Required Practicals · P2: Electricity · Required Practicals
Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 1 (Foundation), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.