AQA GCSE Physics Physics Paper 1 (Higher), June 2023: Question 6

16 marks · Standard Demand difficulty · Extended Answer

Describe an experimental method to determine the I-V characteristics of a filament lamp, calculate its resistance from the graph, calculate energy transferred over time, and explain non-linear power behaviour.

Practise this question

Question

Figure 6 displays a current-potential difference graph for a filament lamp. The horizontal axis shows potential difference in volts from -6 to +6 V, and the vertical axis shows current in amps from -0.25 to +0.25 A. The characteristic S-shaped curve passes through the origin (0, 0), curving as potential difference increases in both positive and negative directions. Parts 06.1 to 06.4 follow: 06.1 asks for a method including a circuit diagram to obtain these results (6 marks); 06.2 asks to determine the resistance at +3.0 V (3 marks); 06.3 asks to calculate energy transferred in 30 minutes at 6.0 V and 0.21 A (5 marks); 06.4 asks to explain why power would not quadruple if potential difference doubled (2 marks).

Question text

06 A student investigated how the current in a filament lamp varies with the potential

difference across the filament lamp.

Figure 6 shows the results.

Figure 6

06.1 Describe a method the student could use to obtain these results.

You should include a circuit diagram.

[6 marks]

06.2 Determine the resistance of the filament lamp when the potential difference across it

is +3.0 V.

Use the Physics Equations Sheet.

Use Figure 6 on page 18.

[3 marks]

Resistance = Ω

06.3 The current in the lamp is 0.21 A when the potential difference across the lamp

is 6.0 V.

Calculate the energy transferred by the filament lamp in 30 minutes.

Use the Physics Equations Sheet.

[5 marks]

Energy transferred =21 J

06.4 The power output of the lamp is 1.0 W when the potential difference across the lamp

is 5.0 V.

*20* A student predicts that the power output would be 4.0 W if the potential difference

was doubled.

Explain why the student is not correct.

[2 marks]

Mark scheme

Show the mark scheme

Question 6

AO /

Question Answers Mark

Spec. Ref.

06.1 Level 3: The method would lead to the production of a valid 5–6 AO1

outcome. The key steps are identified and logically sequenced. 4.2.1.4

RPA4

Level 2: The method would not necessarily lead to a valid 3–4

outcome. Most steps are identified, but the method is not fully

logically sequenced.

Level 1: The method would not lead to a valid outcome. Some 1–2

relevant steps are identified, but links are not made clear.

No relevant content 0

Indicative content

• ammeter in series with filament lamp

• current measured with an ammeter

• voltmeter in parallel with filament lamp

• p.d. measured with a voltmeter

• variable resistor (or variable power pack or variable number of

cells) used to vary current in and p.d. across filament lamp

• range of p.d. of 0 to 6 V

• interval of p.d. of 1 V

• reverse connections to power supply to obtain negative values

• take repeat readings and calculate a mean

• discard anomalies

Indicative content may be seen in a circuit diagram.

Level 3 answer: needs to include a circuit which would work (if

included) and a method to obtain negative values.

– HYSICS – –

AO /

Question Answers Extra information Mark

Spec. Ref.

06.2 3.0 = 0.16 × R allow a correct substitution of an 1 AO2

incorrect value of I in the range 4.2.1.3

0.15 (A) to 0.17 (A)

3.0 allow a correct rearrangement of 1

R =

0.16 an incorrect value of I in the

range 0.15 (A) to 0.17 (A)

R = 18.75 (Ω) allow 19 (Ω) 1

allow 18.8

AO /

Spec. Ref.

06.3 t = 1800 (s) 1 AO2

4.2.4.2

Q = 0.21 × 1800 all subsequent marks can score 1 4.1.1.4

if an incorrectly / not converted 4.2.1.2

value of t is used 4.2.4.1

Q = 378 (C) 1

E = 378 × 6.0 1

E = 2268 (J) allow an answer to 2 or 3 s.f. 1

P = 0.21 × 6.0 (1)

P = 1.26 (W) (1)

t = 1800 (s) (1) all subsequent marks can score

if an incorrectly / not converted

value of t is used

E = 1.26 × 1800 (1)

E = 2268 (J) (1) allow an answer to 2 or 3 s.f.–HYSICS – –

AO /

Spec. Ref.

06.4 AO3

(for the power to quadruple) the 1 4.2.1.4

current and the p.d. would both 4.2.4.1

need to double

18 (but the current doesn’t double) 1

because the resistance of the

filament lamp increases

(but the current doesn’t double allow the graph does not show

because the graph shows that) direct proportionality

current is not proportional to p.d.

ignore the graph is not a straight

line

ignore the graph is not linear

Total Question 6 16

How to answer it

Investigating Filament Lamp Characteristics & Electrical Power

What This Question Tests

This question assesses practical mastery and core mathematical skills across Electricity (AQA Specification 4.2):

Required Practical 4: Designing and conducting a valid circuit experiment to plot current-voltage (I-V) characteristics across both positive and negative values.
Graphical Analysis: Accurately reading non-linear graphs with fractional grid scales.
Formula Applications: Calculating resistance ( V = I × R ) and electrical energy transfer ( E = P × t or E = Q × V ).
Scientific Reasoning: Explaining why power does not quadruple when potential difference doubles in a non-ohmic conductor.

Question 06.1: Investigating I-V Characteristics

Method & Circuit Design • 6 Marks (Level of Response)

💡 Key Knowledge: The Circuit

A workable circuit must include:

Filament lamp connected in series with an ammeter.
Voltmeter connected in parallel across the lamp.
A method to vary the p.d.: either a variable resistor or a variable DC power supply.
A switch and power source (battery/cells).

🧠 Top-Level (Level 3) Method

Set up circuit with ammeter in series and voltmeter across the lamp.
Adjust the variable resistor/power source to record potential difference ( V ) and current ( I ) at regular intervals (e.g. 0 to 6 V in steps of 1 V).
Reverse the connections at the battery/power supply terminals to obtain negative p.d. and negative current values (0 to -6 V).
Switch off between readings to prevent the lamp overheating.
Repeat the experiment twice more to calculate mean current values and identify anomalies.

✅ What Examiners Look For

To reach Level 3 (5–6 marks), your answer must:

Include a workable circuit diagram (or clear description).
Crucial step: Explicitly describe how negative values are obtained (reversing the cell/battery connections). Without this, you cannot achieve top marks!

❌ Common Errors

Placing the voltmeter in series or ammeter in parallel.
Forgetting to explain how to obtain negative values on the graph.
Omitting a variable resistor or variable power supply, making it impossible to vary the p.d.

Question 06.2: Calculating Resistance from Graph

Data Interpretation & Ohm's Law • 3 Marks

📐 Step-by-Step Calculation

1 Read the graph at V = +3.0 V:
Find +3.0 V on the horizontal axis. Follow vertically to the curve and read horizontally across to the vertical current axis.
Current, I = 0.16 A (acceptable range: 0.15 A to 0.17 A).

2 State the formula and substitute:
V = I × R → 3.0 = 0.16 × R

3 Rearrange and solve:
R = 3.0 / 0.16 = 18.75 Ω

Mark Breakdown:
• 1 mark: Reading I = 0.16 A and substituting correctly into 3.0 = I × R .
• 1 mark: Correct rearrangement: R = 3.0 / 0.16 .
• 1 mark: Final value: 18.75 Ω (allow 18.8 Ω or 19 Ω ).

❌ Common Pitfalls

Misreading the scale: On the current axis, 10 small squares = 0.05 A, meaning each single square is 0.005 A . At 3.0 V, the line passes 2 small squares above 0.15 A: 0.15 + (2 × 0.005) = 0.16 A .

Question 06.3: Energy Transferred by the Lamp

Multi-Step Energy Calculation • 5 Marks

📐 Method A: Using Power (P = V × I then E = P × t)

1 Convert time into seconds:
t = 30 × 60 = 1800 s

2 Calculate electrical power:
P = V × I = 6.0 × 0.21 = 1.26 W

3 Calculate total energy transferred:
E = P × t = 1.26 × 1800
E = 2268 J (or 2300 J )

📐 Method B: Using Charge (Q = I × t then E = Q × V)

1 Convert time into seconds:
t = 30 × 60 = 1800 s

2 Calculate total charge flown:
Q = I × t = 0.21 × 1800 = 378 C

3 Calculate energy transferred:
E = Q × V = 378 × 6.0
E = 2268 J (or 2300 J )

❌ Major Calculation Trap

Unit Conversion: Time must always be converted from minutes to seconds ( 30 × 60 = 1800 s ). Using 30 s or leaving it as 30 will cost you at least 2 marks!

Question 06.4: Power Doubling Prediction

Resistance & Non-Ohmic Behaviour • 2 Marks

✅ Model Answer

For power to quadruple from 1.0 W to 4.0 W ( P = V × I ), both the potential difference and current would need to double.
However, the current does not double because the resistance of the filament increases as temperature rises (or because current is not directly proportional to potential difference).

🧠 Examiner Insight

Why did the student think power would quadruple? Because mathematically, if resistance stayed constant, P = V² / R means doubling V quadruples P .
The catch: A filament lamp is non-ohmic. Resistance is NOT constant.
Mark scheme warning: Do not just say "the graph is not a straight line" or "it is not linear". You must refer specifically to increasing resistance or lack of direct proportionality.

Topics

Physics · Required Practicals · Required Practicals · P2: Electricity

Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 1 (Higher), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.