AQA GCSE Physics Physics Paper 1 (Foundation), June 2024: Question 8
7 marks · Standard Demand difficulty · Short Answer
Calculate the mean current, calculate the power of the component at 3.0 V using an I-V graph, describe the gradient trend, and identify the electrical component.
Practise this questionQuestion
Question text
08 A student had an unknown electrical component inside a sealed box.
Figure 14 shows the circuit the student used to identify the component.
Figure 14
The student varied the potential difference across the component and measured the
current in the component.
Table 3 shows the results when the potential difference across the component
was 6.0 V.
Table 3
Potential Current in amps
difference
in volts 1st reading 2nd reading 3rd reading Mean
6.0 0.26 0.21 0.25 X
08.1 Calculate value X in Table 3.
[2 marks]
X = A
Figure 15 shows the results.
Figure 15
08.2 Calculate the power of the component when the potential difference across the
component is 3.0 V.
Use Figure 15 and the equation:
power = potential difference × current
[3 marks]
Power = W
08.3 Complete the sentence.
Choose the answer from the box.
[1 mark]
decreases stays the same increases
As the potential difference across the component increases, the gradient of
the graph .
08.4 What is the component in the sealed box?
[1 mark]
Tick ( ) one box.
Diode
Filament lamp
Resistor at constant temperature
Mark scheme
Show the mark scheme
Question 8
AO /
Question Answers Extra information Mark
Spec. Ref.
08.1 0.26 + 0.21 + 0.25 1 AO2
X =
3 4.2.1.4
X = 0.24 (A) 1
0.26 + 0.25
allow X = = 0.255
for 2 marks
AO /
Spec. Ref.
08.2 current = 0.17 (A) 1 AO2
4.2.4.1
power = 3.0 × 0.17 allow a correct substitution using 1
a value of I in the range 0.16 to
0.18 A
power = 0.51 (W) allow an answer consistent 1
using a value of I in the range
0.16 to 0.18 A
answers of 0.456, 5.1 or 51
score 2 marks
AO /
Spec. Ref.
08.3 decreases 1 AO2
4.2.1.4
AO /
Spec. Ref.
08.4 filament lamp 1 AO3
4.2.1.4
Total Question 8 7
How to answer it
Electrical Circuits and Component Characteristics
What this question tests
This question assesses your ability to calculate means from experimental data, read values accurately from current-potential difference graphs, apply electrical power equations, interpret graph gradients, and recognize the characteristic I-V curve of a standard circuit component (a filament lamp).
Calculate value X in Table 3
✅ Correct Answer
Mean current X = 0.24 A
📐 Step-by-Step Calculation
- Identify the relevant data values in Table 3: 0.26 , 0.21 , and 0.25 .
- Add them together: 0.26 + 0.21 + 0.25 = 0.72
- Divide by the number of valid readings ( 3 ): 0.72 / 3 = 0.24
❌ Common Errors
Some students mistakenly include anomalous data or average only two values. Note that the mark scheme also accepts 0.255 if a student legitimately chose to average just the first and third readings (ignoring the 0.21 anomaly), but standard practice is to use all three unless instructed otherwise.
🧠 Exam Technique
Always show your working out! Even if you press buttons incorrectly on your calculator, writing down the addition step can secure you a method mark.
Calculate power when potential difference is 3.0 V
✅ Correct Answer
Power = 0.51 W
📐 Step-by-Step Calculation
- Find 3.0 V on the x-axis of Figure 15.
- Read across to the curve and down to find the current: I = 0.17 A .
- Recall the equation: power = potential difference × current
- Substitute values: power = 3.0 × 0.17 = 0.51 W
💡 Key Knowledge
Examiners allow a tolerance window for reading graphs. Any current value between 0.16 A and 0.18 A was accepted, leading to corresponding power marks (e.g., 0.456 W or 0.54 W ).
❌ Common Errors
Students often misread the fine grid lines on the y-axis. Each small square on the vertical current axis represents 0.005 A . Double-check your scale reading before calculating!
Complete the sentence about the gradient
✅ Correct Answer
As potential difference increases, the gradient of the graph decreases.
💡 Key Knowledge
The gradient of an I-V graph represents current divided by voltage ( I / V ), which is equal to 1 / resistance (or conductance). As a filament lamp gets hotter with increasing voltage, its resistance increases, meaning its conductance and the graph's gradient decrease (the curve gets shallower).
Identify the component in the sealed box
✅ Correct Answer
Tick: Filament lamp
💡 Key Knowledge
You must commit standard I-V (Current-Voltage) graphs to memory:
- Resistor at constant temperature: Straight line through the origin (Ohm's Law obeyed).
- Filament lamp: S-shaped curve starting at the origin that bends outwards/flattens out due to heating.
- Diode: Current only flows in one direction after a threshold voltage is reached.
Topics
Physics · Required Practicals · P2: Electricity · Required Practicals
Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 1 (Foundation), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.