AQA GCSE Combined Science: Trilogy Physics Paper 1 (Foundation), 2024: Question 4

10 marks · Standard Demand difficulty · Short Answer

Complete a circuit diagram, calculate mean current, plot a graph, and identify characteristics of a diode.

Practise this question

Question

A series of physics exam questions about a circuit containing a cell, switch, variable resistor, diode, and ammeter. Figure 6 shows an incomplete circuit diagram. Question 4.1 asks to add a voltmeter symbol. Question 4.2 asks which component (variable resistor, ammeter, or cell) adjusts potential difference. Question 4.3 asks to identify 'random', 'systematic', or 'zero' errors for varying measurements. Question 4.4 asks for a mean calculation of three current values: 0.27 A, 0.32 A, and 0.31 A. Figure 7 is a graph of mean current in amps versus potential difference in volts, and Table 2 provides two additional data points (1.25V, 0.60A and 1.50V, 1.00A) to be plotted. Question 4.6 asks if the relationship is 'directly proportional', 'inversely proportional', or 'non-linear'. Question 4.7 asks what happens to the current when diode connections are reversed.
Question text

04 A student investigated how the current in a diode varies with the potential difference

across the diode.

Figure 6 shows an incomplete diagram of the circuit used.

Figure 6

04.1 The student measured the potential difference across the diode.

Complete Figure 6 by adding the symbol for a voltmeter in the correct position.

[2 marks]

04.2 Which component should the student adjust to change the potential difference across

the diode?

[1 mark]

Tick ( ) one box.

The student measured the current three times for each value of potential difference.

04.3 Complete the sentence.

Choose the answer from the box.

[1 mark]

random systematic zero

When the potential difference was 1.50 V the current measurements varied between

0.95 A and 1.08 A. This was caused by errors.

04.4 For one value of potential difference, the measurements of current were:

0.27 A 0.32 A 0.31 A

Calculate the mean current.

[2 marks]

Mean current = A

Figure 7 shows some of the results.

Figure 7

04.5 Table 2 shows the results when the potential difference was greater than 1.00 V.

Table 2

Potential difference in volts Mean current in amps

1.25 0.60

1.50 1.00

Complete Figure 7.

You should:

• plot the results from Table 2

• draw a line of best fit.

[2 marks]

04.6 Complete the sentence.

Choose the answer from the box.

*18* [1 mark]

directly proportional inversely proportional non-linear

Figure 7 shows that the relationship between potential difference and current for the

diode is .

04.7 The student adjusted the circuit so that the current in the diode was 1.00 A.

The student then reversed the connections to the diode.

What happened to the current in the diode when the connections were reversed?

[1 mark]

Tick ( ) one box.

The current decreased to 0.00 A.

The current remained at 1.00 A.

The current increased to 2.00 A.

Mark scheme

Show the mark scheme The mark scheme for Question 4, totaling 10 marks. 4.1: Voltmeter symbol correct and connected across the diode (2 marks). 4.2: Variable resistor symbol ticked (1 mark). 4.3: 'random' error (1 mark). 4.4: Calculation (0.27 + 0.32 + 0.31) / 3 = 0.3 A (2 marks). 4.5: Both points from Table 2 plotted correctly and a line of best fit drawn (2 marks). 4.6: 'non-linear' relationship (1 mark). 4.7: 'the current decreased to 0.00 A' (1 mark).

Question 4

AO /

Question Answers Extra information Mark

Spec. Ref.

04.1 voltmeter symbol correct 1 AO1

6.2.1

voltmeter connected across allow voltmeter connected 1 6.2.1.4

diode across diode and ammeter RPA16

AO /

Spec. Ref.

04.2 1 AO1

6.2.1

6.2.1.4

RPA16

AO /

Spec. Ref.

04.3 random 1 AO3

6.2.1.4

RPA16

AO /

Spec. Ref.

04.4 0.27 + 0.32 + 0.31 1 AO2

3 6.2.1.4

RPA16

0.3 (A) allow 0.30 (A) 1

AO /

Spec. Ref.

04.5 both points plotted correctly allow a tolerance of ± ½ small 1 AO2

square 6.2.1.4

RPA16

line of best fit – ignore line before 0.5 V – 1 –

AO /

Spec. Ref.

04.6 non-linear 1 AO2

6.2.1.4

6.2.1.4

RPA16

AO /

Spec. Ref.

04.7 the current decreased to 0.00 A 1 AO1

6.2.1.4

6.2.1.4

RPA16

Total Question 4 10

How to answer it

Investigating Diode I-V Characteristics

What this question tests

This question focuses on Required Practical 16. You need to know how to set up a circuit to measure current and potential difference, identify circuit symbols, handle experimental errors, calculate means, and interpret non-linear graphs.

Part 04.1 & 04.2

Circuit Setup and Symbols

Correct Answer (04.1)

Draw a circle with a 'V' inside. It must be connected in parallel (across) the diode.

Note: The mark scheme also allows the voltmeter to be connected across both the diode and the ammeter.

Correct Answer (04.2)

Tick the first box: The Variable Resistor symbol (a rectangle with a diagonal arrow through it).

Key Knowledge

  • Voltmeters always go in parallel.
  • Ammeters always go in series.
  • A Variable Resistor is used to change the total resistance of the circuit, which allows you to vary the current and the potential difference across the component you are testing.

Exam Technique

When drawing circuit symbols, ensure the lines of the circuit go right up to the symbol but not through it. A voltmeter "jumps" over the component it is measuring.

Part 04.3 & 04.4

Data Handling and Errors

Types of Error

Random errors cause readings to spread about the true value. They are usually caused by things like human reaction time or slight fluctuations in the environment.

Systematic errors cause readings to differ from the true value by a consistent amount each time (e.g., a zero error on a meter).

Calculation (04.4)

  1. Sum the values: 0.27 + 0.32 + 0.31 = 0.90
  2. Divide by the count: 0.90 / 3 = 0.30
  3. Final Answer: 0.3 A

Common Errors

Students often forget to check if there is an anomaly (outlier) before calculating a mean. In this case, the three values are close together, so all are used. If one value was very different (e.g., 0.85 A), you would exclude it from the mean.

Part 04.5 & 04.6

Graphing and Relationships

Plotting the Graph

  • Point 1: (1.25, 0.60). Go across to 1.25 (halfway between 1.0 and 1.5) and up to 0.60.
  • Point 2: (1.50, 1.00). Go across to 1.50 and up to 1.00.
  • Line of Best Fit: Draw a smooth curve that passes through the origin (0,0) and follows the trend of the points.

Relationship (04.6)

The relationship is non-linear.

Marking tip: Because the line is a curve and not a straight line through the origin, it cannot be "directly proportional".
Part 04.7

The Diode Property

Key Fact: One-Way Street

A diode has a very high resistance in the reverse direction. This means it only allows current to flow in one direction.

Correct Answer

Tick: The current decreased to 0.00 A.

Common Misconception

Some students think reversing the battery just makes the current "negative". While the potential difference becomes negative, the diode's high resistance in reverse bias effectively stops the current entirely in a standard school circuit.

Topics

Physics · P2: Electricity

Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Physics Paper 1 (Foundation), 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.