AQA GCSE Physics Physics Paper 1 (Higher), June 2025: Question 3

8 marks · Standard Demand difficulty · Short Answer

Analyze an investigation of the I–V characteristics of an LED, including determining resistance from a graph and explaining reverse bias behavior.

Practise this question

Question

Question 3 displays a circuit diagram in Figure 5 containing a battery, open switch, ammeter, LED with voltmeter in parallel, and a variable resistor in series. Figure 6 shows an I-V graph plotting Current in amps against Potential difference in volts for an LED, showing zero current until approximately 2.0 V, after which it curves upwards to about 0.21 A at 3.4 V. Parts 03.1 to 03.4 ask how to vary potential difference, to identify the equation V = I x R, to determine resistance at 3.0 V using the graph, and to explain why the ammeter reads 0.0 A when connections are reversed.

Mark scheme

Show the mark scheme Mark scheme for Question 3 detailing: 03.1 allows changing resistance of variable resistor or number of cells (1 mark); 03.2 indicates V = I x R (1 mark); 03.3 awards 1 mark for reading I = 0.125 A, 1 mark for substituting into 3.0 = 0.125 x R, 1 mark for rearrangement R = 3.0 / 0.125, and 1 mark for 24 ohms; 03.4 awards 1 mark for large resistance and 1 mark for in the reverse direction.

How to answer it

Investigating the I-V Characteristics of an LED

📋 What This Question Tests

This question focuses on AQA Required Practical 4: Investigating I-V characteristics of circuit components, specifically a Light Emitting Diode (LED).

  • Circuit control: Explaining how to alter potential difference across a component using a variable resistor or cell configuration.
  • Equation recall: Identifying the relationship between potential difference, current, and resistance ( V = I × R ).
  • Graph interpretation & calculations: Reading values from an I-V characteristic curve and carrying out a 4-mark resistance calculation.
  • Diode theory: Explaining why reverse-biased diodes prevent current flow using the concept of resistance.

Question 03.1: Varying Potential Difference

1 Mark • AO1

✅ Acceptable Answers (Any One)

  • Change the resistance of the variable resistor (or "adjust the variable resistor").
  • Change the number of cells in the battery / add another battery.
  • Change the potential difference / voltage of the power pack.
  • Add extra resistors or components to the circuit.

🧠 Exam Technique: Look at the Diagram

Figure 5 clearly shows a variable resistor (a rectangle with a diagonal arrow through it) connected in series with the LED.

The primary job of a variable resistor in this required practical is to smoothly alter total circuit resistance, thereby changing the p.d. shared across the test component.

Mark Scheme Note: 1 mark for identifying any valid method to change circuit potential difference.

Question 03.2: Equation Recall

1 Mark • AO1

✅ Correct Answer

V = I × R

Tick the second box.

💡 Key Knowledge: Ohm's Law

  • V: Potential difference (volts, V)
  • I: Current (amperes / amps, A)
  • R: Resistance (ohms, Ω)

❌ Common Errors

Confusing this equation with the power equation: P = I² × R (the first option in the list). Always check the symbols carefully.

Question 03.3: Resistance Calculation from Graph

4 Marks • AO2

📐 Step-by-Step Calculation

  1. Step 1: Read current from Figure 6 at 3.0 V
    Find 3.0 V on the horizontal x-axis. Move vertically up to meet the plotted line, then move horizontally across to the vertical y-axis.
    The plotted point sits exactly between 0.12 A and 0.13 A .
    Current, I = 0.125 A (allowable range: 0.12 A to 0.13 A) [1 mark]
  2. Step 2: Substitute values into the equation
    V = I × R
    3.0 = 0.125 × R [1 mark]
  3. Step 3: Rearrange to solve for R
    R = 3.0 / 0.125 [1 mark]
  4. Step 4: Calculate the final answer
    R = 24 Ω [1 mark]
    (If you used 0.12 A: R = 25 Ω; if you used 0.13 A: R = 23.1 Ω)

❌ Common Traps & Lost Marks

  • Inverting the formula: Calculating R = I / V ( 0.125 / 3.0 = 0.042 Ω ). If resistance is tiny when voltage is high, recheck your algebra!
  • Misreading the scale: On the y-axis, 10 small squares = 0.05 A, so 1 small square = 0.005 A. Don't rush reading graph gridlines!

🧠 Top-Tier Exam Technique

Always draw straight construction lines with a pencil on the exam graph to demonstrate exactly where you read your values. Even if your final arithmetic has an error, examiners can award method marks!

Question 03.4: Reversing the Power Supply (Reverse Bias)

2 Marks • AO1

✅ Full Mark Model Answer

The LED has a very large / high resistance [1 mark] in the reverse direction [1 mark] .

💡 Diode Characteristics

  • Forward bias: Beyond the threshold voltage (~2.0 V here), resistance drops very low, allowing high current.
  • Reverse bias: When connected backwards, a diode/LED has extremely high resistance, meaning virtually zero current ( 0.0 A ) can flow.

❌ Why Students Lost Marks Here

Many students simply wrote: "An LED only lets current flow in one direction."

While true, this is an incomplete physics explanation. The mark scheme states that this only earns 1 consolation mark. To get both marks, you must explain the mechanism: high resistance in reverse.

🧠 Examiner Tip on Dependent Marks

Mark 2 is dependent on Mark 1: You cannot get the mark for "in reverse direction" unless you explicitly mention the large/high resistance!

Summary of Marks (Part 03.4):
• Mark 1: (The LED / circuit has a) large resistance.
• Mark 2: In the reverse direction (dependent on Mark 1).

Topics

Physics · Required Practicals · P2: Electricity · Required Practicals

Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 1 (Higher), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.