AQA GCSE Physics Physics Paper 1 (Foundation), June 2025: Question 10

8 marks · Standard Demand difficulty · Short Answer

Investigate the current-potential difference characteristics of an LED, identify the equation linking potential difference, current, and resistance, calculate the LED's resistance from a graph, and explain the effect of reversing the power supply connections.

Practise this question

Question

Question 10 shows Figure 18, a circuit diagram containing a battery, open switch, ammeter, variable resistor, and a light-emitting diode (LED) connected in parallel with a voltmeter. Figure 19 shows a graph of current in amps on the y-axis (from 0.00 to 0.25 A) against potential difference in volts on the x-axis (from 0.0 to 4.0 V). The curve remains at 0.00 A until approximately 2.0 V, after which it rises steeply to about 0.21 A at 3.4 V. Parts 10.1 to 10.4 ask how to vary the potential difference, to select the correct equation linking V, I, and R, to calculate resistance at 3.0 V from the graph, and to explain why current is 0.0 A when connections are reversed.

Mark scheme

Show the mark scheme Mark scheme table for Question 10. 10.1 gives 1 mark for changing variable resistor resistance or cell number. 10.2 gives 1 mark for V = I × R. 10.3 awards 4 marks: 1 for current I = 0.125 A (range 0.12 to 0.13 A), 1 for substitution into formula, 1 for rearrangement R = 3.0 / 0.125, and 1 for 24 ohms. 10.4 awards 2 marks for stating the LED has a large/high resistance in the reverse direction. Total: 8 marks.

How to answer it

I–V Characteristics & Resistance of an LED

SPECIFICATION: AQA 4.2.1.3 & 4.2.1.4 (RPA 4)

What this question tests

This question assesses practical circuit skills and theoretical understanding of diode/LED behaviour:

  • Methods for varying potential difference in a test circuit.
  • Recalling and applying Ohm's law equation: V = I × R .
  • Reading values accurately from an I–V graph and calculating electrical resistance.
  • Explaining diode behaviour under reverse bias in terms of high resistance.
PART 10.1 • 1 MARK

Controlling Potential Difference in a Circuit

Give one way the student could have varied the potential difference across the LED.

✅ Acceptable Answers (Any One)

  • Adjust / change the resistance of the variable resistor (rheostat).
  • Change the number of cells in the battery (or add another cell/battery).
  • Change the potential difference setting on a variable power supply.
  • Add extra resistors into the circuit.

🧠 Exam Technique

Look at Figure 18 carefully before answering. A variable resistor is already drawn in series with the LED. Mentioning that component directly is the fastest, safest route to the mark!

Mark allocation: [1 mark] for any clear method of adjusting circuit pd or total resistance.
PART 10.2 • 1 MARK

Selecting the Correct Equation

Which equation links current (I), potential difference (V), and resistance (R)?

✅ Correct Answer

V = I × R

Tick the second box from the top.

❌ Common Errors

  • Confusing Ohm's law with electrical power: P = I² × R .
  • Incorrect rearrangements such as V = I / R or V = R / I . Always check against the standard formula sheet provided in the exam.
PART 10.3 • 4 MARKS

Calculating LED Resistance from the Graph

Determine the resistance of the LED when the potential difference across the LED was 3.0 V. Use Figure 19.

📐 Step-by-Step Calculation

Step 1: Read current from graph at V = 3.0 V
Locate 3.0 V on the horizontal axis. Follow up to the plotted curve, then read across to the vertical axis.
Current, I = 0.125 A (Acceptable range: 0.12 A to 0.13 A)
Award 1 mark [AO2]
Step 2: Substitute into formula
3.0 = 0.125 × R
Award 1 mark [AO2]
Step 3: Rearrange for Resistance (R)
R = 3.0 / 0.125
Award 1 mark [AO2]
Step 4: Calculate final answer with units
R = 24 Ω
Award 1 mark [AO2]

❌ Common Traps on this Graph

  • Scale misreading: On the y-axis, 10 small squares = 0.05 A. This means each small square equals 0.005 A . At 3.0 V, the point is exactly 5 small squares above 0.10 A:
    0.10 + (5 × 0.005) = 0.125 A .
  • Gradient trap: Do NOT try to take the gradient of the curve to find resistance directly. Resistance is calculated using point values ( R = V / I ), not ΔV / ΔI .
  • Inverting the division: Dividing current by voltage ( 0.125 / 3.0 = 0.042 ) gives conductance, not resistance!
PART 10.4 • 2 MARKS

Explaining LED Behaviour in Reverse Bias

Explain why the ammeter displayed a value of 0.0 A for all values of potential difference when connections were reversed.

✅ Model Answer (Full 2 Marks)

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

Marking note: The second mark is strictly dependent on mentioning high resistance first.

💡 Key Knowledge: How Diodes & LEDs Work

  • Forward bias: Low resistance once the threshold voltage (around 2.0 V in this graph) is exceeded; current flows freely and light is emitted.
  • Reverse bias: Extremely high resistance prevents current from flowing through the circuit, resulting in an ammeter reading of 0.0 A .

❌ Why Students Lose Marks Here

Many students write: "Because LEDs only let current flow one way."

While this statement is true, the question asks you to explain why. Simply stating it only flows one way scores only 1 consolation mark. To get both marks, you must explain that it blocks current by having a very high resistance in reverse.

Topics

Physics · Required Practicals · P2: Electricity · Required Practicals

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