AQA GCSE Combined Science: Trilogy Physics Paper 1 (Higher), 2022: Question 5

14 marks · Standard Demand difficulty · Short Answer

Explain how to correct an LED circuit, calculate potential difference and current in lighting devices, and identify what happens when xenon emits light.

Practise this question

Question

The question page shows Question 05 about lighting in cameras. Figure 9 is a circuit diagram containing a cell battery, switch, resistor, ammeter in series, and an LED with a voltmeter connected in parallel across the LED; the first part states the voltmeter reads 5.0 V and the ammeter reads 0 mA and asks how to change the circuit so the LED emits light. The next part states the LED emits light, the current is 290 mA and the power is 0.98 W, and asks for the potential difference across the LED to 2 significant figures. Figure 10 is a labelled photograph of a traditional camera with an external flash unit, followed by a multiple-choice item asking what happens when a xenon atom emits light, and a final calculation asking for mean current when the flash tube has a mean potential difference of 200 V, flashes for 2.8 × 10^-4 s, and transfers 1.4 J of energy.
Question text

05 The camera in a mobile phone uses an LED to provide light when

taking a photograph.

A student investigated how the potential difference across an LED varies with the

current in it.

Figure 9 shows the circuit used.

Figure 9

05.1 The student closed the switch. The voltmeter gave a reading of 5.0 V

The ammeter gave a reading of 0 mA

The LED did not emit any light.

Explain how the student should have changed the circuit to make the LED emit light.

[2 marks]

05.2 The student changed the circuit so that the LED emitted light.

The current in the circuit was 290 mA

The power of the LED was 0.98 W

Calculate the potential difference across the LED.

Use the Physics Equations Sheet.

Give your answer to 2 significant figures.

[5 marks]

Potential difference (2 significant figures) = V

A traditional camera uses a flash unit to provide light.

Figure 10 shows a flash unit on a traditional camera.

Figure 10

05.3 The flash unit emits light from xenon gas in a fluorescent tube.

What happens when a xenon atom emits light?

[1 mark]

Tick ( ) one box.

Electrons in the atom fall to a lower energy level.

Electrons in the atom move to a higher energy level.

Electrons leave the atom, causing ionisation.

Electrons transfer to the atom from the electrical circuit.23

05.4 When the flash unit is used there is a mean potential difference of 200 V across the

fluorescent tube.

The flash of light lasts for 2.8 × 10–4 s

1.4 J of energy is transferred.

Calculate the mean current.

Use the Physics Equations Sheet.

[6 marks]

Mean current = A

Mark scheme

Show the mark scheme The mark scheme is presented in tables for parts 05.1 to 05.4 with answer, extra information, mark, and AO/specification reference columns. For 05.1 it awards marks for reversing the LED or battery connections and stating that an LED or diode only allows current in one direction. For 05.2 it shows converting 290 mA to 0.29 A and using P = VI to find V = 0.98/0.29 = 3.4 V to 2 significant figures; for 05.3 the correct response is that electrons in the atom fall to a lower energy level. For 05.4 it shows either using E = QV then Q = 1.4/200 = 0.0070 C and I = Q/t to get 25 A, or using P = E/t then P = VI to reach the same answer; total marks for Question 5 are 14.

Question 5

AO /

Question Answers Extra information Mark

Spec. Ref.

05.1 reverse the connections to the allow reverse the potential 1 AO3

LED / battery difference across the LED / 6.2.1.4

diode

because an LED / diode only allow because an LED / diode 1

allows current through in one has a large resistance in the

reverse direction

direction

AO /

Spec. Ref.

05.2 290 mA = 0.29 A 1 AO2

6.2.4.1

0.98 = V × 0.29 allow a correct substitution of an 1

incorrectly / not converted

or current

0.98

R = 2 allow R = 11.652…

0.29

ignore 0.98 = (0.29)2 × 𝑅𝑅

allow a correct substitution of an

incorrectly / not converted

current

0.98

V = allow 𝑉𝑉 = 11.65 × 0.29 1

0.29

V = 3.379… allow a correct rearrangement 1

using an incorrectly / not

converted current

V = 3.4 (V) allow a correct calculation using 1

an incorrectly / not converted

current

allow a correctly rounded

answer to 2 sig figs consistent

with their calculated value of V

16 using numbers from the

question

AO /

Spec. Ref.

05.3 electrons in the atom fall to a 1 AO1

lower energy level 6.4.1.1

AO /

Spec. Ref.

05.4 1.4 = Q × 200 1 AO2

6.2.4.2

1.4

Q = 1 6.2.1.2

Q = 0.0070 (C) 1

0.0070 = I × 2.8 × 10-4 allow a correct substitution of 1

their calculated value of Q

0.0070

I = -4 allow a correct re-arrangement 1

2.8×10

using their value of Q

I = 25 (A) allow an answer consistent with 1

their value of Q

OR

1.4 = P × 2.8 × 10-4 (1)

1.4

P = -4 (1)

2.8×10

P = 5000 (W) (1)

5000 = 200 × I (1) allow a correct substitution of

their calculated value of P

5000 allow a correct re-arrangement

I = (1)

200 using their value of P

allow an answer consistent with

I = 25 (A) (1)

their value of P

Total Question 5 14

How to answer it

Low Demand

LEDs and Flash Units: Circuits, Energy and Current

What this question tests

Focus: using circuit knowledge, simple equations and particle ideas to explain and calculate what is happening in an LED and a flash tube.

  • Know that an LED is a diode and current only flows in one direction.
  • Use P = IV , E = VQ , and Q = It correctly.
  • Convert units such as mA to A.
  • Explain fluorescence in a xenon tube using electron energy levels.
  • Show calculations clearly, with units and final answers rounded correctly.
Question 05.1

How should the circuit be changed so the LED emits light?

✅ Correct answer

Reverse the connections to the LED or battery.
The LED must be connected the correct way round so it is forward-biased and current can flow.

💡 Key knowledge

  • An LED is a diode.
  • A diode only allows current to flow in one direction.
  • If it is reversed, it has a very large resistance and no current flows.

🧠 Exam technique

  • To get both marks, give the change and the reason.
  • Use the phrase reverse the LED/battery or reverse the potential difference.
  • Link the change to current flow and the LED lighting.

❌ Common errors

  • Just saying “increase the current” is too vague.
  • Writing “make the battery stronger” does not explain the physics.
  • Some students forgot the key idea that a diode only works in one direction.
Mark breakdown: 1 mark for reversing the connections, 1 mark for explaining that an LED/diode only allows current in one direction.
Question 05.2

Calculate the potential difference across the LED

✅ Correct answer

3.4 V (to 2 significant figures)

📐 Calculations: step by step

  1. Use the power equation:
    P = IV
  2. Rearrange to make voltage the subject:
    V = P ÷ I
  3. Convert the current from mA to A:
    290 mA = 0.29 A
  4. Substitute values:
    V = 0.98 ÷ 0.29
  5. Calculate:
    V = 3.379...
  6. Round to 2 significant figures:
    V = 3.4 V

💡 Key knowledge

  • Power, current and voltage are linked by P = IV .
  • Current must be in amps, not milliamps, when using SI units.
  • 2 significant figures means the answer should match the given data.

🧠 Exam technique

  • Show the equation first, then rearrange clearly.
  • Always convert 290 mA to 0.29 A before substituting.
  • Include units in the final answer: V.

❌ Common errors

  • Using 290 instead of 0.29 gives the wrong answer by a factor of 1000.
  • Using the wrong rearrangement, such as V = 0.98 × 0.29 .
  • Forgetting to round to 2 significant figures.
  • Mixing up power and energy.
Mark breakdown: 1 mark for converting 290 mA to 0.29 A, 1 mark for a correct substitution, 1 mark for correct rearrangement, 1 mark for correct calculation, 1 mark for a correctly rounded answer.
Question 05.3

What happens when a xenon atom emits light?

✅ Correct answer

Electrons in the atom fall to a lower energy level.

💡 Key knowledge

  • In a fluorescent tube, electrons in the atom become excited first.
  • When they return to a lower energy level, energy is released as light.
  • This is the key idea behind fluorescence in xenon gas.

🧠 Exam technique

  • Pick the option about falling to a lower energy level.
  • The mark scheme wants the precise physics idea, not a general description of “giving off light”.

❌ Common errors

  • Choosing “move to a higher energy level” is the opposite of emission.
  • “Electrons leave the atom” is ionisation, not light emission.
  • “Electrons transfer from the electrical circuit” is not the explanation here.
Mark breakdown: 1 mark for the correct statement only.
Question 05.4

Calculate the mean current in the flash tube

✅ Correct answer

25 A

📐 Calculations: method 1 using charge

  1. Use the equation:
    E = VQ
  2. Rearrange to find charge:
    Q = E ÷ V
  3. Substitute values:
    Q = 1.4 ÷ 200
  4. Calculate charge transferred:
    Q = 0.0070 C
  5. Use:
    Q = It
  6. Rearrange for current:
    I = Q ÷ t
  7. Substitute values:
    I = 0.0070 ÷ 2.8 × 10⁻⁴
  8. Calculate:
    I = 25 A

📐 Calculations: method 2 using power

  1. First find power using:
    P = E ÷ t
  2. Substitute values:
    P = 1.4 ÷ 2.8 × 10⁻⁴
  3. Calculate:
    P = 5000 W
  4. Use:
    P = IV
  5. Rearrange:
    I = P ÷ V
  6. Substitute values:
    I = 5000 ÷ 200
  7. Calculate:
    I = 25 A

💡 Key knowledge

  • E = VQ links energy, potential difference and charge.
  • Q = It links charge, current and time.
  • P = E ÷ t gives power from energy transferred in time.
  • Use standard units: V, C, A, s, W.

🧠 Exam technique

  • Underline the given numbers and units before starting.
  • Be careful with the time: 2.8 × 10⁻⁴ s is very small.
  • Give the answer with the correct unit: A.
  • Either method is acceptable if every step is shown clearly.

❌ Common errors

  • Not using the time in seconds.
  • Forgetting to divide by the time after finding charge.
  • Using I = V ÷ Q or another incorrect rearrangement.
  • Mixing up energy and power.
Mark breakdown: marks are awarded for using the correct equation, correct substitution, correct rearrangement, and correct final answer. Consistent working using a valid alternative method also scores full marks.

What top-level responses did well

💡 Strong features

  • They used the right physics idea, not just the right number.
  • They converted units correctly, especially mA to A.
  • They showed working in a logical order.
  • They gave answers with correct units and sensible rounding.

🧠 Quick checklist for full marks

  • Read the command word carefully: explain, calculate, or tick.
  • Use one equation at a time.
  • Check whether current must be converted.
  • Look back at the question to see how many significant figures are needed.

Final revision points

❌ Biggest traps to avoid

  • Thinking an LED works both ways round.
  • Using mA in equations without converting to A.
  • Confusing energy, power and charge equations.
  • Forgetting that light is emitted when electrons fall to a lower energy level.

Topics

Physics · P2: Electricity

Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Physics Paper 1 (Higher), 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.