AQA AS Level Physics Paper 1, June 2018: Question 6

10 marks · Medium difficulty · Short Answer

Determine which two lamps are at their normal brightness using calculations, explain observations when a fifth lamp E is added, and deduce the effect on battery current if lamp B fails.

Practise this question

Question

Three parts of a physics question about circuits containing lamps. Figure 6 shows a 9.0 V battery connected to four lamps A, B, C, and D arranged in a bridge-like configuration. Table 1 lists operating voltages and powers for lamps A and C (6.0 V, 6.0 W) and B and D (3.5 V, 4.1 W). Figure 7 shows a fifth lamp E added between the middle junctions of the circuit. Subsequent sub-questions deal with lamp B failing and require calculations of potential differences, resistances, and current.
Question text

06 A student connects four lamps A, B, C and D in the circuit shown in Figure 6.

The battery has an emf of 9.0 V and negligible internal resistance.

Figure 6

06.1 Table 1 shows the operating conditions for the lamps when they are at normal

brightness.

Table 1

Lamps Operating voltage / V Power / W

A and C 6.0 6.0

B and D 3.5 4.1

The student observes that two of the lamps are at their normal brightness.

Assume that any changes in resistance of the lamps are negligible.

Determine which two lamps are at their normal brightness.

Use calculations to support your answer.

[4 marks]

06.2 The student connects another lamp E in the circuit as shown in Figure 7.

Lamp E is identical to lamps A and C.

Figure 7

Explain what the student would observe regarding the brightness of the lamps.

Refer to potential differences across lamp E in your answer.

[3 marks]

06.3 Lamp B in Figure 7 fails so that it no longer conducts. This change does not affect

the resistance of the other lamps.

*20* Deduce the effect on the current in the battery.

Use calculations to support your answer.

[3 marks]

Mark scheme

Show the mark scheme Mark scheme for three sub-questions. Question 06.1 awards marks for calculating resistances of lamps and potential differences across them to show lamps A and C are at normal brightness. Question 06.2 awards marks for identifying that potential difference across lamp E is 0V so it does not light, leaving other lamps unaffected. Question 06.3 awards marks for calculating initial and final total currents from the battery and showing that the current decreases.

Question Answers Additional Comments/Guidelines Mark

resistance of lamp B and D = 3.52/4.1 = 3.0 (2.98)( )

resistance of lamp A and C = 6.02/6.0 = 6.0 ( )

pd across lamp B and lamp D = 3/9 × 9.0 = 3.0 (V) OR pd across

lamp A and C = 6.0 (V) can justify in terms of current i.e. current needed by

06.1

hence A and C normal brightness A and C is 1 A provided resistance values

calculated

must have some correct working for conclusion

mark

the pd across new lamp = 0 / E does not light 2nd and 3rd marks conditional on 1st mark 1

no current in E 1

06.2

other lamps are not affected 1

because the current in the lamps/pd across lamps does not change (MAX 3)

in first circuit current in battery = 9.0/4.5 = 2.0 A allow ecf from 06.1

in second circuit current in battery = 9.0/7 = 1.2857 A original current = 2A can come from 06.1 and score

hence current in battery decreases here

06.3 1

if say circuit resistance increases so current

decreases and no other marks awarded score 1

mark

How to answer it

Analysis of Complex Lamp Bridge Circuits

📌 What this question tests

This question assesses your ability to apply Kirchhoff's laws, potential divider principles, and power equations (P = V²/R) to complex DC networks. You must combine resistance calculations with circuit topology to determine operating conditions, potential differences, and total source currents.

Question 06.1

Determine which two lamps are at normal brightness with calculations.

📐 Step-by-Step Calculations

  1. Find resistance of lamps B and D:
    Using R = V²/P → 3.5²/4.1 = 2.98 Ω (or ~3.0 Ω)
  2. Find resistance of lamps A and C:
    Using R = V²/P → 6.0²/6.0 = 6.0 Ω
  3. Calculate potential difference (pd) across branch AB:
    Top branch has resistance R_A + R_B = 6.0 + 3.0 = 9.0 Ω .
    pd across lamp B = (3.0 / 9.0) × 9.0 V = 3.0 V .
  4. Conclusion:
    Lamps A and C get 6.0 V across them, matching their operating voltage. Therefore, lamps A and C are at normal brightness.

✅ Correct Answer & Mark Scheme

  • Resistance of B & D = 3.0 Ω (1 mark)
  • Resistance of A & C = 6.0 Ω (1 mark)
  • pd across B & D = 3.0 V OR pd across A & C = 6.0 V (1 mark)
  • Valid conclusion stating A and C are at normal brightness with correct working (1 mark)

❌ Common Errors & Examiner Commentary

  • No working shown: Simply guessing "A and C" scores 0 marks. The conclusion mark is strictly conditional on having correct numerical backing.
  • Current method confusion: Some students tried calculating currents instead of pd sharing; this is fully accepted by the mark scheme provided resistances are correctly calculated first.

Question 06.2

Explain observations when lamp E is added between parallel branches.

✅ Correct Answer & Mark Scheme

  • pd across new lamp E = 0 V / E does not light (1 mark)
  • No current flows through E (1 mark)
  • Other lamps are not affected because current/pd across them does not change (1 mark)

Note: Marks 2 and 3 are strictly conditional on achieving the first mark.

🧠 Exam Technique & Insight

Always check for symmetry in bridge circuits. Lamp E is connected directly between two points that sit at identical electrical potentials. Since there is no potential difference, electrons experience no driving force across E, resulting in zero current and zero light emission.

Question 06.3

Deduce the effect on current in the battery if lamp B fails.

📐 Step-by-Step Calculations

  1. Initial total circuit resistance (Figure 7):
    Branch ACB resistance = 6.0 + 3.0 = 9.0 Ω .
    Branch ADB resistance = 6.0 + 3.0 = 9.0 Ω .
    Total resistance R_total = 9.0 / 2 = 4.5 Ω .
  2. Initial battery current:
    I = V / R_total = 9.0 / 4.5 = 2.0 A (ecf from 06.1).
  3. New resistance with lamp B failed (open circuit):
    Branch ADB becomes 6.0 + ∞ = ∞ (no current in ADB branch).
    Only branch ACB carries current, so new total resistance = 9.0 Ω .
  4. New battery current:
    I_new = 9.0 / 9.0 = 1.29 A (or 1.2857 A ).
  5. Conclusion: Current in the battery decreases.

✅ Correct Answer & Mark Scheme

  • Initial current = 2.0 A (1 mark)
  • New current = 1.29 A (1 mark)
  • Explicit statement that current decreases (1 mark)

❌ Common Errors

A common trap is stating "resistance increases therefore current decreases" without providing numerical proof. The mark scheme explicitly limits answers lacking clear numerical calculations to a maximum of 1 mark.

Topics

Physics · 3.5 Electricity

Question and mark scheme from the AQA AS Level Physics examination, Paper 1, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.