AQA A-Level Physics Paper 1, June 2022: Question 4
11 marks · Medium difficulty · Short Answer
Calculate the resistance, emf, wire resistivity, and explain circuit changes involving a lamp, parallel resistors, and internal resistance.
Practise this questionQuestion
Question text
04 A student assembles the circuit in Figure 6.
Figure 6
The battery has an internal resistance of 2.5 Ω.
04.1 Show that the resistance of the 6.2 V, 4.5 W lamp at its working potential difference
(pd) is about 9 Ω.
[1 mark]
04.2 The terminal pd across the battery is 6.2 V.
Calculate the emf of the battery.
[3 marks]
emf = V
The student makes a variable resistor to control the brightness of the lamp.
Figure 7 shows her circuit.
*14* Figure 7
04.3 She uses a resistance wire with a diameter of 0.19 mm to make the variable resistor.
A 5.0 m length of this wire has a resistance of 9.0 Ω.
Calculate the resistivity of the wire.
[3 marks]
16 resistivity = Ω m
04.4 Figure 8 shows the 5.0 m length of wire wrapped around a tube to make the variable
resistor.
Figure 8
Two plugs connect the variable resistor into the circuit. A moveable copper contact is
used to vary the length of wire in series with the lamp.
When the contact is placed on the tube at one particular position, the lamp is dim.
The contact is then moved slowly to the right as shown in Figure 8.
Explain, without calculation, what happens to the brightness of the lamp as the
contact is moved.
[2 marks]
04.5 The student now makes a different circuit by connecting the variable resistor
in parallel with the lamp.
The contact is returned to its original position on the tube as shown in Figure 8 and
the lamp is dim. The contact is again slowly moved to the right.
Explain, without calculation, what happens to the brightness of the lamp as the
contact is moved.
[2 marks]
Mark scheme
Show the mark scheme
Question Answers Additional comments/Guidelines Mark AO
Condone use of W for P.
04.1 1 AO1
𝑉2 6.22
𝑅 = = = 8.5(4) ( )✓
𝑃 4.5
– A-LEVEL PHYSICS – –
04.2 3 3 x AO2
Calculation of current in lamp (0.73 A) Allow ecf from 04.1
OR Allow alternative methods
Calculation of current in 12 resistor (0.52 A)
OR
Calculation of parallel pair resistance (5.0 Ω) ✓
Calculation of total circuit current (1.2(4) A) Give full credit to answers that use 9 :
Expected values for this method are
OR
Lamp current = 0.69 A
Calculation of total circuit resistance (7.5 ) Current in 12 resistor = 0.52 A
OR Parallel pair resistance = 5.1(4)
Total circuit resistance = 7.6(4)
Expression of potential divider arrangement
Total circuit current = 1.2(1) A
𝜀−6.2 6.2
= OR
𝑟 external 𝑅 emf = 9.2(1) V
14 𝜀 6.2
= ✓
total circuit restance external R
(emf = terminal pd + Ir = 6.2 + (1.24 × 2.5))
– A-LEVEL PHYSICS – –
9.3(1) V ✓
Question Answers Additional comments/Guidelines Mark AO 15
04.3 Evidence of calculation of A ( = π (d / 2)2 = 2.84 × 10−8 ) ✓ Allow POT errors in MP1 and MP2 3 3 x AO2
Use of their A in the resistivity equation = RA/l ✓
−8 Allow answers that round to 5.10× 10−8 ( m)
To give 5.1× 10 ( m)✓
04.4 Resistance increases ✓ 2 2 x AO3
Reduces current through lamp Do not condone explanations that confuse
current and potential difference.
and lamp dimmer
Do not condone “current across” or “pd
OR through”.
Greater pd across plugs as potential divider
and lamp dimmer✓
– A-LEVEL PHYSICS – –
04.5 (Resistance increases) 2 2 x AO3
Reduces current in circuit / battery Award MAX 1 for arguments dealing with
16 initial dimming of bulb when wire attached.
OR
Increases (external) circuit resistance ✓
Condone “pd across lamp and resistor /
Reduces pd dropped across internal resistance of cell / parallel section” for “terminal pd”.
increases terminal pd so lamp brighter. ✓ Condone “lost volts”.
Total 11
How to answer it
DC Circuits, Internal Resistance & Resistivity
What this question tests
This comprehensive circuit analysis question tests your mastery of electrical power, resistance calculations, internal resistance, terminal potential difference, and emf equations (P = V²/R, ε = V + Ir). It also assesses practical skills including determining material resistivity from dimensions and analyzing qualitative changes in series and parallel circuits when variable resistors are adjusted.
Resistance of the Lamp
✅ Correct Answer
R = 8.5 Q (or 8.54 Q)
💡 Key Knowledge
- Power equations: P = IV , P = V²/R , and P = I²R .
- Rearranging for resistance: R = V²/P .
📐 Calculation Steps
- Identify values: V = 6.2 V, P = 4.5 W.
- Substitute into R = V²/P -> 6.2² / 4.5 .
- Evaluate to give 8.54 Q (condone use of 'W' for power symbols).
❌ Common Errors
Using incorrect power formulas or failing to square the potential difference value.
Calculating the EMF of the Battery
✅ Correct Answer
emf = 9.3 V (Accept 9.2 V to 9.3 V depending on whether unrounded values or the rounded 9 Q from 04.1 are used).
💡 Key Knowledge
- Emf formula: ε = V + Ir or ε = I(R + r) .
- Kirchhoff's Second Law applied to complete loops containing internal resistance.
📐 Calculation Steps
- Find current through lamp: I_lamp = 4.5 / 6.2 = 0.726 A (or use R = 9 Q -> I = 6.2/9 = 0.689 A).
- Find current through 12 Q resistor: I_12 = 6.2 / 12 = 0.517 A.
- Find total circuit current: I_total = 0.726 + 0.517 = 1.24 A.
- Calculate 'lost volts': Ir = 1.24 × 2.5 = 3.1 V .
- Add terminal pd: emf = 6.2 + 3.1 = 9.3 V .
🧠 Exam Technique
Allow error carried forward (ecf) from part 04.1. Full credit is given whether students use the exact value (8.54 Q) or the prompted approximation (9 Q).
Resistivity of the Wire
✅ Correct Answer
resistivity = 5.1 × 10⁻⁸ Q m (Accept 5.10 × 10⁻⁸ Q m)
💡 Key Knowledge
- Resistivity equation: p = RA / l where A = π(d/2)² .
- Unit conversions: mm to m ( ×10⁻³ ).
📐 Calculation Steps
- Convert diameter: d = 0.19 mm = 0.19 × 10⁻³ m.
- Calculate cross-sectional area: A = π × (0.19 × 10⁻³ / 2)² = 2.835 × 10⁻⁸ m².
- Rearrange and calculate resistivity: p = (9.0 × 2.835 × 10⁻⁸) / 5.0 = 5.10 × 10⁻⁸ Q m.
❌ Common Errors
Forgetting to halve the diameter when finding the radius, or missing the unit conversion from millimeters to meters (power of 10 errors).
Variable Resistor in Series (Qualitative)
✅ Correct Answer
The lamp becomes dimmer (resistance increases, reducing current through the lamp).
💡 Key Knowledge
- In a series circuit, adding more length of resistance wire increases total circuit resistance.
- An increase in resistance decreases the overall current.
🧠 Examiner Commentary
To gain both marks, students must link the movement of the contact to an increase in resistance, and then explain the consequence on current or potential difference. Do not condone terms like "current across" or "pd through".
❌ Common Errors
Confusing current and potential difference, or incorrectly stating that moving the contact decreases resistance.
Variable Resistor in Parallel (Qualitative)
✅ Correct Answer
The lamp becomes brighter (increasing external resistance reduces current drawn from the cell, lowering 'lost volts' and thus increasing terminal pd).
💡 Key Knowledge
- ">
- Adding resistance in parallel with a branch can counterintuitively increase total *external* circuit resistance if configured as a potential divider/rheostat control.
- Formula link: V_terminal = ε - Ir . Less current drawn means smaller Ir drop, increasing terminal pd.
🧠 Examiner Commentary
Top-level responses clearly identified that increasing the resistance of the added parallel branch increases the total external resistance, reduces the total current from the supply, decreases lost volts, and consequently raises the terminal potential difference across the lamp.
❌ Common Errors
Assuming that adding any component in parallel always decreases total circuit resistance without analyzing how the sliding contact alters the lengths of the resistive paths.
Topics
Physics · 3.5 Electricity
Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.