AQA GCSE Combined Science: Trilogy Physics Paper 1 (Foundation), November 2020: Question 1

12 marks · Low Demand difficulty · Short Answer

Investigate how the resistance of a wire varies with its length, identifying circuit symbols, independent and dependent variables, calculating mean potential difference, resistance, power, charge flow, and sketching the effect of wire thickness on the resistance graph.

Practise this question

Question

Question 01 consists of multiple subquestions based on a circuit diagram (Figure 1) investigating wire resistance versus length. Subquestion 01.1 asks to complete the voltmeter and ammeter symbols in Figure 1. Subquestions 01.2 and 01.3 require ticking the independent and dependent variables from multiple-choice lists. Subquestion 01.4 gives three potential difference readings (0.16 V, 0.17 V, 0.15 V) to calculate the mean. Subquestions 01.5, 01.6, and 01.7 provide equations to calculate resistance, power dissipated, and charge flow using given values of current, potential difference, and time. Subquestion 01.8 shows Figure 2, a sketch graph of resistance against length, asking to draw a line showing how resistance varies with length for a thicker wire.
Question text

01 A student investigated how the resistance of a wire varies with the length of the wire.

Figure 1 shows the circuit used.

Figure 1

01.1 The symbols for the voltmeter and ammeter in Figure 1 are not complete.

Complete the symbols for the voltmeter and ammeter in Figure 1.

[1 mark]

01.2 Which variable is the independent variable?

[1 mark]

Tick ( ) one box.

The current in the wire

The length of the wire being tested

The resistance of the wire

The thickness of the wire 3

01.3 Which variable is the dependent variable?

[1 mark]

Tick ( ) one box.

*02* The current in the wire

The length of the wire being tested

The resistance of the wire

The thickness of the wire

01.4 The student took repeat readings of potential difference for a 30 cm length of the wire.

The readings were:

0.16 V 0.17 V 0.15 V

Calculate the mean potential difference.

[2 marks]

Mean potential difference = V

The length of the wire was increased to 60 cm

The current in the wire was 0.50 A

The mean potential difference across the wire was 0.32 V

01.5 Calculate the resistance of the 60 cm length of wire.

Use the equation:

potential difference

resistance =

current

[2 marks]

Resistance = Ω

01.6 Calculate the power dissipated in the 60 cm length of wire.

Use the equation:

power = potential difference × current

[2 marks]

5 Power = W

01.7 Calculate the charge flow when there is a current of 0.50 A in the wire for 17 s

Use the equation:

charge flow = current × time

[2 marks]

Charge flow = C

01.8 Figure 2 is a sketch graph of the results.

Figure 2

The student repeated the investigation using a thicker wire made from the same

metal. For the same length, the thicker wire has a lower resistance.

Draw a line on Figure 2 to show how the resistance of the thicker wire varies

with length.

[1 mark]

Mark scheme

Show the mark scheme Mark scheme for question 01 detailing marks for each sub-question: 01.1 requires the ammeter symbol 'A' in series and voltmeter 'V' in parallel across the wire (1 mark). 01.2: 'the length of the wire being tested' (1 mark). 01.3: 'the resistance of the wire' (1 mark). 01.4: (0.16 + 0.17 + 0.15) / 3 = 0.16 (V) (2 marks). 01.5: 0.32 / 0.50 = 0.64 (Ω) (2 marks). 01.6: 0.32 × 0.50 = 0.16 (W) (2 marks). 01.7: 0.50 × 17 = 8.5 (C) (2 marks). 01.8: a straight line through the origin with a lower gradient (1 mark). Total 12 marks.

AO /

Question Answers Extra information Mark

Spec. Ref.

01.1 both symbols correct and in the correct position 1 AO1

6.2.1.3

6.2.1.4

RPA 15

01.2 the length of the wire being 1 AO1

tested 6.2.1.3

RPA 15

01.3 the resistance of the wire 1 AO1

6.2.1.3

RPA 15

0.16 + 0.17 + 0.15

01.4 mean p.d. = 1 AO2

6.2.1.3

mean p.d. = 0.16 (V) 1 RPA 15

01.5 R = 0.32/0.50 1 AO2/1

6.2.1.3

R = 0.64 (Ω) 1 6.2.1.3

RPA 15

01.6 power = 0.32 × 0.50 1 AO2

7 6.2.4.1

power = 0.16 (W) 1

01.7 charge flow = 0.50 × 17 1 AO2

6.2.1.2

8.5 (C) 1

01.8 a straight line through the origin 1 AO1/2

with a lower gradient 6.2.1.3

RPA 15

Total 12

How to answer it

Investigating Resistance of a Wire (Required Practical 15)

What this question tests

This question assesses foundational skills from AQA GCSE Physics / Combined Science Topic 2 (Electricity), specifically the required practical on factor affecting resistance:

  • Circuit Symbols & Set-up: Correct placement of an ammeter (in series) and voltmeter (in parallel).
  • Variables: Differentiating between independent and dependent variables in experimental design.
  • Data Handling: Calculating the mean of repeated trials.
  • Formula Application: Using Ohm's Law ( R = V / I ), electrical power ( P = V × I ), and charge flow ( Q = I × t ).
  • Graph Skills: Predicting the gradient of a relationship when wire diameter increases.
Part 01.1 • 1 Mark

Completing Circuit Symbols in Figure 1

Ammeter and Voltmeter Placement

✅ Correct Answer

  • Top circle (in series with battery and resistor): A (Ammeter)
  • Bottom circle (in parallel across the test wire): V (Voltmeter)
Both symbols and their positions must be correct to secure the 1 mark.

💡 Key Knowledge

Remember meter placement rules:

  • Ammeter (A): Measures current through a component → connected in series.
  • Voltmeter (V): Measures potential difference across a component → connected in parallel.
Parts 01.2 & 01.3 • 2 Marks Total

Independent and Dependent Variables

Identifying Variables from the Investigation Aim

✅ Correct Answers

01.2 Independent Variable (1 mark):

  • [✓] The length of the wire being tested

01.3 Dependent Variable (1 mark):

  • [✓] The resistance of the wire

🧠 Exam Technique

Look back at the first sentence of the question prompt:

"A student investigated how the resistance varies with the length of the wire."

  • Independent (I change): Length of the wire (altered using the crocodile clip).
  • Dependent (Data measured/calculated): Resistance of the wire.
  • Control: Thickness of the wire, material, and temperature.

❌ Common Errors

  • Selecting current or thickness as the dependent variable. Although current is read off the meter, the final value under test is resistance.
  • Mixing up "independent" (what you alter) with "dependent" (the outcome).
Part 01.4 • 2 Marks

Calculating Mean Potential Difference

Repeat Readings: 0.16 V, 0.17 V, 0.15 V

📐 Step-by-Step Calculation

  1. Sum the repeat values:
    0.16 + 0.17 + 0.15 = 0.48 V
  2. Divide by the number of readings (3):
    Mean = 0.48 / 3 = 0.16 V

Mean potential difference = 0.16 V

Mark 1: Working showing addition and division by 3 | Mark 2: Correct answer (0.16)

❌ Common Errors

  • Calculator syntax error: Typing 0.16 + 0.17 + 0.15 / 3 into a calculator without brackets gives 0.38 because only 0.15 gets divided by 3! Always press equals before dividing.
Part 01.5 • 2 Marks

Calculating Resistance

Data: Current = 0.50 A, Mean p.d. = 0.32 V (for 60 cm length)

📐 Step-by-Step Calculation

Given equation: resistance = potential difference / current

  1. Substitute the values:
    R = 0.32 / 0.50
  2. Calculate the result:
    R = 0.64 Ω

Resistance = 0.64 Ω

Mark 1: Correct substitution (0.32 / 0.50) | Mark 2: Correct evaluation (0.64)

🧠 Top Tip

Dividing by 0.50 is mathematically identical to multiplying by 2:

0.32 × 2 = 0.64

Always check that your answer makes physical sense: resistance is positive and reasonable for a short length of metal wire.

Part 01.6 • 2 Marks

Calculating Power Dissipated

Data: p.d. = 0.32 V, Current = 0.50 A

📐 Step-by-Step Calculation

Given equation: power = potential difference × current

  1. Substitute known values:
    P = 0.32 × 0.50
  2. Calculate the power:
    P = 0.16 W

Power = 0.16 W

Mark 1: Correct substitution (0.32 × 0.50) | Mark 2: Correct evaluation (0.16)

💡 Key Knowledge

Power is the rate of energy transfer, measured in Watts (W) where 1 W = 1 J/s .

Alternative formulas for electrical power on the physics equation sheet:

  • P = I² × R
  • P = V² / R
Part 01.7 • 2 Marks

Calculating Charge Flow

Data: Current = 0.50 A, Time = 17 s

📐 Step-by-Step Calculation

Given equation: charge flow = current × time

  1. Substitute known values:
    Q = 0.50 × 17
  2. Calculate the charge:
    Q = 8.5 C

Charge flow = 8.5 C

Mark 1: Correct substitution (0.50 × 17) | Mark 2: Correct evaluation (8.5)

❌ Common Errors

  • Dividing current by time instead of multiplying.
  • Forgetting that time must always be in seconds ( s ). Here time was already in seconds, so no conversion was needed!
Part 01.8 • 1 Mark

Sketching the Graph for a Thicker Wire

Effect of Wire Thickness on Resistance

✅ What to Draw on Figure 2

Draw a single straight line that:

  • Starts at the origin (0, 0) .
  • Has a lower gradient (shallower slope) than the existing line, remaining underneath it across the entire graph.
1 mark: A straight line through the origin with a lower gradient.

💡 Why is the Gradient Lower?

The gradient of a Resistance vs. Length graph represents resistance per unit length:

  • A thicker wire has a larger cross-sectional area, allowing electrons to flow through more easily.
  • Because it has lower resistance for any given length, the line must be shallower (below the original line).
  • Resistance is still directly proportional to length, so it must remain a straight line starting at (0, 0) .

❌ Common Errors

  • Drawing a curved line (thinking thickness causes non-linear behaviour).
  • Drawing the line above the original line (confusing higher thickness with higher resistance).
  • Not starting the line directly at the origin (0, 0) .

Topics

Physics · P2: Electricity

Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Physics Paper 1 (Foundation), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.