AQA GCSE Combined Science: Trilogy Physics Paper 1 (Higher), 2019: Question 1
8 marks · Standard Demand difficulty · Short Answer
Answer questions about measuring current and potential difference in a resistor circuit, the effect of changing resistance, the current–potential difference relationship, and calculate resistance using V = IR.
Practise this questionQuestion
Question text
01 A student investigated how the current in a resistor varies with the potential difference
across the resistor.
Figure 1 shows part of the circuit used.
Figure 1
01.1 The student connected an ammeter and a voltmeter into the circuit.
What is the correct way to connect the ammeter and the voltmeter into the circuit?
[1 mark]
Tick ( ) one box.
Ammeter Voltmeter
In parallel with the resistor In series with the resistor
In parallel with the cell In series with the resistor
In series with the resistor In parallel with the resistor
In series with the resistor In parallel with the cell
01.2 The student increased the resistance of the variable resistor.
How did increasing the resistance affect the current in the circuit?
[1 mark]
01.3 How should the student change the circuit to give negative values for current and
potential difference?
[1 mark]
01.4 Name the type of relationship between current and potential difference for a resistor at
constant temperature.
[1 mark]
01.5 Write the equation which links current, potential difference and resistance.
[1 mark]
01.6 The current in the resistor was 0.12 A when the potential difference across the resistor
was 3.0 V
Calculate the resistance of the resistor.
[3 marks]
Resistance = Ω
Mark scheme
Show the mark scheme
AO /
Question Answers Extra information Mark ID
Spec. Ref.
01.1 ammeter in series with the 1 AO1/1 A
resistor, voltmeter in parallel
with the resistor 6.2.1.4
RP 16
WS 2.4
01.2 current decreased ignore slows down 1 AO1/1 E
6.2.1.3
RP 16
WS 3.6
01.3 reverse the connections to the allow battery for cell 1 AO1/2 E
cell
allow reverse the cell 6.2.1.3
RP 16
WS 2.2
01.4 (directly) proportional do not allow inversely 1 AO1/2 G
proportional
6.2.1.3
do not allow indirectly RP 16
proportional WS 3.5
01.5 potential difference = current × allow voltage for potential 1 AO1/1 E
resistance difference
6.2.1.3
or RP 16
WS 3.3
V=IR
allow any correct
re-arrangement
01.6 an answer of 25 (Ω) scores 3 AO2/1 E
marks
6.2.1.3
3.0 = 0.12 × R 1 RP 16
WS 3.3
3.0
R = 1
0.12
R = 25 (Ω) 1
Total 8
How to answer it
Current, potential difference and resistance in a resistor circuit
You need to know how to connect an ammeter and voltmeter correctly, how changing resistance affects current, how to reverse a cell to get negative readings, the meaning of a directly proportional relationship, and how to use V = IR to calculate resistance.
Measuring current, voltage and resistance in a resistor circuit
Focus: circuit symbols, graph relationship, and one resistance calculation
💡 Key knowledge
- An ammeter measures current and must be connected in series.
- A voltmeter measures potential difference and must be connected in parallel across the component.
- For a fixed resistor at constant temperature, current and potential difference are directly proportional.
- Use the equation V = IR .
🧠 Exam technique
- Read the circuit carefully: series means the component is in the same loop.
- When asked for a relationship, use the exact wording the mark scheme wants: directly proportional.
- For calculations, always write the equation first, then substitute numbers, then rearrange.
- Include units in the final answer: Ω for resistance.
❌ Common errors
- Putting the ammeter in parallel — this is wrong.
- Putting the voltmeter in series — this is wrong.
- Saying current “slows down” instead of decreases; the mark scheme allows “decreased”.
- Saying “inversely proportional” for a resistor at constant temperature.
Part 01.1
How should the ammeter and voltmeter be connected?
✅ Correct answer
Ammeter in series with the resistor; voltmeter in parallel with the resistor.
💡 Key knowledge
- The ammeter measures the current flowing through the resistor, so it must be part of the same loop.
- The voltmeter measures the potential difference across the resistor, so it must be connected across it.
🧠 Exam technique
In multiple choice tables, match both columns correctly. One correct pairing gains the mark only if the full row is right.
Part 01.2
What happens to the current when the variable resistance is increased?
✅ Correct answer
The current decreased.
💡 Key knowledge
Increasing resistance makes it harder for charge to flow, so the current gets smaller.
❌ Common errors
Some students wrote “slows down”. The mark scheme says this is not needed; the marking point is that the current decreased.
Part 01.3
How can the circuit be changed to give negative values for current and potential difference?
✅ Correct answer
Reverse the connections to the cell.
💡 Key knowledge
- Negative readings happen when the direction of current is reversed relative to the meter connections.
- The mark scheme also accepts reverse the cell or reverse the battery.
🧠 Exam technique
If a question asks for negative values, think about reversing polarity rather than changing the meters.
Part 01.4
Name the type of relationship between current and potential difference for a resistor at constant temperature.
✅ Correct answer
Directly proportional.
💡 Key knowledge
For an ohmic resistor at constant temperature, if potential difference increases, current increases by the same proportion.
❌ Common errors
- Do not say inversely proportional.
- Do not say indirectly proportional.
- For 1 mark, the exact phrase matters.
Part 01.5
Write the equation linking current, potential difference and resistance.
✅ Correct answer
potential difference = current × resistance
or V = IR
💡 Key knowledge
Potential difference can also be called voltage, and both mean the same thing in this context.
🧠 Exam technique
You can earn the mark with the word equation or the symbol equation. Make sure it is written correctly and clearly.
Part 01.6
Calculate the resistance of the resistor.
📐 Calculations — step by step
- Write the equation: V = IR
- Substitute the values: 3.0 = 0.12 × R
- Rearrange: R = 3.0 ÷ 0.12
- Calculate: R = 25
- Include the unit: 25 Ω
✅ Correct answer
Resistance = 25 Ω
❌ Common errors
- Using the wrong equation, such as I = VR .
- Forgetting to rearrange before calculating.
- Missing the unit Ω .
- Writing the answer as 0.25 Ω or 250 Ω due to a calculator slip.
🧠 How the marks are awarded
- 1 mark for substituting correctly: 3.0 = 0.12 × R
- 1 mark for rearranging: R = 3.0 ÷ 0.12
- 1 mark for the correct final answer: 25 Ω
The examiner note says an answer of 25 Ω scores full marks, so even if working is brief, the final value must be right.
Quick recall checklist
💡 Remember these facts
- Ammeter: series
- Voltmeter: parallel
- Resistance increased → current decreased
- Negative readings → reverse the cell
- Ohmic resistor: directly proportional
- V = IR
🧠 Best exam habit
When stuck, ask: “Is this about the circuit setup, the direction of current, the relationship, or the calculation?” That helps you choose the right GCSE physics idea quickly.
Topics
Physics · Required Practicals · P2: Electricity · Physics Required Practicals
Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Physics Paper 1 (Higher), 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.