AQA GCSE Combined Science: Trilogy Physics Paper 1 (Foundation), 2023: Question 6
15 marks · Standard Demand difficulty · Extended Answer
Analyze electrical resistance in parallel circuits and use graphical data to evaluate body water percentage.
Practise this questionQuestion
Question text
06 Body analysis scales use the electrical resistance of a person’s legs to estimate the
percentage of water in the person’s body.
Figure 8 shows body analysis scales.
Figure 8
The person’s legs contain both solid tissue and water.
A student used resistors to model the solid tissue and water.
The student connected a 20 Ω resistor in parallel with a resistor, R.
Figure 9 shows the circuit diagram.
Figure 9
06.1 To determine the total resistance of both resistors, a voltmeter must be connected into
the circuit.
Complete Figure 9 to show where the voltmeter should be connected.
27 [1 mark]
06.2 The student calculated the total resistance of the two resistors.
The student’s answer was 26 Ω.
Explain why the student’s answer cannot be correct.
[2 marks]
Use the Physics Equations Sheet to answer questions 06.3 and 06.4.
06.3 Write down the equation that links current (I), resistance (R) and
potential difference (V).
[1 mark]
06.4 When the total resistance of the resistors was 7.5 Ω the current in the circuit
was 480 mA.
Calculate the potential difference across the two resistors.
[3 marks]
Potential difference = V
The student investigated how the resistance of R affected the total resistance of
the circuit.
Table 2 shows the results.
Table 2
Resistance of R Total resistance of the circuit
in ohms in ohms
5.0 4.0
10.0 6.7
15.0 8.6
20.0 10.0
25.0 11.1
Some of the results are plotted in Figure 10.
Figure 10
06.5 Complete Figure 10.
You should:
• label both axes
• plot the two remaining values from Table 2
• draw the line of best fit.
[3 marks]
06.6 What resistance of R would give a total resistance of 4.4 Ω?
Use Figure 10.
[1 mark]
Resistance of R = Ω
The body analysis scales initially show a reading of 0.0 kg.
When the student steps onto the scales the reading is 64.8 kg.
The student steps off the scales and then immediately steps back on.
The scales now show a reading of 64.1 kg.
06.7 Complete the sentence.
[1 mark]
The difference between the two values given by the scales is due
to a error.
06.8 The height of the student is programmed into the scales.
The scales place the student into a category, A, B or C, based on height and mass.
Figure 11 shows how the scales use the category and the total resistance of the legs
to determine the body water percentage.
Figure 11
The total resistance of the student’s legs is 600 kΩ. A healthy body water percentage
is between 45% and 65%.
*30* The different measurements of the mass of the student mean that the student could
be in either category A or category B.
Evaluate if the student has a healthy body water percentage.
[3 marks]
Mark scheme
Show the mark scheme
Question 6
AO /
Question Answers Extra information Mark
Spec. Ref.
06.1 voltmeter symbol correct and 1 AO1
connected across the resistors 6.2.1.3
6.2.1.1
6.2.1.4
RPA15
AO /
Spec. Ref.
06.2 the total resistance must be less allow the total resistance cannot 1 AO2
than 20 be more than 20
because the total resistance of allow the total resistance of the 1 AO1
the resistors (in parallel) is less resistors (in parallel) is less than
than the resistance of the either resistor 6.2.2
smallest resistor 6.2.1.3
RPA15
AO /
Spec. Ref.
06.3 potential difference = current × 1 AO1
resistance 6.2.1.3
RPA15
or
V = IR – OMBINED SCIENCE: TRILOGY – 1F –
AO /
Spec. Ref.
06.4 480 mA = 0.48 A 1 AO2
6.2.1.3
V = 0.48 × 7.5 allow a correct substitution of an 1 RPA15
incorrectly / not converted value
for current
V = 3.6 (V) allow an answer consistent with 1
their incorrectly / not converted
value for current
AO /
Spec. Ref.
06.5 x-axis labelled resistance of R in 1 AO2
and y-axis labelled Total 6.2.1.3
resistance (of resistors) in RPA15
both points plotted correctly points must be plotted within ½ 1
small square
curved line of best fit drawn allow a line of best fit which 1
ignores an outlier
AO /
Spec. Ref.
06.6 reading from graph consistent allow an answer within ½ small 1 AO3
with their line of best fit square 6.2.1.3
RPA15
AO /
Spec. Ref.
06.7 random 1 AO3
– OMBINED SCIENCE: TRILOGY – – 6.2.1.3
AO /
Spec. Ref.
06.8 in category A the body water allow a value for A between 1 AO3
percentage is 61% 60% and 62% 6.2.1.3
in category B the body water 1
percentage is 68%
if in category A they have a 1
healthy body water percentage
and if in category B they have
an unhealthy body water
percentage
Total Question 6 15
How to answer it
Resistance and Body Composition Analysis
What this question tests
This question assesses your understanding of parallel circuits, the relationship between potential difference, current, and resistance (Ohm's Law), graphing skills, and the ability to evaluate data to form a conclusion.
Circuit Symbols and Connections
Key Knowledge
- Voltmeters measure potential difference and must always be connected in parallel (across the component).
- Ammeters measure current and must be connected in series.
Correct Answer
Draw a circle with a 'V' inside. Connect the wires from the voltmeter to either side of the parallel resistor block.
Resistance in Parallel
Exam Technique
When resistors are added in parallel, the total resistance is always less than the resistance of the smallest individual resistor. This is because you are providing more "paths" for the current to flow through.
Correct Answer
- The total resistance must be less than 20 Ω. [1 mark]
- Because the total resistance of resistors in parallel is less than the resistance of the smallest resistor. [1 mark]
Common Error
Students often mistakenly add the resistances together (20 + R). This only applies to series circuits. In parallel, the resistance always drops.
Calculating Potential Difference
The Equation
Potential Difference (V) = Current (I) × Resistance (R)
Step-by-Step Calculation
- Convert Units: Current is given as 480 mA. You must convert this to Amps (A).
480 / 1000 = 0.48 A - Substitute Values: Use the total resistance (7.5 Ω) and the converted current.
V = 0.48 × 7.5 - Final Answer: V = 3.6 V
Calculation Trap
Unit Conversion: Forgetting to convert mA to A is the most common way to lose marks here. Always check if the prefix 'm' (milli) is present!
Graphing and Data Analysis
Graph Skills
- Labels: Include the variable name AND the unit (e.g., Resistance in Ω).
- Plotting: Use a sharp pencil. Points must be within half a small square of the correct value.
- Line of Best Fit: For this data, it should be a smooth curve, not a straight line.
Reading the Graph
To find the resistance of R for a total resistance of 4.4 Ω:
- Find 4.4 on the y-axis (Total Resistance).
- Move horizontally to your line of best fit.
- Move vertically down to the x-axis to read the value (should be approx 6 Ω).
Types of Error
Correct Answer
The difference is due to a random error.
Why?
Because the student stepped off and back on and got a slightly different result, it shows a lack of precision in the measurement process that varies each time. If the scales always showed 0.1 kg too much, it would be a systematic error.
Evaluating Conclusions
Data Extraction (from Figure 11)
At a resistance of 600 kΩ:
- Category A: Body water percentage = 61%
- Category B: Body water percentage = 68%
The Evaluation [3 marks]
- If the student is in Category A, they have a healthy percentage (61% is between 45% and 65%).
- If the student is in Category B, they have an unhealthy percentage (68% is higher than 65%).
- Conclusion: It is impossible to say for sure if the student is healthy without knowing their correct category.
Topics
Physics · P2: Electricity
Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Physics Paper 1 (Foundation), 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.