Physics 9702/36 — October/November 2023

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and EvaluationFree sample

In this experiment, you will investigate an electrical circuit.

(a)

● Connect the circuit shown in Fig. 1.1.

Fig. 1.1

● Ensure that the polarities of the two power supplies and the voltmeter are as shown in Fig. 1.1.

● Connect one of the labelled resistors into the component holder as resistor Y, as shown in Fig. 1.1. Record the resistance $R$ of resistor Y.

$R$ = ______

● The voltmeter reading should be approximately 3V.

Record the voltmeter reading $V$ .

$V$ = ______

DifficultyMedium-Easy

Worked solution

Answer

Record the labelled value of resistor $Y$ as $R$ (in $\Omega$ ) and record the voltmeter reading $V$ (in $\text{V}$ , approximately $3\,\text{V}$ ).

Example (typical):

R = 100\,\Omega

V = 3.02\,\text{V}

Final answer

R and V recorded (student-dependent)

Detailed explanation

Background Concept

A voltmeter measures the potential difference (p.d.) between two points in a circuit. To obtain a meaningful reading, it must be connected in parallel with the part of the circuit whose p.d. is required, and its polarity must match the circuit polarity (positive terminal at the higher potential).

A resistor value may be given by its label (or colour code) or measured with an ohmmeter. In practical exams, you must record values with units and with sensible precision consistent with the instrument.

Understanding the Question

You are told to build the circuit in Fig. 1.1, ensuring the polarities of both power supplies and the voltmeter match the diagram. You then:

place one labelled resistor into the holder as resistor $Y$ ,
record its resistance $R$ ,
record the voltmeter reading $V$ (expected to be about $3\,\text{V}$ ).

The actual numerical readings depend on which labelled resistor you choose and the real circuit components.

Approach

Assemble the circuit exactly as shown (especially the + and − terminals).
Insert one labelled resistor into the holder for $Y$ .
Record $R$ from the resistor label (or meter reading, if used).
Switch on and allow the reading to settle.
Record $V$ from the voltmeter scale/digital display with appropriate precision.

Step-by-Step Reasoning

Correct polarity matters because reversing the voltmeter leads would give a negative reading (or a reversed needle deflection), and reversing a supply changes the potentials in the circuit.
Record $R$ in ohms, $\Omega$ . If the resistor is labelled, you copy the stated value; if measured, you quote the reading to the meter’s resolution.
Record $V$ in volts, $\text{V}$ . For a digital meter, you copy all displayed digits; for an analogue meter, you estimate to about half a smallest division.

Key Takeaways

Voltmeter in parallel, correct polarity.
Record measured quantities with correct units and appropriate precision.

Common Mistakes

Swapping polarities of a supply or the voltmeter, leading to incorrect/negative readings.
Writing $R$ with no unit or writing $V$ in the wrong unit.
Over-rounding a digital voltmeter reading (losing significant information).

Things to Be Careful About

Ensure firm connections in the component holder; loose contacts cause fluctuating readings.
Do not confuse $R$ (in $\Omega$ ) with $1/R$ (in $\Omega^{-1}$ ), which is only needed later.
If using an analogue voltmeter, choose the correct range so that readings are around mid-scale for best precision.

Techniques used

set up the circuit with correct polaritiesread a resistor value from its label (or measure with an ohmmeter if provided)take a voltmeter reading with correct unit and suitable precision

(b)

Change Y and record $R$ and $V$ . Repeat until you have six sets of values of $R$ and $V$ .

Record your results in a table. Include values of $\frac{1}{R}$ in your table.

10M

DifficultyMedium

Worked solution

Answer

Take six different resistors for $Y$ . For each, record $R$ and the corresponding $V$ , then calculate $\dfrac{1}{R}$ .

Record in one table with headings (including units), e.g.

$R\ /\ \Omega$
$V\ /\ \text{V}$
$1/R\ /\ \Omega^{-1}$

Calculate each value using

\frac{1}{R}

(using consistent s.f. in the $1/R$ column).

Example of one row:

R = 100\,\Omega,\quad V = 3.02\,\text{V},\quad \frac{1}{R} = 0.0100\,\Omega^{-1}

Final answer

Table of six values of R, V and 1/R (student-dependent)

Detailed explanation

Background Concept

In Paper 3, marks for tables are awarded for clear presentation and correct processing:

all readings in one table,
clear column headings with quantity and unit,
consistent precision within a column,
correct calculation of any derived quantity (here, $1/R$ ).

The quantity $1/R$ is often included because it allows a linear graph when the underlying relationship involves $R$ in the denominator.

Understanding the Question

You must replace resistor $Y$ with different labelled resistors and obtain six pairs of readings $(R, V)$ . Then you must add a calculated column for $1/R$ .

So the table must contain three columns: measured $R$ , measured $V$ , and calculated $1/R$ .

Approach

Choose six different resistors that give a good spread of $R$ values (not all similar).
For each resistor:
- record $R$ and $V$ ,
- compute $1/R$ .
Present everything in a single, clearly ruled table with proper headings and units.

Step-by-Step Reasoning

Since $R$ is being changed, it is your independent variable. Using a wide range of $R$ values helps the graph cover a wide range of $1/R$ values, improving the gradient determination.
Enter raw values of $R$ and $V$ directly from the instruments/labels.
Compute the derived quantity for each row:

\frac{1}{R}

Example: if $R = 220\,\Omega$ then

\frac{1}{R} = \frac{1}{220} = 4.55\times 10^{-3}\,\Omega^{-1}

Precision: it is common to quote $1/R$ to 3 significant figures (or consistent with how precisely $R$ is known). Keep the number of decimal places consistent down a column where appropriate.

Key Takeaways

A good table is about structure: headings, units, consistent precision.
Derived quantities must be calculated correctly and recorded for each set of readings.

Common Mistakes

Missing units in headings (e.g. writing just $R$ instead of $R/\Omega$ ).
Writing $1/R$ with incorrect units (must be $\Omega^{-1}$ ).
Inconsistent rounding (e.g. mixing 0.01, 0.0100, 0.010 in the same column without reason).
Splitting results across multiple tables.

Things to Be Careful About

Use the same resistor as $Y$ for the whole measurement of each row; do not change anything else.
Ensure the voltmeter reading is stable before recording.
Avoid transcription errors when calculating $1/R$ (especially powers of ten for large $R$ ).

Techniques used

collect a suitable range of values of the independent variablerecord results in a single table with correct headings and unitscalculate a derived quantity for each rowkeep consistent significant figures within each column

(c)

(i)

Plot a graph of $V$ on the $y$ -axis against $\frac{1}{R}$ on the $x$ -axis.

DifficultyMedium-Easy

Worked solution

Answer

Plot a graph with:

$y$ -axis: $V\ /\ \text{V}$
$x$ -axis: $1/R\ /\ \Omega^{-1}$

Use a sensible scale and plot all six points accurately.

Final answer

Graph of V (y) against 1/R (x) plotted

Detailed explanation

Background Concept

A graph is used to display how one quantity depends on another and to allow parameters (such as gradient) to be found. Good graph technique in Cambridge practical papers includes:

correct axis choice (dependent variable on $y$ , independent on $x$ ),
correct axis labels with units,
scales that use a large fraction of the grid and are easy to read,
accurate plotting.

Understanding the Question

You already have a results table containing $V$ and $1/R$ . This part asks you to plot $V$ against $1/R$ with $V$ on the vertical axis.

Approach

Draw axes with enough space.
Choose scales that cover the full range of your data and use at least half the grid in both directions.
Label axes as $V/\text{V}$ and $1/R\ /\ \Omega^{-1}$ .
Plot all six points using small, neat crosses.

Step-by-Step Reasoning

The instruction “ $V$ on the $y$ -axis against $1/R$ on the $x$ -axis” fixes the variables. Do not swap them.
Decide the minimum and maximum values of $V$ and $1/R$ from your table, then choose scale steps like 0.1 V, 0.2 V, or convenient multiples of $10^{-3}\,\Omega^{-1}$ .
Plot each pair $(1/R,\ V)$ carefully. If you use crosses, the intersection is the exact point.

Key Takeaways

Correct labels and scales are essential for graph marks.
Accurate plotting is needed for a reliable best-fit line and gradient.

Common Mistakes

Missing units or writing units incorrectly (e.g. writing $1/R$ in $\Omega$ instead of $\Omega^{-1}$ ).
Using awkward scales (e.g. 3 squares = 1 unit) or scales that use only a small part of the grid.
Plotting blobs/dots too large to judge the line of best fit.

Things to Be Careful About

If you use standard form on the axis (e.g. label $1/R$ in $10^{-3}\,\Omega^{-1}$ ), make it explicit on the axis label.
Check each plotted point corresponds to the correct row (mixing rows is a common error).

Techniques used

choose appropriate axes variables and orientationlabel axes with quantity and unituse a sensible scale that uses at least half the graph gridplot points accurately from a table

(ii)

Draw the straight line of best fit.

DifficultyEasy

Worked solution

Answer

Draw a single straight line of best fit through the plotted points (balanced about the line).

Final answer

Straight best-fit line drawn

Detailed explanation

Background Concept

A best-fit line represents the overall trend in data when you expect a linear relationship. It should not be forced through every point; instead, it should be positioned so that the scatter of points is roughly balanced above and below the line.

Understanding the Question

You have plotted six points of $V$ against $1/R$ . You must now draw the straight line that best represents the trend.

Approach

Use a ruler.
Place the line so that (approximately) the same number of points lie above and below it, and the deviations are of similar size.

Step-by-Step Reasoning

Do not join points with a zig-zag; that is not a best-fit line.
If one point is clearly anomalous compared with the general trend, you still draw the best-fit line through the main trend rather than bending the line to include it.

Key Takeaways

Best-fit means “overall trend”, not “connect-the-dots”.

Common Mistakes

Joining points sequentially.
Forcing the line through the origin without evidence.

Things to Be Careful About

Use a long line (extend across most of the plotted range) to make gradient/intercept readings more accurate.

Techniques used

draw a straight best-fit line with balanced scatteravoid joining point-to-point

(iii)

Determine the gradient and $y$ -intercept of this line.

gradient = ______
$y$ -intercept = ______

DifficultyMedium

Worked solution

Working

Choose two well-separated points on the best-fit line and find

\text{gradient} = \frac{\Delta V}{\Delta\left(1/R\right)}

Read the $y$ -intercept where $1/R = 0$ .

Example (typical):

\text{gradient} = 18.0\,\text{V}\,\Omega

\text{$y$-intercept} = 1.50\,\text{V}

Answer

gradient $= \dfrac{\Delta V}{\Delta(1/R)}$ (in $\text{V}\,\Omega$ )

y-intercept $=$ value of $V$ at $1/R = 0$ (in $\text{V}$ )

Final answer

Gradient and y-intercept obtained from best-fit line (student-dependent)

Detailed explanation

Background Concept

For a straight-line graph of the form

y = mx + c

the gradient is $m = \Delta y / \Delta x$ ,
the $y$ -intercept is $c$ (the value of $y$ when $x=0$ ).

Here $y \equiv V$ and $x \equiv 1/R$ . So the gradient has units

\frac{\text{V}}{\Omega^{-1}} = \text{V}\,\Omega

Understanding the Question

After plotting $V$ vs $1/R$ and drawing the best-fit line, you must extract two numerical values from the line:

the gradient,
the $y$ -intercept.

These will be used in part (d).

Approach

Use a large triangle on the best-fit line to reduce percentage reading uncertainty.
Compute gradient as $\Delta V / \Delta(1/R)$ .
Find intercept by extending the best-fit line to meet the $V$ axis (where $1/R = 0$ ).

Step-by-Step Reasoning

Pick two points on the line, not necessarily measured data points. They should be far apart so that $\Delta V$ and $\Delta(1/R)$ are large.
Read the coordinates carefully from the axes.
Calculate

\text{gradient} = \frac{V_2 - V_1}{(1/R)_2 - (1/R)_1}

Units: if $V$ is in volts and $1/R$ is in $\Omega^{-1}$ , then the gradient is in $\text{V}\,\Omega$ .
For the intercept, go to $1/R = 0$ on the $x$ -axis and read the value of $V$ where the best-fit line crosses the $y$ -axis.

Key Takeaways

Use a large triangle: it is the main practical skill for reliable gradients.
Always compute gradient as “change in $y$ divided by change in $x$ ”.

Common Mistakes

Using a small triangle, giving a very uncertain gradient.
Taking $\Delta x / \Delta y$ instead of $\Delta y / \Delta x$ .
Using two experimental points that are not on the best-fit line.
Quoting gradient without units.

Things to Be Careful About

Make sure you read the axes correctly, especially if you used a scale factor like $\times 10^{-3}$ .
Intercept may be outside the plotted region; extend the line neatly with a ruler to reach the axis.

Techniques used

determine gradient using a large triangle on the best-fit lineread the y-intercept by extending the best-fit line to the axisuse consistent units when computing a gradient

(d)

It is suggested that the quantities $V$ and $R$ are related by the equation

V = \frac{a}{R} + b

where $a$ and $b$ are constants.

Using your answers in (c)(iii), determine the values of $a$ and $b$ .
Give appropriate units.

$a$ = ______
$b$ = ______

DifficultyMedium-Easy

Worked solution

Working

Given

V = \frac{a}{R} + b

Let $x = 1/R$ . Then

V = a\left(\frac{1}{R}\right) + b = ax + b

So gradient $= a$ and $y$ -intercept $= b$ .

Units:

[a] = \frac{\text{V}}{\Omega^{-1}} = \text{V}\,\Omega, \qquad [b] = \text{V}

Answer

a = \text{gradient}\ (\text{V}\,\Omega)

b = \text{$y$-intercept}\ (\text{V})

Final answer

a = gradient (V Ω), b = y-intercept (V)

Detailed explanation

Background Concept

Many practical investigations aim to test whether a relationship is linear. If you can rewrite an equation into the form

y = mx + c

then plotting $y$ against $x$ produces a straight line where:

$m$ is the gradient,
$c$ is the $y$ -intercept.

Understanding the Question

You are told that the circuit may obey

V = \frac{a}{R} + b

You have already plotted $V$ against $1/R$ and found the gradient and intercept in part (c)(iii). This question asks you to use those graph values to determine $a$ and $b$ , including units.

Approach

Identify what you used for the $x$ -axis: $x = 1/R$ .
Rewrite the given equation in terms of $x$ .
Compare with $V = mx + c$ .
Read off $a$ and $b$ , then assign units from the axes units.

Step-by-Step Reasoning

Start with

V = \frac{a}{R} + b

Since your graph uses $1/R$ on the horizontal axis, define

x = \frac{1}{R}

Then

V = ax + b

Comparing with $V = mx + c$ :

gradient $m = a$ ,
intercept $c = b$ .

Units:

The gradient unit is (unit of $V$ )/(unit of $1/R$ ):

\frac{\text{V}}{\Omega^{-1}} = \text{V}\,\Omega

so $a$ is in $\text{V}\,\Omega$ .

The intercept is a voltage reading, so $b$ is in $\text{V}$ .

Key Takeaways

If you plot the variables exactly as suggested by the equation, constants drop straight out as gradient and intercept.
Units come directly from axis units.

Common Mistakes

Saying $a = 1/(\text{gradient})$ (wrong for this linearisation).
Giving $a$ in volts rather than $\text{V}\,\Omega$ .
Confusing the resistor value $R$ with $1/R$ when matching to $y=mx+c$ .

Things to Be Careful About

Use your own gradient and intercept values from (c)(iii); do not recalculate them from raw data here.
If your $x$ -axis used a scale factor (e.g. $1/R$ in $10^{-3}\,\Omega^{-1}$ ), ensure your gradient (and hence $a$ ) is adjusted accordingly.

Techniques used

match a given equation to the straight-line form y = mx + cidentify constants from gradient and interceptdeduce units of constants from the plotted variables

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Q2Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation · Presentation of Data and Observations17M

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