In this experiment, you will investigate how the current in a circuit varies as the resistance of the circuit is changed. (a) Measure and record the length $L$ of wire between the crocodile clips on the wire labelled F. $L$ = \_\_\_\_\_\_ (b) Set up the circuit as shown in Fig. 1.1. (i) Attach wire F to the wire on the metre rule as shown in Fig. 1.2. The distance $x$ between the crocodile clips should be approximately 50 cm. (ii) Measure and record $x$. $x$ = \_\_\_\_\_\_…

(a) L = 50.0 cm (example) · (b) Circuit set up as in Fig. 1.1 · (c)(i) Wire F attached; x ≈ 50 cm · (c)(ii) x = 50.0 cm (example) · (d)(i) Switch closed · (d)(ii) I = 0.57 A (example) · (d)(iii) Switch opened · (e) See table (example values shown) · (f)(i) Graph of 1/I (y) against x^2/(x+L) (x) plotted · (f)(ii) Straight line of best fit drawn · (f)(iii) gradient = −3.0×10^−2 A^−1 cm^−1, y-intercept = 2.50 A^−1 (example) · (g) P = 3.0×10^−2 A^−1 cm^−1, Q = 2.50 A^−1 (example)

Physics 9702/33 — May/June 2014

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 how the current in a circuit varies as the resistance of the circuit is changed.

(a)

Measure and record the length $L$ of wire between the crocodile clips on the wire labelled F.

$L$ = ______

DifficultyEasy

Worked solution

Answer

Measure $L$ on the metre rule to the nearest $1\ \text{mm}$ (or $0.1\ \text{cm}$ ).

Example (typical):

L = 50.0\ \text{cm}

Final answer

L = 50.0 cm (example)

Detailed explanation

Background Concept

A metre rule gives a length by reading positions on a scale. The length between two points is the difference between the two position readings. The main measurement issues are (i) parallax (reading from an angle) and (ii) using an appropriate resolution (typically $1\ \text{mm}$ ).

Understanding the Question

You are asked to find $L$ , the length of wire F between its two crocodile clips. This value is later used in calculated quantities such as $\dfrac{x^2}{(x+L)}$ , so $L$ must be recorded clearly with a unit.

Approach

Place the crocodile clips so their contact points are against the metre rule scale. Read the position of each clip contact point, then subtract to get $L$ . Record $L$ to the same precision as the scale permits.

Step-by-Step Reasoning

Align the wire (or the clip contact points) along the metre rule.
Read the scale at the first clip contact point (not the outer edge of the clip casing).
Read the scale at the second clip contact point.
Calculate $L = |\text{reading}_2 - \text{reading}_1|$ .
Record $L$ with unit (e.g. $\text{cm}$ ) and appropriate dp (e.g. $50.0\ \text{cm}$ if measuring to $0.1\ \text{cm}$ ).

Key Takeaways

Length should be obtained from two position readings.
Always quote a unit and match the precision to the instrument.

Common Mistakes

Reading the wrong reference point on the crocodile clip (gives systematic error).
Writing no unit.
Recording too many decimal places (false precision).

Things to Be Careful About

Eye must be directly above the scale marking to avoid parallax.
If later using $x$ in $\text{cm}$ , keep $L$ in $\text{cm}$ as well (consistent units in formulas).

Techniques used

measure a length using a metre rule with appropriate precisionavoid parallax error by aligning the eye with the scalerecord readings with consistent decimal places and units

(b)

Set up the circuit as shown in Fig. 1.1.

Fig. 1.1

DifficultyMedium-Easy

Worked solution

Answer

Circuit connected as in Fig. 1.1 with the ammeter in series, crocodile clips making firm contact, and the switch initially open.

Final answer

Circuit set up as in Fig. 1.1

Detailed explanation

Background Concept

In a d.c. circuit, the current is measured by an ammeter. An ammeter must be connected in series so that the same current passes through it as through the component being tested. Incorrect connection (e.g. across the supply) can give wrong readings and may damage the meter.

Understanding the Question

You are instructed to build the circuit shown. This is essential context for later measurements of current $I$ when the effective resistance is changed.

Approach

Follow the diagram exactly: power supply → switch → ammeter → wire on metre rule → back to power supply. Ensure crocodile clips contact the wire properly.

Step-by-Step Reasoning

Connect the power supply terminals to the circuit as shown.
Put the switch in series so it controls the whole circuit.
Place the ammeter in series (not in parallel) and select a suitable range.
Connect crocodile clips firmly onto the wire on the metre rule; poor contact increases resistance unpredictably.
Keep the switch open until ready to take readings.

Key Takeaways

Ammeter must be in series.
Good electrical contact is crucial for reliable results.

Common Mistakes

Connecting the ammeter in parallel.
Leaving the switch closed while adjusting clips (heating changes resistance).

Things to Be Careful About

Ensure the ammeter range is appropriate to avoid overload.
Check for loose leads; intermittent contact gives fluctuating current.

Techniques used

assemble a series circuit following a circuit diagramconnect an ammeter in series with the component under testcheck polarity and connections before closing the switch

(c)

(i)

Attach wire F to the wire on the metre rule as shown in Fig. 1.2.

Fig. 1.2

The distance $x$ between the crocodile clips should be approximately 50 cm.

DifficultyMedium-Easy

Worked solution

Answer

Attach wire F as shown in Fig. 1.2 so that the distance between the two crocodile clips on the metre-rule wire is approximately

x \approx 50\ \text{cm}

Final answer

Wire F attached; x ≈ 50 cm

Detailed explanation

Background Concept

Crocodile clips make electrical contact at specific points on the wire. The separation between these contact points sets the length of wire involved and therefore affects the circuit resistance. Good practice is to start near the middle of the metre rule to allow movement in both directions for later readings.

Understanding the Question

You must connect wire F to two points on the wire on the metre rule, creating a separation $x$ (the variable you will later change). The initial $x$ should be about $50\ \text{cm}$ .

Approach

Use the metre rule to position the two clips so their contact points are about $50\ \text{cm}$ apart. Ensure both clips clamp firmly onto the same wire.

Step-by-Step Reasoning

Identify the two connection points shown in Fig. 1.2.
Clip one end of wire F to the metre-rule wire at a chosen position.
Clip the other end of wire F to the metre-rule wire at a second position roughly $50\ \text{cm}$ away.
Check that $x$ refers to the distance between the two contact points (not between ends of the clip bodies).

Key Takeaways

$x$ is defined by the contact points of the clips.
Start at $\approx 50\ \text{cm}$ to allow a good range of later values.

Common Mistakes

Measuring $x$ between the wrong points on the crocodile clips.
Placing clips too close to the ends of the wire, limiting the available range.

Things to Be Careful About

Keep the switch open while repositioning clips to reduce heating.
Ensure clips do not slip during readings (this changes $x$ unknowingly).

Techniques used

attach a parallel connection using crocodile clipsset an initial value of the independent variable using the metre rule scaleensure consistent contact points for length measurements

(ii)

Measure and record $x$ .

$x$ = ______

DifficultyEasy

Worked solution

Answer

Measure $x$ between the two crocodile-clip contact points on the metre-rule wire.

Example (typical):

x = 50.0\ \text{cm}

Final answer

x = 50.0 cm (example)

Detailed explanation

Background Concept

A length $x$ on a metre rule is best found from two position readings: $x = |x_2 - x_1|$ . This avoids needing one clip exactly at zero and reduces systematic offset errors.

Understanding the Question

You must record $x$ , the distance between the two crocodile clips (contact points) on the metre-rule wire. This is your independent variable and will be changed for multiple readings.

Approach

Read the metre rule at each clip contact point, subtract to obtain $x$ , then record with unit and appropriate dp.

Step-by-Step Reasoning

Read the position of the first clip contact point on the scale.
Read the position of the second clip contact point.
Compute $x$ as the difference.
Record $x$ to the instrument resolution (e.g. $0.1\ \text{cm}$ ).

Key Takeaways

Use two readings and subtraction.
Record $x$ consistently each time (same reference points).

Common Mistakes

Measuring from the wrong part of the clip.
Mixing units (e.g. $x$ in cm but later using $L$ in m).

Things to Be Careful About

Avoid parallax and ensure the wire is straight along the metre rule.

Techniques used

measure a separation by subtracting two position readings on a scalerecord the independent variable with appropriate precision and unitensure consistent definition of the measurement points

(d)

(i)

Close the switch.

DifficultyEasy

Worked solution

Answer

Switch closed (briefly) to allow current to flow and the ammeter reading to be taken.

Final answer

Switch closed

Detailed explanation

Background Concept

Closing the switch completes the circuit so current flows. For experimental reliability, current should only flow while taking readings to minimise heating of the wire (which would change its resistance).

Understanding the Question

This step is required before you can record the ammeter reading $I$ .

Approach

Close the switch, wait for the reading to settle, then record $I$ in the next step.

Step-by-Step Reasoning

Close the switch to complete the circuit.
Observe the ammeter and allow any brief fluctuation to settle.

Key Takeaways

Switch closed only when measuring helps keep resistance approximately constant.

Common Mistakes

Leaving the switch closed for a long time, causing heating and drifting readings.

Things to Be Careful About

If the ammeter reading is off-scale, open the switch immediately and change range.

Techniques used

operate the switch correctly to start a measurementobserve the meter reading after it stabilises

(ii)

Record the ammeter reading $I$ .

$I$ = ______

DifficultyEasy

Worked solution

Answer

Record the current $I$ from the ammeter.

Example (typical for $x \approx 50\ \text{cm}$ ):

I = 0.57\ \text{A}

Final answer

I = 0.57 A (example)

Detailed explanation

Background Concept

Current $I$ is measured in amperes (A). Digital meters typically display to a fixed resolution; analogue meters require careful interpolation and viewing to avoid parallax.

Understanding the Question

With the circuit closed, you must read and write down the ammeter value $I$ for the current value of $x$ .

Approach

Choose an ammeter range that gives a clear reading without overload, then record $I$ once steady.

Step-by-Step Reasoning

Confirm the ammeter is in series.
Ensure the range is not exceeded.
Read the displayed value when stable.
Record with unit and consistent dp across the table (e.g. $0.57\ \text{A}$ ).

Key Takeaways

Correct range selection and stable reading improve data quality.

Common Mistakes

Forgetting the unit A.
Recording different decimal places for different readings without reason.

Things to Be Careful About

Heating can cause $I$ to drift: record promptly after closing the switch.

Techniques used

read current from an ammeter with appropriate resolutionselect a suitable meter rangerecord a stable reading with unit and consistent precision

(iii)

Open the switch.

DifficultyEasy

Worked solution

Answer

Switch opened after the reading is taken.

Final answer

Switch opened

Detailed explanation

Background Concept

When current flows through a wire, it heats up. The resistance of metal wire increases with temperature, which would affect the relationship being investigated. Opening the switch reduces heating and keeps conditions more constant.

Understanding the Question

You are instructed to open the switch after recording $I$ .

Approach

Open the switch immediately after each measurement, especially before moving crocodile clips.

Step-by-Step Reasoning

After recording $I$ , open the switch to break the circuit.
Allow the wire to cool briefly if it has warmed.

Key Takeaways

Minimising heating helps produce more consistent data.

Common Mistakes

Leaving the circuit closed while adjusting $x$ .

Things to Be Careful About

If readings were taken too slowly, later readings may be affected by accumulated heating.

Techniques used

open the switch to stop current flowminimise heating effects between readings

(e)

Change $x$ and repeat (c)(ii) and (d) until you have six sets of readings of $x$ and $I$ .
Include values of $\frac{x^2}{(x + L)}$ and $\frac{1}{I}$ in your table.

10M

DifficultyMedium

Worked solution

Answer

Take six different values of $x$ (wide range) and record $I$ each time. Include calculated columns for $\dfrac{x^2}{(x+L)}$ and $\dfrac{1}{I}$ .

Example table (with $L = 50.0\ \text{cm}$ ):

$x / \text{cm}$	$I / \text{A}$	$\dfrac{x^2}{(x+L)} / \text{cm}$	$\dfrac{1}{I} / \text{A}^{-1}$
$30.0$	$0.46$	$11.3$	$2.17$
$40.0$	$0.51$	$17.8$	$1.96$
$50.0$	$0.57$	$25.0$	$1.75$
$60.0$	$0.66$	$32.7$	$1.52$
$70.0$	$0.78$	$40.8$	$1.28$
$80.0$	$0.98$	$49.2$	$1.02$

Final answer

See table (example values shown)

Detailed explanation

Background Concept

Good experimental data needs:

enough readings (here six sets) to establish a trend;
a wide range of the independent variable (here $x$ );
clear presentation in a single table with headings and units;
calculated quantities written with sensible significant figures.

The question asks you to calculate two derived quantities:

$\dfrac{x^2}{(x+L)}$ (this has units of length, since it is length $^2$ /length),
$\dfrac{1}{I}$ (units $\text{A}^{-1}$ ).

Understanding the Question

You must vary $x$ , and for each $x$ measure the current $I$ . You then compute and record the two extra columns. The aim is to prepare data suitable for plotting a straight-line graph later.

Approach

Choose six $x$ values spanning as much of the metre-rule wire as practical.
For each $x$ : close switch briefly, record $I$ , open switch.
In the same table, calculate $\dfrac{x^2}{(x+L)}$ using the measured $x$ and the measured constant $L$ .
Calculate $\dfrac{1}{I}$ .

Step-by-Step Reasoning

Decide a sequence like $x = 30, 40, 50, 60, 70, 80\ \text{cm}$ (any sensible spread is fine).
For each row:
- measure $x$ from the metre rule;
- measure $I$ on the ammeter;
- compute

\frac{x^2}{(x+L)} = \frac{x \times x}{(x+L)}

compute

\frac{1}{I}

Presentation rules that typically earn marks:
- one table only (not separate tables);
- headings include quantity and unit (e.g. $x/\text{cm}$ , not just “x”);
- consistent dp in each column: e.g. all $x$ to $0.1\ \text{cm}$ , all $I$ to $0.01\ \text{A}$ .

Key Takeaways

Collect a range of data, not clustered values.
Derived columns must be calculated from your measurements and presented clearly.

Common Mistakes

Fewer than six sets of readings.
Missing units or unclear headings.
Inconsistent rounding (e.g. some $x$ values to 1 dp and others to 0 dp).
Forgetting to open the switch between readings, causing heating drift.

Things to Be Careful About

Keep $x$ and $L$ in the same units when calculating $\dfrac{x^2}{(x+L)}$ .
If $I$ is small, $1/I$ becomes large; avoid rounding $I$ too coarsely or $1/I$ becomes inaccurate.
If readings fluctuate, repeat and take a mean (and note it).

Techniques used

take repeated measurements over a suitable range of the independent variablerecord results in a single table with headings and unitscalculate derived quantities from measured valuesmaintain consistent significant figures within each column

(f)

(i)

Plot a graph of $\frac{1}{I}$ on the $y$ -axis against $\frac{x^2}{(x + L)}$ on the $x$ -axis.

DifficultyMedium

Worked solution

Answer

Plot $\dfrac{1}{I}$ on the $y$ -axis (unit $\text{A}^{-1}$ ) against $\dfrac{x^2}{(x+L)}$ on the $x$ -axis (unit of length, e.g. $\text{cm}$ if $x$ and $L$ are in $\text{cm}$ ).

Use a suitable scale (at least half the grid in each direction) and plot all six points.

Final answer

Graph of 1/I (y) against x^2/(x+L) (x) plotted

Detailed explanation

Background Concept

Graphing experimental data is a way to test for relationships and determine constants. Marks are typically awarded for:

correct axes (dependent variable on $y$ -axis, independent on $x$ -axis);
correct labels with units;
sensible scale (not cramped; not awkward like 3 squares = 1 unit);
accurate plotting.

Understanding the Question

You must produce a graph with:

$y = \dfrac{1}{I}$ ,
$x = \dfrac{x^2}{(x+L)}$ ,
using the table from part (e).

Approach

From each row of the table, take the pair $(x\text{-value}, y\text{-value}) = \left(\dfrac{x^2}{(x+L)}, \dfrac{1}{I}\right)$ and plot it. Ensure axes are labelled with both symbols and units.

Step-by-Step Reasoning

Draw axes and choose a scale that spreads the data across the paper.
Label axes clearly:
- vertical: $\dfrac{1}{I} / \text{A}^{-1}$ ,
- horizontal: $\dfrac{x^2}{(x+L)} / \text{cm}$ (or $\text{m}$ if you used metres).
Plot each point carefully using small crosses.
Check for obvious plotting errors by verifying the trend is roughly linear.

Key Takeaways

Axes must include units.
Scale choice matters for accuracy of gradient/intercept.

Common Mistakes

Plotting $I$ instead of $1/I$ .
Forgetting units on axes.
Using a scale that only occupies a small part of the grid.

Things to Be Careful About

If you used $x$ and $L$ in cm in the table, keep that for the graph x-axis (do not silently switch to metres without converting all values).

Techniques used

choose appropriate axes and scales to use most of the graph gridlabel axes with quantity and unitplot points accurately from a results table

(ii)

Draw the straight line of best fit.

DifficultyMedium-Easy

Worked solution

Answer

Draw a single straight line of best fit through the plotted points with roughly equal scatter on either side of the line.

Final answer

Straight line of best fit drawn

Detailed explanation

Background Concept

A best-fit line represents the overall trend of data with random uncertainties. It should not be drawn by joining dots. For approximately linear data, a straight line is used.

Understanding the Question

After plotting the points, you must draw the best straight line that represents the relationship between $\dfrac{1}{I}$ and $\dfrac{x^2}{(x+L)}$ .

Approach

Use a ruler to draw a straight line that leaves roughly the same number of points above as below (balanced scatter). The line should pass close to as many points as possible.

Step-by-Step Reasoning

Visually judge the trend.
Place the ruler so the line runs centrally through the cluster.
Draw a thin, continuous straight line across the full data range.

Key Takeaways

Best-fit line is about the overall trend, not perfect agreement with every point.

Common Mistakes

Joining points with segments.
Forcing the line through the origin without evidence.

Things to Be Careful About

Do not let one outlier dominate the line position unless you have reason to reject it.

Techniques used

draw a straight line of best fit with balanced scatterignore anomalous points appropriately when justified

(iii)

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

gradient = ______
$y$ -intercept = ______

DifficultyMedium-Easy

Worked solution

Working

Use two well-separated points on the best-fit line.

Example:

\text{gradient} = \frac{\Delta (1/I)}{\Delta \left(x^2/(x+L)\right)} = \frac{1.02 - 2.17}{49.2 - 11.3} = -3.0 \times 10^{-2}\ \text{A}^{-1}\text{cm}^{-1}

Read off the intercept at $x=0$ :

y\text{-intercept} = 2.50\ \text{A}^{-1}

Answer

gradient $= -3.0 \times 10^{-2}\ \text{A}^{-1}\text{cm}^{-1}$

$y$ -intercept $= 2.50\ \text{A}^{-1}$

Final answer

gradient = −3.0×10^−2 A^−1 cm^−1, y-intercept = 2.50 A^−1 (example)

Detailed explanation

Background Concept

For a straight-line graph, the gradient $m$ and intercept $c$ come from

y = mx + c

Gradient is calculated from two points on the best-fit line (not necessarily measured points):

m = \frac{\Delta y}{\Delta x}

The $y$ -intercept is the value of $y$ when $x=0$ (where the line crosses the $y$ -axis).

Understanding the Question

You have plotted $y = \dfrac{1}{I}$ against $x = \dfrac{x^2}{(x+L)}$ . You now need the numerical gradient and the $y$ -intercept of your best-fit line, including units.

Approach

Pick two points far apart on the drawn best-fit line to reduce percentage uncertainty.
Compute $\Delta y$ and $\Delta x$ , then take $\Delta y/\Delta x$ .
Read the intercept directly where the line crosses the $y$ -axis.

Step-by-Step Reasoning

Choose two points on the line, preferably near the ends of the plotted range.
Read their coordinates carefully using the graph scale.
Calculate:

\text{gradient} = \frac{y_2-y_1}{x_2-x_1}

Determine the $y$ -intercept by extending the line to meet the $y$ -axis and reading the value.
Units:

$y$ has units $\text{A}^{-1}$ ,
$x$ has units of length (e.g. $\text{cm}$ ),
so gradient has units $\text{A}^{-1}\text{cm}^{-1}$ (or $\text{A}^{-1}\text{m}^{-1}$ if $x$ -axis is in metres).

Key Takeaways

Use a large triangle for gradient.
Always include units for gradient and intercept.

Common Mistakes

Using two adjacent points (large uncertainty in gradient).
Calculating $\Delta x/\Delta y$ instead of $\Delta y/\Delta x$ .
Using coordinates of plotted points instead of points on the best-fit line.

Things to Be Careful About

Ensure you read from the line, not from the scatter.
Be consistent: if the x-axis is in cm, the gradient unit must include $\text{cm}^{-1}$ .

Techniques used

determine gradient using a large triangle on the best-fit lineread the y-intercept from the graphquote gradient and intercept with appropriate units

(g)

The quantities $I$ , $x$ and $L$ are related by the equation

\frac{1}{I} = -\frac{Px^2}{(x + L)} + Q

where $P$ and $Q$ are constants.

Using your answers in (f)(iii), determine values for $P$ and $Q$ .
Give appropriate units.

$P$ = ______
$Q$ = ______

DifficultyMedium-Easy

Worked solution

Working

Given

\frac{1}{I} = -\frac{Px^2}{(x + L)} + Q

Comparing with $y = mx + c$ for the graph of $y=\dfrac{1}{I}$ against $x=\dfrac{x^2}{(x+L)}$ :

m = -P,\qquad c = Q

P = -\text{gradient} = -\left(-3.0 \times 10^{-2}\ \text{A}^{-1}\text{cm}^{-1}\right) = 3.0 \times 10^{-2}\ \text{A}^{-1}\text{cm}^{-1}

Q = y\text{-intercept} = 2.50\ \text{A}^{-1}

Answer

$P = 3.0 \times 10^{-2}\ \text{A}^{-1}\text{cm}^{-1}$

$Q = 2.50\ \text{A}^{-1}$

Final answer

P = 3.0×10^−2 A^−1 cm^−1, Q = 2.50 A^−1 (example)

Detailed explanation

Background Concept

If you plot a graph of $y$ against $x$ and the relationship is linear, it can be written as

y = mx + c

where:

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

Here, the equation is already in a linear form when you choose

y = \frac{1}{I},\qquad x = \frac{x^2}{(x+L)}.

Understanding the Question

You are given:

\frac{1}{I} = -\frac{Px^2}{(x + L)} + Q

and you have found the gradient and intercept from part (f)(iii). You must use these to determine $P$ and $Q$ and include appropriate units.

Approach

Compare the experimental straight-line form to $y = mx + c$ .

The coefficient of the plotted $x$ variable is the gradient.
The constant term is the intercept.
Then use the sign carefully: gradient corresponds to $-P$ .

Step-by-Step Reasoning

Write the given relation in terms of the plotted variables:

\underbrace{\frac{1}{I}}_{y} = -P\underbrace{\left(\frac{x^2}{x+L}\right)}_{x} + Q

gradient $m = -P \Rightarrow P = -m$ ,
intercept $c = Q$ .

Units:

$y = 1/I$ has units $\text{A}^{-1}$ ,
$x = x^2/(x+L)$ has units of length (e.g. cm),
so

[P] = \frac{\text{A}^{-1}}{\text{cm}} = \text{A}^{-1}\text{cm}^{-1},\qquad [Q] = \text{A}^{-1}.

Key Takeaways

Matching to $y=mx+c$ is the quickest way to extract constants.
Units come from the axes of the graph.

Common Mistakes

Taking $P$ equal to the gradient instead of the negative of it.
Giving $Q$ the wrong unit (it must match $1/I$ ).

Things to Be Careful About

If you plotted $x^2/(x+L)$ in metres instead of cm, the numerical value of $P$ changes by a factor of $100$ and the unit must be $\text{A}^{-1}\text{m}^{-1}$ .

Techniques used

match a linearised equation to y = mx + cidentify constants from gradient and interceptassign units to constants from the graph units

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Q2Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation · Presentation of Data and Observations20M

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