Physics 9702/35 — May/June 2012

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

You may not need to use all of the materials provided.

In this experiment, you will investigate how the current in a circuit depends on the resistance of the circuit.

(a)

Set up the circuit as shown in Fig. 1.1. The crocodile clip should be positioned so that three of the resistors from the chain are included in the circuit.

Fig. 1.1

All the resistors have the same value of resistance $R$ .

DifficultyMedium-Easy

Worked solution

Answer

Set up the circuit as in Fig. 1.1 with the cell, switch and ammeter in series, and position the crocodile clip so that three resistors from the chain are included (so $n=3$ ).

Final answer

Circuit set up with three resistors included (n = 3).

Detailed explanation

Background Concept

In a simple d.c. circuit, the current depends on the total resistance in the circuit. To investigate this experimentally, we change the resistance in a controlled way and measure the resulting current with an ammeter.

An ammeter measures current and must be connected in series so that the same current flows through it as through the components.

A crocodile clip can be used as a movable connection point, allowing different numbers of identical resistors to be included in the circuit.

Understanding the Question

You are given a chain of identical resistors (each has resistance $R$ ). The circuit diagram shows that a crocodile clip picks a connection point on the chain. You must set the clip so that three resistors from the chain are included in the circuit to begin with.

The key success criteria are:

ammeter in series,
correct selection of three resistors using the crocodile clip,
secure connections so readings are stable.

Approach

Build the series circuit (cell–switch–ammeter–resistor chain).
Use the crocodile clip to choose the junction such that the current path includes exactly three resistors.
Check that the switch can safely open/close the circuit before taking readings.

diagram-1

Step-by-Step Reasoning

Connect the cell, switch, and ammeter in series.
Connect one end of the resistor chain into the circuit.
Attach the crocodile clip to the junction after the third resistor (counting along the chain from the connected end) so that exactly three resistors lie in the conducting path.
Ensure the crocodile clip grips metal firmly (poor contact increases resistance unpredictably).
Confirm the ammeter reads approximately zero when the switch is open.

Key Takeaways

Ammeter is always connected in series.
The crocodile clip changes how many identical resistors are included, changing the circuit resistance.
Good electrical contact is essential for reliable data.

Common Mistakes

Connecting the ammeter in parallel (can damage the meter and gives wrong readings).
Miscounting resistors so the wrong $n$ is used.
Loose crocodile clip contact causing fluctuating current readings.

Things to Be Careful About

Count included resistors carefully: the conducting path must pass through exactly three resistors.
Avoid short circuits: ensure the clip does not bypass resistors unintentionally.
Check ammeter range is appropriate before closing the switch.

Techniques used

assemble a series circuit including a cell, switch and ammeteruse a crocodile clip to select a required number of resistorscheck correct polarity and secure electrical connections

(b)

(i)

Close the switch.

DifficultyEasy

Worked solution

Answer

Close the switch.

Final answer

Switch closed.

Detailed explanation

Background Concept

Closing the switch completes the circuit so that a potential difference from the cell is applied across the external resistance, producing a current.

Understanding the Question

You are instructed to close the switch so that current flows and the ammeter can give a reading.

Approach

Simply close the switch and allow the reading to settle before recording it in the next part.

Step-by-Step Reasoning

Close the switch to complete the circuit.
Watch the ammeter needle/digital display.
Wait briefly for the reading to become steady (a fluctuating reading often indicates poor contact).

Key Takeaways

Switch closed → circuit complete → current flows.

Common Mistakes

Reading the ammeter before the switch is closed.
Closing the switch only momentarily so the reading cannot stabilise.

Things to Be Careful About

If the current is unexpectedly large, open the switch and check connections and meter range.

Techniques used

operate the switch to complete the circuit safelyobserve the ammeter for a steady reading

(ii)

Record the ammeter reading $I$ and the number $n$ of resistors from the chain included in the circuit.

$I$ = ______
$n$ = ______

DifficultyMedium-Easy

Worked solution

Answer

$I = 0.22\ \text{A}$ (example reading)

$n = 3$

Final answer

I = 0.22 A, n = 3 (example).

Detailed explanation

Background Concept

Current $I$ is measured using an ammeter in series. The number of resistors included, $n$ , is a dimensionless count of identical resistors in the conducting path.

Understanding the Question

With the crocodile clip set so that three resistors are in circuit, you must record:

the ammeter reading $I$ (include unit A),
the number of included resistors $n$ (should be $3$ here).

Approach

Read the ammeter once it is steady.
Record $n$ by counting the resistors actually in the current path (as selected by the crocodile clip position).

Step-by-Step Reasoning

With the switch closed, observe the ammeter reading.
Record $I$ to the meter’s resolution (e.g. to 0.01 A on a digital meter if that is what it displays).
Count the number of resistors between the fixed end connection and the crocodile-clip junction; record this as $n$ .
Enter both values clearly.

Key Takeaways

$I$ needs a unit; $n$ does not.
A correct $n$ value is essential because it is the independent variable later.

Common Mistakes

Omitting the unit for $I$ .
Recording the wrong $n$ because of miscounting or misunderstanding which resistors are included.
Recording an unstable reading caused by a loose crocodile clip.

Things to Be Careful About

Make sure the ammeter is on the correct range.
If the reading drifts, improve contact and re-check connections before recording.

Techniques used

read the ammeter to appropriate resolution and record with unitcount the number of resistors included in the circuitrecord raw data clearly

(iii)

Open the switch.

DifficultyEasy

Worked solution

Answer

Open the switch.

Final answer

Switch opened.

Detailed explanation

Background Concept

Opening the switch breaks the circuit so current stops. This reduces heating of resistors and avoids draining the cell between readings.

Understanding the Question

You are instructed to open the switch after taking the measurement so the circuit is not left conducting continuously.

Approach

Open the switch immediately after recording the reading.

Step-by-Step Reasoning

After recording $I$ and $n$ , open the switch.
Confirm the ammeter returns to (approximately) zero.

Key Takeaways

Open switch between readings to keep conditions more consistent.

Common Mistakes

Leaving the switch closed so the resistors warm up and resistance changes slightly.

Things to Be Careful About

If the ammeter does not return near zero when open, check for an unintended alternative current path.

Techniques used

break the circuit safely using the switchprevent heating of resistors between readings

(c)

By attaching the crocodile clip to different junctions and terminals on the chain of resistors, repeat (b) until you have six sets of readings of $I$ and $n$ .

Include values of $\frac{(n + 1)}{I}$ in your table.

10M

DifficultyMedium-Hard

Worked solution

Answer

Six sets of readings of $I$ and $n$ recorded, with calculated values of $\dfrac{(n+1)}{I}$ .

Example table:

$n$	$I / \text{A}$	$\dfrac{(n+1)}{I} / \text{A}^{-1}$
1	0.273	7.33
2	0.237	12.7
3	0.222	18.0
4	0.214	23.4
5	0.209	28.7
6	0.206	34.0

Final answer

Six readings of I and n with calculated (n+1)/I in a single table (student-dependent).

Detailed explanation

Background Concept

To investigate how current depends on resistance, you must vary the circuit resistance systematically and measure the resulting current. Good experimental data requires:

an adequate range of the independent variable,
sufficient number of data points,
consistent measurement technique,
clear presentation with units.

A derived quantity like $\dfrac{(n+1)}{I}$ must be calculated from measured values and recorded with correct units. Since $n$ is dimensionless and $I$ is in amperes, $\dfrac{(n+1)}{I}$ has unit $\text{A}^{-1}$ .

Understanding the Question

You must move the crocodile clip to select different numbers of resistors $n$ , take six readings of current $I$ , and tabulate these. You are explicitly told to include a column for $\dfrac{(n+1)}{I}$ .

So the table must include at least:

$n$ ,
$I$ with unit A,
$\dfrac{(n+1)}{I}$ with unit $\text{A}^{-1}$ ,
and it must contain six rows of data covering different $n$ values.

Approach

Choose six different values of $n$ (e.g. $n=1$ to $6$ if available).
For each $n$ $n$ :
- close switch, wait for steady ammeter reading, record $I$ ,
- open switch,
- compute $\dfrac{(n+1)}{I}$ .
Keep presentation consistent (same decimal places where appropriate).

Step-by-Step Reasoning

Start from an initial setting (e.g. $n=3$ as in part b).
Move the crocodile clip to a different junction to change $n$ .
Each time:
- close the switch, read $I$ steadily,
- open the switch to minimise heating,
- calculate

\frac{(n+1)}{I}

using your recorded $n$ and $I$ .

Record all results in a single clear table.
Use consistent significant figures: typically $I$ to the meter resolution, and the calculated column to 2–3 s.f. (or matching the uncertainty implied by $I$ ).

Key Takeaways

Use a sensible range of $n$ values and at least six data points.
Always include units in table headings for measured and derived quantities.
Calculated quantities should be recorded to sensible significant figures.

Common Mistakes

Not taking six different $n$ values (repeating the same $n$ ).
Forgetting to include the derived column $\dfrac{(n+1)}{I}$ .
Writing units in the body of the table rather than in the heading.
Inconsistent significant figures/decimal places within a column.

Things to Be Careful About

Ensure $n$ is correct each time you move the clip.
Avoid leaving the switch closed for long periods (temperature rise changes resistance).
Watch for poor contact at the crocodile clip; it can add extra resistance and spoil the trend.

Techniques used

vary the independent variable by repositioning the crocodile cliprecord multiple sets of readings over an appropriate rangecalculate a derived quantity from each pair of raw readingspresent results in a single table with correct headings and units

(d)

(i)

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

DifficultyMedium

Worked solution

Answer

Plot $y = \dfrac{(n+1)}{I}$ (in $\text{A}^{-1}$ ) on the $y$ -axis against $n$ on the $x$ -axis, using suitable scales and plotting all six points.

Final answer

Graph of (n+1)/I (A^-1) against n plotted.

Detailed explanation

Background Concept

A graph is used to reveal relationships between variables. To gain marks in practical graphing:

axes must be labelled with the correct quantities and units,
scales should be simple and use a large fraction of the available grid,
points must be plotted accurately.

Here, $n$ is dimensionless and $\dfrac{(n+1)}{I}$ has unit $\text{A}^{-1}$ .

Understanding the Question

You are told explicitly what to plot:

$y$ -axis: $\dfrac{(n+1)}{I}$
$x$ -axis: $n$

So you must transfer your table values to a graph with correct axis labels and scales.

Approach

Draw axes and label them:
- horizontal: $n$
- vertical: $\dfrac{(n+1)}{I} / \text{A}^{-1}$
Choose scales that spread the points out (at least half the grid in each direction).
Plot each point carefully.

diagram-1

Step-by-Step Reasoning

Put $n$ on the horizontal axis because it is the independent variable you set by moving the crocodile clip.
Put $\dfrac{(n+1)}{I}$ on the vertical axis because it is calculated from measurements.
Use a simple scale (e.g. 1 or 2 units per large square) and avoid awkward scales (like 3 per square).
Plot all six points using small, neat crosses.

Key Takeaways

Correct axes and units are essential.
A good scale and accurate plotting are what earn the practical marks.

Common Mistakes

Plotting $I$ against $n$ instead of $\dfrac{(n+1)}{I}$ against $n$ .
Missing units on the $y$ -axis label.
Using a tiny graph area so points are bunched up.

Things to Be Careful About

Do not join point-to-point; you will draw a best-fit line in the next part.
Ensure the axis label includes the unit format, e.g. $\dfrac{(n+1)}{I} / \text{A}^{-1}$ .

Techniques used

choose appropriate axes variables from the instructionlabel axes with quantity and unitselect a suitable scale using at least half the graph gridplot data points accurately

(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 (balanced with points roughly evenly scattered about the line).

Final answer

Straight line of best fit drawn.

Detailed explanation

Background Concept

A best-fit line represents the overall linear trend in experimental data when there is scatter. It should not be drawn as a join-the-dots curve unless the relationship is clearly non-linear.

Understanding the Question

You have already plotted $\dfrac{(n+1)}{I}$ against $n$ . You must now draw the straight line that best represents the trend.

Approach

Use a ruler to draw one straight line so that the points are balanced: similar numbers of points above and below the line, and the line runs through the middle of the scatter.

Step-by-Step Reasoning

Place the ruler so that the line passes centrally through the cluster of points.
Adjust by eye to balance the residuals (vertical distances) of the points from the line.
Draw the line across most of the graph area (not just between the first and last points).

Key Takeaways

Best-fit line is about the trend, not passing through every point.

Common Mistakes

Joining points one-by-one.
Forcing the line through the origin when not justified.
Drawing a line only through two points (ignoring the others).

Things to Be Careful About

Use a sharp pencil and a ruler.
Extend the line across the full range used for gradient/intercept measurements.

Techniques used

draw a single straight line representing the overall trendbalance the distribution of points about the line

(iii)

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

gradient = ______
$y$ -intercept = ______

DifficultyMedium

Worked solution

Working

Using two points on the best-fit line, e.g. $(n, y) = (1, 7.3)$ and $(6, 34.0)$ :

\text{gradient} = \frac{\Delta y}{\Delta x} = \frac{34.0 - 7.3}{6 - 1} = 5.34\ \text{A}^{-1}

$y$ -intercept from the line at $n=0$ :

\text{intercept} \approx 2.0\ \text{A}^{-1}

Answer

gradient $= 5.34\ \text{A}^{-1}$

$y$ -intercept $= 2.0\ \text{A}^{-1}$

Final answer

gradient = 5.34 A^-1, y-intercept = 2.0 A^-1 (example).

Detailed explanation

Background Concept

For a straight-line graph, the gradient (slope) is

\text{gradient} = \frac{\Delta y}{\Delta x}

and the $y$ -intercept is the value of $y$ when $x=0$ .

Here:

$x = n$ (dimensionless)
$y = \dfrac{(n+1)}{I}$ (unit $\text{A}^{-1}$ )

So the gradient has unit $\text{A}^{-1}$ per unit $n$ , i.e. $\text{A}^{-1}$ .

Understanding the Question

You must extract two quantities from your best-fit line:

the gradient,
the $y$ -intercept.

These will be used later to find constants $P$ and $Q$ .

Approach

Choose two well-separated points on the best-fit line (not necessarily plotted points).
Read their coordinates accurately.
Compute $\Delta y$ and $\Delta x$ , then gradient $= \Delta y / \Delta x$ .
Read the intercept where the line crosses the $y$ -axis (at $n=0$ ).

diagram-1

Step-by-Step Reasoning

Pick two points far apart to reduce percentage reading error (a small triangle gives a large fractional uncertainty).
Suppose points on the line are approximately $(1, 7.3)$ and $(6, 34.0)$ .
Then:

\Delta y = 34.0 - 7.3 = 26.7\ \text{A}^{-1}

\Delta x = 6 - 1 = 5

So:

\text{gradient} = \frac{26.7}{5} = 5.34\ \text{A}^{-1}

For the $y$ -intercept, extend the best-fit line to $n=0$ and read the value of $y$ there (e.g. $2.0\ \text{A}^{-1}$ ).

Key Takeaways

Use a large triangle for gradient.
Use the best-fit line, not individual noisy points.
Include units with gradient and intercept.

Common Mistakes

Calculating gradient as $\Delta x / \Delta y$ .
Using two adjacent points (small triangle).
Reading the intercept from the first plotted point rather than from $n=0$ .

Things to Be Careful About

Ensure your chosen points lie on the line you drew.
Read values to the precision allowed by the graph scale.
Do not round too early; round at the end to sensible s.f.

Techniques used

use two widely spaced points on the best-fit linecalculate gradient as \u0394y/\u0394x with correct unitsdetermine the y-intercept from the line at n = 0

(e)

It is suggested that the relationship between $I$ and $n$ is

\frac{(n + 1)}{I} = Pn + Q

where $P$ and $Q$ are constants.

Use your answers in (d)(iii) to determine values for $P$ and $Q$ .

$P$ = ______
$Q$ = ______

DifficultyMedium-Easy

Worked solution

Working

Given

\frac{(n + 1)}{I} = Pn + Q

Comparing with $y = mx + c$ for a graph of $y = \dfrac{(n+1)}{I}$ against $x=n$ :

P = \text{gradient},\qquad Q = y\text{-intercept}

Answer

$P = 5.34\ \text{A}^{-1}$

$Q = 2.0\ \text{A}^{-1}$

Final answer

P = gradient, Q = y-intercept (e.g. P = 5.34 A^-1, Q = 2.0 A^-1).

Detailed explanation

Background Concept

If a relationship can be written in the straight-line form

y = mx + c

then a graph of $y$ against $x$ is a straight line with:

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

Understanding the Question

You are given:

\frac{(n + 1)}{I} = Pn + Q

and you have plotted $\dfrac{(n+1)}{I}$ on the $y$ -axis against $n$ on the $x$ -axis. You must use the gradient and intercept from part (d)(iii) to identify $P$ and $Q$ .

Approach

Match the given equation directly to $y = mx + c$ by identifying:

$y \equiv \dfrac{(n+1)}{I}$ ,
$x \equiv n$ ,
$m \equiv P$ ,
$c \equiv Q$ .

Step-by-Step Reasoning

Rewrite the comparison clearly:

Your graph is $y$ vs $x$ where

y = \frac{(n+1)}{I},\quad x = n

The equation is

y = P x + Q

So the gradient of the graph equals $P$ and the intercept equals $Q$ .

Use your numerical values from (d)(iii) directly:

$P =$ (your gradient)
$Q =$ (your intercept)

Key Takeaways

Choosing the correct axes makes the constants drop straight out as gradient and intercept.

Common Mistakes

Swapping $P$ and $Q$ .
Using a graph of $I$ vs $n$ and then trying to read off $P$ and $Q$ .

Things to Be Careful About

Ensure you are using the gradient/intercept of the best-fit line, not from two raw points.
Keep the units consistent: both $P$ and $Q$ have unit $\text{A}^{-1}$ here because $n$ is dimensionless.

Techniques used

compare the plotted straight-line form with y = mx + cidentify the gradient as the coefficient of xidentify the y-intercept as the constant term

(f)

Disconnect the circuit.
Connect the voltmeter across the cell.
Measure and record the voltage $V$ across the cell.

$V$ = ______ $\text{V}$

DifficultyMedium-Easy

Worked solution

Answer

$V = 1.50\ \text{V}$ (example reading)

Final answer

V = 1.50 V (example).

Detailed explanation

Background Concept

A voltmeter measures potential difference and must be connected in parallel with the component whose voltage you want to measure. A cell’s terminal p.d. can differ from its emf if current is flowing; measuring it with the circuit disconnected reduces this issue.

Understanding the Question

You must:

disconnect the circuit,
connect the voltmeter across the cell terminals,
measure and record $V$ in volts.

Approach

Open the circuit so no current flows, then place the voltmeter leads directly across the cell terminals and read $V$ .

Step-by-Step Reasoning

Ensure the switch is open and/or remove a lead so the circuit is disconnected.
Connect the voltmeter across the cell: one lead to each terminal.
Select an appropriate voltmeter range.
Record the voltage reading with unit V and to the meter resolution.

Key Takeaways

Voltmeter is always connected in parallel.
Recording with unit and sensible precision is essential in practical work.

Common Mistakes

Connecting the voltmeter in series.
Measuring voltage while current is flowing (terminal p.d. may be lower than emf).
Omitting the unit V.

Things to Be Careful About

Ensure firm contact at the cell terminals.
If using an analogue meter, avoid parallax error and read from the correct scale.

Techniques used

connect a voltmeter in parallel across a componentread and record potential difference with correct unitdisconnect circuit safely before reconfiguring connections

(g)

The constant $P$ is related to $R$ and $V$ by

P = \frac{2R}{V}

Using your answers in (e) and (f), calculate a value for $R$ .

$R$ = ______

DifficultyMedium-Easy

Worked solution

Working

Given

P = \frac{2R}{V}

Rearrange:

R = \frac{PV}{2}

Substitute $P = 5.34\ \text{A}^{-1}$ and $V = 1.50\ \text{V}$ :

R = \frac{(5.34)(1.50)}{2} = 4.01\ \Omega

Answer

$R = 4.0\ \Omega$

Final answer

R = 4.0 Ω (example).

Detailed explanation

Background Concept

When a constant is related to measured quantities by an equation, you can determine the unknown by rearranging algebraically and substituting your experimental values.

Here the given relationship is:

P = \frac{2R}{V}

where $P$ and $V$ are known from earlier parts, so $R$ can be found.

Understanding the Question

You have obtained $P$ from the graph and $V$ from a voltmeter reading. You must calculate the resistance $R$ using the provided formula.

Approach

Rearrange to make $R$ the subject.
Substitute your values of $P$ and $V$ .
Quote $R$ with unit $\Omega$ and sensible significant figures.

Step-by-Step Reasoning

Start with:

P = \frac{2R}{V}

Multiply both sides by $V$ :

PV = 2R

Divide by 2:

R = \frac{PV}{2}

Now substitute your measured/derived values (example):

$P = 5.34\ \text{A}^{-1}$
$V = 1.50\ \text{V}$

R = \frac{(5.34)(1.50)}{2} = 4.01\ \Omega

So $R \approx 4.0\ \Omega$ to 2–3 s.f.

Key Takeaways

Rearranging equations cleanly avoids substitution mistakes.
Always include the final unit ( $\Omega$ for resistance).

Common Mistakes

Rearranging incorrectly (e.g. using $R = \dfrac{2P}{V}$ ).
Forgetting to divide by 2.
Omitting the unit or rounding too aggressively.

Things to Be Careful About

Use your own experimental values for $P$ and $V$ (not the example values).
Keep enough significant figures in intermediate steps; round at the end.

Techniques used

rearrange a formula to make the required variable the subjectsubstitute experimental values consistently with unitsquote a final value to appropriate significant figures

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Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation20M

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