In this experiment, you will investigate the potential difference between two points in a circuit. (a) Assemble the circuit of Fig. 1.1. (i) Connect the voltmeter across the power supply. Record the voltmeter reading $E$. $E$ = \_\_\_\_\_\_ (ii) Disconnect the voltmeter from the power supply. (i) Position the voltmeter leads on the wires at distance $x$ from the zero ends of both rules as shown in Fig. 1.2, where $x$ is approximately 20 cm. (ii) Record $x$ and record the…

(a) Circuit assembled (see diagram/description). · (b)(i) E = 2.00 V · (b)(ii) Voltmeter disconnected from the power supply. · (c)(i) Voltmeter contacts positioned at x ≈ 20 cm from each rule’s zero end. · (c)(ii) x = 20.0 cm, V = +0.40 V · (c)(iii) x = 50.0 cm (at V = 0) · (d) Six sets of x and V recorded with a V/E column (see table). · (e)(i) Graph of V/E (y) against x (x) plotted with correct labels and scales. · (e)(ii) Straight line of best fit drawn. · (e)(iii)…

Physics 9702/36 — October/November 2013

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 the potential difference between two points in a circuit.

(a)

Assemble the circuit of Fig. 1.1.

Fig. 1.1

DifficultyMedium-Easy

Worked solution

diagram-1

Answer

Circuit assembled as in Fig. 1.1 with secure connections: d.c. supply connected to the resistor and the two parallel wires (on metre rules) using crocodile clips, and with the voltmeter available for connection in parallel across the required two points.

Final answer

Circuit assembled (see diagram/description).

Detailed explanation

Background Concept

In circuit practical work, the key ideas are:

Potential difference (p.d.) is measured between two points and must be measured with a voltmeter connected in parallel with those two points.
Good experimental results require correct circuit assembly, secure electrical contact, and correct placement of measuring instruments.

Understanding the Question

You are given a circuit diagram (Fig. 1.1) showing a d.c. supply, a resistor, and two parallel wires mounted along metre rules. Crocodile clips make electrical contact to the wires. You must build exactly this arrangement so that later parts (measuring $E$ , then measuring $V$ at different positions $x$ ) are possible.

Approach

Identify the main series loop (supply, resistor, wire path).
Identify where the voltmeter will be connected (always across two points, not in series).
Assemble with secure connections and ensure the metre rules are oriented so the zero ends are opposite as shown.

Step-by-Step Reasoning

diagram-1

Connect the d.c. supply terminals to the circuit as shown.
Ensure the resistor is in the correct part of the circuit (as in the diagram).
Connect each wire (mounted on a metre rule) into the circuit using crocodile clips so that the current passes through the intended wire sections.
Check the metre rule orientation: the top wire’s zero end and the bottom wire’s zero end are at opposite ends (this matters because later $x$ is measured from each wire’s own zero end).
Keep the voltmeter ready to be placed across either the power supply or across two points on the wires in later steps.

Key Takeaways

Voltmeters measure p.d. and must be connected in parallel.
Correct physical layout (including which end is the “zero” on each rule) is essential for meaningful position measurements.

Common Mistakes

Connecting the voltmeter in series (gives incorrect readings and can affect the circuit).
Using loose crocodile clips (intermittent contact causes fluctuating readings).
Mixing up the zero ends of the metre rules (makes $x$ inconsistent).

Things to Be Careful About

Confirm the voltmeter is on the correct d.c. voltage range before connecting.
Avoid short circuits when moving clips.
Ensure the wires are not touching each other electrically unless intended.

Techniques used

assemble the circuit following a circuit diagramcheck correct polarity and secure connectionsuse crocodile clips to make contact at chosen points

(b)

(i)

Connect the voltmeter across the power supply.
Record the voltmeter reading $E$ .

$E$ = ______

DifficultyEasy

Worked solution

Answer

Voltmeter connected across the power supply.

Example reading (to appropriate resolution):

E = 2.00\ \text{V}

Final answer

E = 2.00 V

Detailed explanation

Background Concept

The electromotive force (e.m.f.)/supply potential difference $E$ is measured using a voltmeter connected across the two terminals of the supply.

A voltmeter has a high resistance and is designed to be connected in parallel.
The reading should be recorded to the instrument’s resolution (e.g. 0.01 V for a digital meter on a 2 V/20 V range, or to the smallest division for an analogue meter).

Understanding the Question

You are asked to connect the voltmeter across the power supply and record the reading $E$ . This value will later be used to compute the ratio $V/E$ .

Approach

Put the voltmeter on a suitable d.c. voltage range.
Connect it directly across the supply terminals (parallel connection).
Wait for a steady reading and record $E$ with unit and appropriate decimal places.

Step-by-Step Reasoning

Ensure the voltmeter is set to measure d.c. voltage.
Choose a range that prevents over-ranging but gives good resolution.
Connect the red lead to the positive terminal and the black lead to the negative terminal.
Read the value and record it as $E$ in volts.

Key Takeaways

Voltmeters are connected in parallel.
Record values with sensible precision and include units.

Common Mistakes

Connecting the voltmeter in series.
Using the wrong meter setting (a.c. instead of d.c., or wrong range).
Omitting the unit or recording too many/few decimal places.

Things to Be Careful About

If the reading fluctuates, check for loose connections.
If the meter shows a negative sign, the leads are reversed; swap them for a positive reading of $E$ .

Techniques used

connect a voltmeter in parallel across two pointsselect an appropriate voltmeter range and read a digital/analogue scalerecord a reading to appropriate resolution

(ii)

Disconnect the voltmeter from the power supply.

DifficultyEasy

Worked solution

Answer

Disconnect the voltmeter leads from the power supply terminals.

Final answer

Voltmeter disconnected from the power supply.

Detailed explanation

Background Concept

In multi-step circuit experiments, the same voltmeter is moved to measure different potential differences. The voltmeter must only be connected where you intend to measure, otherwise it may:

give the wrong measurement for the next step, or
unintentionally provide an additional current path.

Understanding the Question

After measuring $E$ across the supply, you must remove the voltmeter from the supply because the next measurements require the voltmeter to be connected to the wires at positions $x$ .

Approach

Carefully remove the voltmeter connections from the supply terminals while leaving the rest of the circuit intact.

Step-by-Step Reasoning

Remove one voltmeter lead from the supply terminal.
Remove the second lead.
Keep the supply and other circuit connections unchanged.

Key Takeaways

Always connect the voltmeter across the two points of interest only.

Common Mistakes

Leaving one voltmeter lead attached to the supply while moving the other lead elsewhere (this changes what p.d. is being measured).

Things to Be Careful About

Avoid shorting the supply terminals with meter leads or crocodile clips while disconnecting.

Techniques used

isolate a measuring instrument from a circuit without disturbing the rest of the circuitfollow procedural instructions to change measurement configuration

(c)

(i)

Position the voltmeter leads on the wires at distance $x$ from the zero ends of both rules as shown in Fig. 1.2, where $x$ is approximately 20 cm.

Fig. 1.2

DifficultyMedium-Easy

Worked solution

diagram-2

Answer

Place the two voltmeter contacts on the wires so that each contact is at a distance $x \approx 20\ \text{cm}$ from the zero end of its own rule (with the zeros at opposite ends as shown).

Final answer

Voltmeter contacts positioned at x ≈ 20 cm from each rule’s zero end.

Detailed explanation

Background Concept

A potential difference can be measured between any two points in a circuit by connecting a voltmeter between those points. In this experiment, the points are chosen by placing contacts on two wires.

Because each wire is mounted on a separate metre rule whose zero ends are at opposite ends, “distance $x$ from the zero end” means:

measure $x$ from the left on one rule, but
measure $x$ from the right on the other rule (if its zero is at the right).

Understanding the Question

You must put the voltmeter leads onto the two wires at positions that correspond to the same distance $x$ from each wire’s zero end, with $x$ about $20\ \text{cm}$ . Fig. 1.2 shows the intended geometry.

Approach

Identify the zero end on each metre rule.
Use the scale to locate $x \approx 20\ \text{cm}$ from each zero end.
Place the voltmeter lead contacts exactly at those positions.

Step-by-Step Reasoning

diagram-2

Look for the “0 cm” marking on each metre rule; note they are at opposite ends.
On the first rule, measure $20\ \text{cm}$ from its 0 end and place the contact.
On the second rule, measure $20\ \text{cm}$ from its 0 end (which is at the opposite end) and place the other contact.
Ensure the contacts press firmly onto the wire (poor contact causes unstable voltmeter readings).

Key Takeaways

Position measurements depend on the defined zero; when zeros are opposite, you must measure $x$ from different physical ends.

Common Mistakes

Measuring $x$ from the same physical end on both rules even though the zero ends are opposite.
Touching the contact to tape/insulation rather than the conducting wire.

Things to Be Careful About

Read the metre rule at eye level to reduce parallax (especially for analogue scales).
Keep the contacts perpendicular and consistent to avoid changing contact resistance.

Techniques used

set contact positions using a metre rule scaleensure both contacts are at matching distances from their respective zero endsconnect a voltmeter between two points in a circuit

(ii)

Record $x$ and record the voltmeter reading $V$ .
Include the sign (+ or $-$ ) of $V$ .

$x$ = ______
$V$ = ______

DifficultyMedium-Easy

Worked solution

Answer

Example (appropriate precision):

x = 20.0\ \text{cm}

V = +0.40\ \text{V}

Final answer

x = 20.0 cm, V = +0.40 V

Detailed explanation

Background Concept

$x$ is a position measured along each metre rule from its zero end.
$V$ is the potential difference between the two chosen points; voltmeters display a sign depending on which lead is at the higher potential.

If the voltmeter shows a negative reading, it means the voltmeter’s “positive” terminal is at a lower potential than the “negative” terminal for that connection.

Understanding the Question

You must record:

the value of $x$ used (around 20 cm for this first set), and
the voltmeter reading $V$ , including whether it is $+$ or $-$ .

Approach

Read $x$ from the metre rule scale (to the nearest mm or 0.1 cm, depending on how you record).
Read the voltmeter value and keep the sign.
Record both with units.

Step-by-Step Reasoning

Use the rule markings to determine $x$ from the appropriate zero end.
Record $x$ with consistent precision (e.g. $20.0\ \text{cm}$ if using 0.1 cm resolution).
Observe the voltmeter display:
- if it shows a minus sign, record $V$ as negative.
- otherwise record it as positive.
Include units: $x$ in cm (or m) and $V$ in volts.

Key Takeaways

The sign of $V$ contains information about which point is at higher potential.
Consistent precision helps later processing and graph plotting.

Common Mistakes

Omitting the sign of $V$ .
Writing $x$ without units.
Recording $x$ to inconsistent decimal places across repeated readings.

Things to Be Careful About

Ensure contacts are on the metal wire, not the tape.
Let the voltmeter reading settle before recording.

Techniques used

read a length from a metre rule to appropriate precisionmeasure a potential difference and record its signrecord results with units and consistent significant figures

(iii)

By moving both contacts, change $x$ until the voltmeter reads zero.
Record $x$ .

$x$ = ______

DifficultyMedium-Easy

Worked solution

Answer

Move both contacts together until the voltmeter reads $0.00\ \text{V}$ .

Example recorded value:

x = 50.0\ \text{cm}

Final answer

x = 50.0 cm (at V = 0)

Detailed explanation

Background Concept

A null method is used when you adjust a variable until a measured quantity becomes zero. It is often accurate because you are looking for a zero reading rather than estimating a non-zero value.

Here, you adjust $x$ until the p.d. between the two contact points is zero, meaning both points are at the same potential.

Understanding the Question

You must slide both voltmeter contacts to new positions (keeping them at the same distance $x$ from each wire’s zero end) until the voltmeter reads zero, then record the corresponding $x$ .

Approach

Move the contacts in small steps.
Watch the voltmeter reading change sign (if it does).
Find the position where the reading is as close to $0.00\ \text{V}$ as possible.
Record $x$ .

Step-by-Step Reasoning

Start from the initial $x$ and move both contacts by the same amount.
If $V$ is positive, move in the direction that reduces the magnitude; if it becomes negative, you have passed the balance point.
Narrow down by smaller adjustments around the point where $V$ changes sign.
Record $x$ at the best null you can achieve.

Key Takeaways

Adjusting for $V = 0$ can give a well-defined reference point for the later graph.

Common Mistakes

Moving only one contact (then the two points no longer correspond to the same $x$ definition).
Recording $x$ when $V$ is small but not actually zero (when a better null is achievable).

Things to Be Careful About

Contact resistance can cause unstable readings; keep the contact pressure consistent.
Record $x$ to a precision consistent with the metre rule scale.

Techniques used

adjust an experimental variable to achieve a null readingidentify the value of a variable when a measured quantity is zerorecord a length reading with appropriate precision

(d)

Repeat (c)(i) and (c)(ii) with different values of $x$ until you have six sets of values of $x$ and $V$ .
Include values of $\frac{V}{E}$ in your table.

10M

DifficultyMedium-Hard

Worked solution

Answer

Record six sets of $x$ and signed $V$ values and calculate $V/E$ for each.

Example of a suitable single table (headings include units):

$x / \text{cm}$	$V / \text{V}$	$V/E$
$20.0$	$+0.60$	$+0.300$
$30.0$	$+0.40$	$+0.200$
$40.0$	$+0.20$	$+0.100$
$50.0$	$0.00$	$0.000$
$60.0$	$-0.20$	$-0.100$
$70.0$	$-0.40$	$-0.200$

(using example $E = 2.00\ \text{V}$ ; $V/E$ calculated for each row).

Final answer

Six sets of x and V recorded with a V/E column (see table).

Detailed explanation

Background Concept

In Paper 3, marks for tables are usually awarded for:

Sufficient number of readings (here: six sets)
Appropriate range of the independent variable ( $x$ )
Clear, single table with correct headings and units
Consistent precision within a column
Correct calculation of any derived quantity (here $V/E$ , which is dimensionless)

Understanding the Question

You must repeat the measurement of $V$ at different positions $x$ until you have six pairs $(x, V)$ . You must also include the calculated ratio $V/E$ for each set.

The independent variable is $x$ and the dependent variable is $V$ (and then $V/E$ ).

Approach

Choose at least six values of $x$ spanning a sensible range (not all clustered together).
For each $x$ , measure $V$ (including sign).
Compute $V/E$ using the previously measured $E$ :

\frac{V}{E} = \frac{V\ (\text{in V})}{E\ (\text{in V})}

Present all data in one clear table with headings and units.

Step-by-Step Reasoning

Select values of $x$ (e.g. in steps of 10 cm) and include values on both sides of the balance point where $V = 0$ if possible.
For each $x$ $x$ :
- Place both contacts at that $x$ from their respective zero ends.
- Read and record $V$ including sign.
Calculate $V/E$ $V / E$ for each row:
- Example: if $V = +0.60\ \text{V}$ and $E = 2.00\ \text{V}$ then

\frac{V}{E} = \frac{+0.60}{2.00} = +0.300

Ensure the derived column has a consistent number of decimal places or significant figures.

Key Takeaways

A good table is about communication: clear headings, units, consistent precision.
Derived quantities must be calculated correctly for every row.

Common Mistakes

Not collecting six complete pairs of $(x, V)$ .
Leaving out the sign of $V$ .
Missing units in headings (e.g. writing just $x$ instead of $x / \text{cm}$ ).
Calculating $E/V$ instead of $V/E$ .

Things to Be Careful About

$V/E$ is dimensionless: do not add a unit to that column.
Use the same $E$ value consistently for all calculations.
Avoid mixing $x$ units (cm in one row, m in another).

Techniques used

collect multiple readings over a suitable range of the independent variablerecord raw measurements with units and consistent precisioncalculate a derived quantity for each row of datause a results table with correct headings and significant figures

(e)

(i)

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

DifficultyMedium

Worked solution

Answer

Plot $y = V/E$ (dimensionless) on the $y$ -axis against $x$ on the $x$ -axis.

Label axes: $x / \text{cm}$ (or $x / \text{m}$ ) and $V/E$ .
Use a sensible scale (at least half the grid in each direction).
Plot all six points accurately.

Final answer

Graph of V/E (y) against x (x) plotted with correct labels and scales.

Detailed explanation

Background Concept

Graph marks typically come from:

Correct choice of variables for axes
Correct axis labels (quantity and unit)
Sensible scales (not cramped; not awkward like 3, 6, 9 per big square)
Accurate plotting (small, neat points)

Here, $V/E$ is a ratio of voltages, so it is dimensionless.

Understanding the Question

You must plot a graph with:

horizontal axis: $x$
vertical axis: $V/E$

using the six data sets from your table.

Approach

Decide whether you will use $x$ in cm or m (either is acceptable as long as you are consistent).
Choose axis limits that comfortably include all your data.
Choose scales that use most of the graph paper.
Plot each point carefully.

Step-by-Step Reasoning

Put $x$ on the horizontal axis and label it $x / \text{cm}$ (or $x / \text{m}$ ).
Put $V/E$ on the vertical axis and label it $V/E$ (no unit).
Choose a scale so that the smallest and largest values of $x$ and $V/E$ are well spread.
Plot points using fine crosses or small dots.

Key Takeaways

Always label axes with quantity and unit (except dimensionless quantities).
Good scales and accurate points are essential for reliable gradients/intercepts.

Common Mistakes

Swapping axes (plotting $x$ on $y$ -axis).
Writing an incorrect label such as $V/E\ (\text{V})$ .
Using a poor scale that uses only a small part of the grid.

Things to Be Careful About

If $V/E$ includes negative values, include negative y-values on the axis.
Keep plotting precision consistent (use a ruler for reading coordinates).

Techniques used

choose suitable axis scales that use most of the gridlabel axes with quantity and unitplot data points accurately from a table

(ii)

Draw the straight line of best fit.

DifficultyMedium-Easy

Worked solution

Answer

Draw a single straight line of best fit with an approximately equal distribution of points on either side (do not join dot-to-dot).

Final answer

Straight line of best fit drawn.

Detailed explanation

Background Concept

A straight line of best fit represents the trend of experimental data when the relationship is expected to be linear. It should reflect overall trend, not pass through every point.

Understanding the Question

You have plotted $V/E$ against $x$ . You must now draw the best-fit straight line that represents the data trend.

Approach

Use a ruler to draw one straight line that:

goes through the middle of the spread of points,
has roughly equal numbers of points above and below,
is not forced through an outlier.

Step-by-Step Reasoning

Visually estimate the trend.
Place the ruler to balance the scatter.
Draw a single thin line across the full range of the data.

Key Takeaways

Best fit is about the overall trend, not connecting points.

Common Mistakes

Dot-to-dot joining.
Forcing the line through the origin when not justified by data.
Drawing a line that only spans a small section of the plotted range.

Things to Be Careful About

If there is an obvious anomalous point, do not bend the line to include it; keep the best fit for the main cluster.

Techniques used

draw a best-fit straight line through scattered databalance points above and below the lineuse a ruler to draw a single thin line

(iii)

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

gradient = ______
$y$ -intercept = ______

DifficultyMedium

Worked solution

Working

Using two well-separated points on the best-fit line, e.g.

(x_1, y_1) = (20.0\ \text{cm},\ 0.30),\quad (x_2, y_2) = (70.0\ \text{cm},\ -0.20)

\text{gradient} = \frac{\Delta y}{\Delta x} = \frac{-0.20 - 0.30}{70.0 - 20.0} = \frac{-0.50}{50.0} = -1.0\times 10^{-2}\ \text{cm}^{-1}

At $x = 0$ ,

y\text{-intercept} = 0.50

Answer

gradient $= -1.0\times 10^{-2}\ \text{cm}^{-1}$

$y$ -intercept $= 0.50$

Final answer

gradient = −1.0×10^−2 cm^−1, y-intercept = 0.50

Detailed explanation

Background Concept

For a straight-line graph of $y$ against $x$ :

y = mx + c

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

On a graph, the gradient should be found using a large triangle to reduce percentage reading error.

Understanding the Question

You have drawn a best-fit straight line on a graph of $V/E$ (y-axis) against $x$ (x-axis). You must now determine:

the gradient of this line, and
the y-intercept.

Approach

Choose two points on the line (not necessarily plotted points) that are far apart.
Read their coordinates accurately.
Compute $\Delta y$ and $\Delta x$ , then calculate the gradient.
Read the intercept where the line crosses the y-axis ( $x=0$ ), or extrapolate if necessary.

Step-by-Step Reasoning

Select two widely separated points on the best-fit line.
Calculate:

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

Keep track of sign: if the line slopes downward as $x$ increases, the gradient is negative.
Find the y-intercept by reading the line at $x = 0$ .

Units:

$V/E$ is dimensionless.
If $x$ is in cm, gradient has units $\text{cm}^{-1}$ .
If $x$ is in m, gradient has units $\text{m}^{-1}$ .

Key Takeaways

Use a large triangle for gradient.
Intercept comes from $x=0$ .
Units of gradient depend on the x-axis unit; intercept is dimensionless here.

Common Mistakes

Using two plotted data points instead of points on the best-fit line (more affected by scatter).
Calculating $\Delta x / \Delta y$ instead of $\Delta y / \Delta x$ .
Forgetting the gradient’s unit.

Things to Be Careful About

Read coordinates from the graph scale correctly.
Do not round intermediate readings too aggressively before calculating the gradient.

Techniques used

determine a gradient using a large triangle on a straight-line graphread two well-separated points from a best-fit linefind the y-intercept by extrapolation or direct reading

(f)

The quantities $V$ , $E$ and $x$ are related by the equation

\frac{V}{E} = ax + b

where $a$ and $b$ are constants.

Use your answers from (e)(iii) to determine the values of $a$ and $b$ .
Give appropriate units.

$a$ = ______
$b$ = ______

DifficultyMedium-Easy

Worked solution

Working

Given

\frac{V}{E} = ax + b

Comparing with $y = mx + c$ for a graph of $y = V/E$ against $x$ :

a = \text{gradient},\quad b = y\text{-intercept}

Using (e)(iii):

a = -1.0\times 10^{-2}\ \text{cm}^{-1}

b = 0.50\ \text{(no unit)}

Answer

a = -1.0\times 10^{-2}\ \text{cm}^{-1}

b = 0.50

Final answer

a = −1.0×10^−2 cm^−1, b = 0.50

Detailed explanation

Background Concept

If an equation has the form

y = mx + c

and you plot $y$ (vertical axis) against $x$ (horizontal axis), then:

the gradient of the straight line is $m$
the y-intercept is $c$

Here the relationship is

\frac{V}{E} = ax + b

So we identify $y = V/E$ , $x = x$ , $m = a$ , and $c = b$ .

Understanding the Question

You already found the gradient and y-intercept from your graph in part (e)(iii). Now you must use these to state the constants $a$ and $b$ , and give the correct units.

Approach

Match the equation to the straight-line form.
Set $a$ equal to the gradient and $b$ equal to the y-intercept.
Determine units:
- $V/E$ is dimensionless
- therefore $b$ is dimensionless
- $a$ must have units of $1/x$ (e.g. $\text{cm}^{-1}$ or $\text{m}^{-1}$ depending on your graph)

Step-by-Step Reasoning

From the plotted graph of $V/E$ $V / E$ against $x$ $x$ :
- gradient $= a$
- intercept $= b$
Units:
- If you used $x$ in cm, then $a$ is in $\text{cm}^{-1}$ .
- If you used $x$ in m, then $a$ is in $\text{m}^{-1}$ .
- $b$ has no unit.

Key Takeaways

Constants in a linear relationship come directly from gradient and intercept.
Always infer units from what is plotted.

Common Mistakes

Giving $b$ a unit (it is dimensionless here).
Using the reciprocal of the gradient.
Quoting $a$ in $\text{m}^{-1}$ when $x$ was plotted in cm (or vice versa) without conversion.

Things to Be Careful About

State units exactly as implied by your axis label.
Keep the sign of $a$ (negative gradient means negative $a$ ).

Techniques used

match an experimental straight-line graph to y = mx + cidentify constants from gradient and interceptdetermine units from the plotted variables

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