Physics 9702/53 — May/June 2025

Cambridge A-Level · Planning, Analysis and Evaluation · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Analysis, Conclusions and Evaluation · Planning

Q115MMedium-HardPlanningAnalysis, Conclusions and Evaluation

Fig. 1.1 shows a thin coil of cross-sectional area $A$ and length $l$ connected to a resistor of resistance $S$ and two terminals.

Fig. 1.1

An alternating voltage is applied to the terminals. The peak value of the alternating voltage is $E$ and the frequency is $f$ . The peak value of the potential difference $V$ across the resistor is determined using an oscilloscope.

It is suggested that $V$ is related to $f$ by the relationship

\frac{ES}{V} = \frac{KAN^2f}{l}

where $N$ is the number of turns on the coil and $K$ is a constant.

Plan a laboratory experiment to test the relationship between $V$ and $f$ .

Draw a diagram showing the arrangement of your equipment.

Explain how the results could be used to determine a value for $K$ .

In your plan you should include:

the procedure to be followed
the measurements to be taken
the control of variables
the analysis of the data
any safety precautions to be taken.

DifficultyMedium-Hard

Worked solution

diagram-1

Apparatus

Signal generator (sine output), thin coil (given), resistor of resistance $S$ , oscilloscope (dual-channel), digital multimeter (to measure $S$ ), connecting leads.

Variables

Independent variable: frequency $f$ of the a.c. supply.
Dependent variable: peak p.d. across resistor, $V$ .
Controlled variables: peak supply voltage $E$ , resistance $S$ , coil geometry ( $A$ , $l$ ) and number of turns $N$ (same coil), waveform shape (sine).

Procedure and measurements

Measure $S$ using a multimeter.
Measure $N$ (count turns), $l$ (axial length of coil using ruler/vernier), and determine $A$ (measure coil radius/diameter with calipers and use $A = \pi r^2$ if circular).
Connect coil in series with resistor $S$ to the signal generator.
Connect oscilloscope:
- Channel 1 across the generator output to measure peak voltage $E$ .
- Channel 2 across the resistor to measure peak voltage $V$ .
Set the signal generator to a sine wave and choose a fixed output amplitude. For each chosen frequency $f$ (at least 6 values across a wide range), adjust amplitude if necessary so that measured $E$ is constant.
For each $f$ , record $f$ , $E$ (peak) and $V$ (peak). Repeat readings and average $V$ (and $E$ if it fluctuates).

Analysis (to test the relationship and find $K$ )

For each reading calculate

Y = \frac{ES}{V}

Plot a graph of $Y$ (y-axis) against $f$ (x-axis).

From

\frac{ES}{V} = \frac{KAN^2}{l} f

the graph should be a straight line through the origin with gradient

m = \frac{KAN^2}{l}

Hence

K = \frac{ml}{AN^2}

(Use the best-fit line gradient; include uncertainty from worst acceptable line if required.)

Safety

Use low a.c. voltages; do not exceed current rating of resistor/coil (to avoid heating). Switch off the signal generator before changing connections; avoid short circuits and trailing leads.

Final answer

See working

Detailed explanation

Background Concept

The circuit contains a coil (inductor) in series with a resistor. For an alternating supply, the inductor has an inductive reactance $X_L$ that increases with frequency:

X_L = \omega L = 2\pi f L

In a series $R$ - $L$ circuit, the peak p.d. across the resistor is related to the peak current by

V = I_0 R

and the current depends on the supply frequency because the total impedance depends on $X_L$ . The question does not require deriving the relationship from first principles; it provides a suggested proportionality:

\frac{ES}{V} = \frac{KAN^2}{l} f

This is already in a linear form $Y = mf$ if we define $Y = ES/V$ . A planning answer must show (i) how to vary $f$ and measure $V$ and $E$ , (ii) how to keep other quantities constant, and (iii) how to use a graph to determine the constant $K$ .

Understanding the Question

You are given a coil of cross-sectional area $A$ , length $l$ and turns $N$ in series with a resistor of resistance $S$ . An a.c. voltage of peak value $E$ and frequency $f$ is applied. The oscilloscope is used to find the peak p.d. $V$ across the resistor.

You must plan an experiment to test whether $V$ depends on $f$ as predicted. The equation shows that if the relationship is correct, then $ES/V$ should be directly proportional to $f$ (a straight-line graph through the origin). You must also explain how the graph gives $K$ .

Approach

Choose the independent variable as $f$ and vary it using a signal generator.
Measure the dependent variable $V$ (peak across the resistor) using an oscilloscope.
Keep the supply amplitude constant so that $E$ is constant (or measure $E$ each time and use it in the calculation).
For each $f$ , compute the derived quantity $ES/V$ .
Plot $ES/V$ against $f$ ; if the relationship is valid you get a straight line through the origin.
Use the gradient and the known values of $A$ , $N$ , $l$ to calculate $K$ .

Step-by-Step Reasoning

diagram-1

1. Set up the circuit

Connect the coil and resistor $S$ in series with a signal generator set to a sine wave output.
Use a dual-channel oscilloscope so that both $E$ and $V$ can be measured as peak values.

2. Measuring $E$ and $V$ with the oscilloscope

Channel 1 across the signal generator output: read peak voltage $E$ (from peak-to-peak divided by 2, or directly if the scope has a peak measurement).
Channel 2 across the resistor: read peak voltage $V$ similarly.
The oscilloscope input resistance is large, so connecting it across components does not significantly change the circuit.

3. Varying $f$ while controlling other variables

Change $f$ on the signal generator in steps (e.g. 6 to 10 values spanning a broad range where the waveform remains stable and not too noisy).
Keep the peak supply voltage constant. Practically, you can do this by:
- setting a fixed output amplitude and checking that $E$ remains the same, or
- adjusting the amplitude at each frequency until the measured $E$ equals the chosen constant value.
Keep the same coil and resistor throughout so that $A$ , $l$ , $N$ and $S$ are constant.

4. Additional measurements needed for determining $K$

Measure $S$ with a multimeter.
Determine $N$ by counting turns.
Measure coil length $l$ with a ruler/vernier (record to appropriate precision, e.g. $\pm 1\ \text{mm}$ ).
Determine cross-sectional area $A$ . If the coil is circular, measure diameter $d$ with calipers and compute

A = \pi \left(\frac{d}{2}\right)^2

(If not circular, measure appropriate dimensions to calculate the area.)

5. Recording data
For each frequency $f$ record: $f$ , $E$ , $V$ . Take repeated readings (or multiple traces) to reduce random variation; use the mean.

6. Linearising and graphing
From the suggested relationship,

\frac{ES}{V} = \frac{KAN^2}{l} f

Let $Y = ES/V$ . Then

Y = mf \quad \text{where} \quad m = \frac{KAN^2}{l}

Calculate $Y$ for each data point.
Plot $Y$ (y-axis) against $f$ (x-axis).
Draw a best-fit straight line. If you are including uncertainties, add error bars (typically from oscilloscope voltage reading uncertainty) and draw a worst acceptable line.

7. Determining $K$

Find the gradient $m$ of the best-fit line using a large triangle:

m = \frac{\Delta Y}{\Delta f}

Rearranging for $K$ :

K = \frac{ml}{AN^2}

If required, estimate uncertainty in $m$ from best-fit and worst-line gradients, then propagate to $K$ (since $K \propto m$ the percentage uncertainty in $K$ is the same as that in $m$ , plus any contributions from $A$ , $l$ and $N$ if they are treated as uncertain).

Key Takeaways

A Paper 5 planning question is mainly about: choosing variables, collecting reliable data, controlling conditions, and using a graph to test a relationship.
Linearisation here is straightforward by defining $Y = ES/V$ so that $Y \propto f$ .
The constant $K$ is obtained from the gradient and measured coil parameters using

K = \frac{ml}{AN^2}.

Common Mistakes

Plotting $V$ against $f$ directly (the given equation predicts linearity for $ES/V$ vs $f$ , not for $V$ vs $f$ ).
Not keeping $E$ constant (or not measuring $E$ each time), which would spoil the proportionality.
Measuring only $V$ and forgetting $S$ , $A$ , $l$ and $N$ , so $K$ cannot be calculated.
Using peak-to-peak voltages inconsistently (mixing peak and peak-to-peak gives a factor-of-2 error).
Failing to state that the graph should pass through the origin (as implied by the relationship).

Things to Be Careful About

Oscilloscope readings: state clearly whether you are using peak or peak-to-peak and convert consistently.
Frequency range: choose enough points (at least 6) and a wide span of $f$ to make a convincing straight-line test.
Heating: at higher currents the resistor value $S$ may change; keep voltage low and take readings quickly or allow cooling.
Geometry measurements: $A$ depends on the square of a length, so measuring diameter/radius carefully matters for the accuracy of $K$ .
Graph technique: use sensible scales, plot points accurately, and calculate gradient as $\Delta y / \Delta x$ using widely spaced points on the best-fit line (not point-to-point).

Techniques used

vary the independent variable over a suitable range while keeping other variables constantuse an oscilloscope to measure peak potential differencesrearrange the given relationship to obtain a straight-line graphcalculate a derived quantity for each data point and plot against the independent variabledetermine a constant from the gradient of a best-fit line

Working

From

E = I\left(\frac{R}{n}+Z\right)

\frac{E}{I}=\frac{R}{n}+Z

\frac{1}{I}=\frac{R}{E}\left(\frac{1}{n}\right)+\frac{Z}{E}

Answer

gradient $= \dfrac{R}{E}$

$y$ -intercept $= \dfrac{Z}{E}$

Final answer

gradient = R/E, y-intercept = Z/E

Detailed explanation

Background Concept

To test a proposed relationship experimentally, we often rearrange it into the straight-line form

y = mx + c

where:

$m$ is the gradient (slope),
$c$ is the $y$ -intercept.

Here, the circuit contains a fixed series resistor $Z$ and a parallel combination of $n$ identical resistors each of resistance $R$ . The equivalent resistance of $n$ identical resistors in parallel is

R_{\text{parallel}} = \frac{R}{n}

So the total resistance is $Z + R/n$ and the current is found using $I = E/R_{\text{total}}$ .

Understanding the Question

You are told the suggested relationship

E = I\left(\frac{R}{n}+Z\right)

and that a graph of $\frac{1}{I}$ (vertical axis) against $\frac{1}{n}$ (horizontal axis) is plotted. The question asks for the gradient and intercept of that graph.

Approach

Rearrange the suggested equation to make $\frac{1}{I}$ the subject.
Write the result in the form

\frac{1}{I} = (\text{something})\left(\frac{1}{n}\right) + (\text{something})

Identify the coefficient of $\frac{1}{n}$ as the gradient, and the constant term as the intercept.

Step-by-Step Reasoning

Start with

E = I\left(\frac{R}{n}+Z\right)

Divide both sides by $I$ :

\frac{E}{I} = \frac{R}{n} + Z

Now divide both sides by $E$ :

\frac{1}{I} = \frac{R}{E}\cdot \frac{1}{n} + \frac{Z}{E}

This matches $y = mx + c$ with:

$y = \frac{1}{I}$
$x = \frac{1}{n}$
$m = \frac{R}{E}$
$c = \frac{Z}{E}$

Key Takeaways

Linearising is done by rearranging to $y = mx + c$ .
The gradient is the coefficient of the chosen $x$ -variable.
The intercept is the constant term when $x=0$ .

Common Mistakes

Rearranging to $I$ instead of $1/I$ and then not getting a straight-line form.
Writing gradient as $E/R$ (inverting the coefficient).
Forgetting that the intercept corresponds to the term independent of $n$ .

Things to Be Careful About

Keep the variable on the $y$ -axis as the subject (here $1/I$ ).
Make sure the coefficient multiplies exactly the $x$ -axis quantity ( $1/n$ ).

Techniques used

rearrange the given equation into the form y = mx + cidentify gradient and intercept by comparing coefficientschoose graph variables to linearise a reciprocal relationship

(b)

Values of $n$ , $\frac{1}{n}$ and $I$ are given in Table 2.1.

Table 2.1

$n$	$\frac{1}{n}$	$I / \mu\text{A}$
5	0.200	$455 \pm 5$
6	0.167	$525 \pm 5$
7	0.143	$580 \pm 5$
8	0.125	$635 \pm 5$
9	0.111	$685 \pm 5$
11	0.0909	$765 \pm 5$

Calculate and record values of $\frac{1}{I} / 10^3\text{ A}^{-1}$ in Table 2.1. Include the absolute uncertainties in $\frac{1}{I}$ .

2M Similar

DifficultyMedium

Worked solution

Working

For $y = \dfrac{1}{I}$ , with $I \pm \Delta I$ :

\Delta\left(\frac{1}{I}\right)=\frac{\Delta I}{I}\left(\frac{1}{I}\right)

(Using $I$ in $\text{A}$ .)

$I=455\ \mu\text{A}=4.55\times10^{-4}\ \text{A}$

\frac{1}{I}=2.198\times10^{3}\ \text{A}^{-1}\Rightarrow 2.20

\Delta\left(\frac{1}{I}\right)=\frac{5}{455}(2.198\times10^{3})=2.42\times10^{1}\ \text{A}^{-1}\Rightarrow 0.024

$I=525\ \mu\text{A}$ : $\dfrac{1}{I}/10^{3}=1.90$ , $\Delta(\dfrac{1}{I})/10^{3}=0.018$
$I=580\ \mu\text{A}$ : $\dfrac{1}{I}/10^{3}=1.72$ , $\Delta(\dfrac{1}{I})/10^{3}=0.015$
$I=635\ \mu\text{A}$ : $\dfrac{1}{I}/10^{3}=1.57$ , $\Delta(\dfrac{1}{I})/10^{3}=0.012$
$I=685\ \mu\text{A}$ : $\dfrac{1}{I}/10^{3}=1.46$ , $\Delta(\dfrac{1}{I})/10^{3}=0.011$
$I=765\ \mu\text{A}$ : $\dfrac{1}{I}/10^{3}=1.31$ , $\Delta(\dfrac{1}{I})/10^{3}=0.0086\,(\approx 0.009)$

Answer

$\dfrac{1}{I}/10^{3}\ \text{A}^{-1}$ (with absolute uncertainties):

$n=5$ : $2.20 \pm 0.024$
$n=6$ : $1.90 \pm 0.018$
$n=7$ : $1.72 \pm 0.015$
$n=8$ : $1.57 \pm 0.012$
$n=9$ : $1.46 \pm 0.011$
$n=11$ : $1.31 \pm 0.009$

Final answer

2.20±0.024, 1.90±0.018, 1.72±0.015, 1.57±0.012, 1.46±0.011, 1.31±0.009 (all in 10^3 A^-1)

Detailed explanation

Background Concept

If a measured quantity $I$ has an absolute uncertainty $\Delta I$ , then for a reciprocal

y = \frac{1}{I}

the fractional (percentage) uncertainty is the same:

\frac{\Delta y}{y} = \frac{\Delta I}{I}

So the absolute uncertainty in $y$ is

\Delta y = \frac{\Delta I}{I}\,y

In this question, $I$ is given in $\mu\text{A}$ , but $1/I$ is in $\text{A}^{-1}$ , so you must convert $\mu\text{A}$ to $\text{A}$ first.

Understanding the Question

You must fill the final column of Table 2.1:

compute $1/I$ for each current reading,
express it as $\dfrac{1}{I}/10^{3}\ \text{A}^{-1}$ (i.e. divide by $10^{3}$ to get numbers around 1 to 2),
and include the absolute uncertainty in $1/I$ coming from $\pm 5\ \mu\text{A}$ in $I$ .

Approach

For each row:

Convert $I$ from $\mu\text{A}$ to $\text{A}$ .
Calculate $1/I$ .
Divide by $10^{3}$ to match the table heading.
Use

\Delta\left(\frac{1}{I}\right)=\frac{\Delta I}{I}\left(\frac{1}{I}\right)

and then also divide the uncertainty by $10^{3}$ to match the scaled column.

Step-by-Step Reasoning

Example (first row):

Convert current:

I = 455\ \mu\text{A} = 455\times 10^{-6}\ \text{A} = 4.55\times 10^{-4}\ \text{A}

Reciprocal:

\frac{1}{I} = \frac{1}{4.55\times 10^{-4}} = 2.198\times 10^{3}\ \text{A}^{-1}

Scale for the table:

\frac{1}{I}/10^{3} = 2.198 \approx 2.20

Uncertainty (fractional uncertainty transfers):

\frac{\Delta I}{I} = \frac{5}{455} = 0.01099

\Delta\left(\frac{1}{I}\right)= 0.01099\times 2.198\times 10^{3} = 24.2\ \text{A}^{-1}

and the scaled uncertainty is

\Delta\left(\frac{1}{I}/10^{3}\right)= \frac{24.2}{10^{3}}=0.0242

Repeat the same process for each row.

Key Takeaways

Always convert to SI before inverting.
For $y=1/I$ , fractional uncertainty in $y$ equals fractional uncertainty in $I$ .
If the table uses a scale factor (here $/10^{3}$ ), apply it to both the value and the uncertainty.

Common Mistakes

Inverting $455$ without converting $\mu\text{A}$ to $\text{A}$ (gives a factor of $10^{6}$ wrong).
Calculating uncertainty in $1/I$ as $1/\Delta I$ (incorrect).
Forgetting to scale the uncertainty when writing $\dfrac{1}{I}/10^{3}$ .

Things to Be Careful About

Quote uncertainties to 1 (sometimes 2) significant figures, and round the value to the same decimal place.
Keep units consistent: $I$ in $\text{A}$ implies $1/I$ in $\text{A}^{-1}$ .

Techniques used

convert microampere readings to amperescalculate reciprocals and apply consistent scaling factorspropagate absolute uncertainty through a reciprocal calculation using fractional uncertainties

(c)

(i)

Plot a graph of $\frac{1}{I} / 10^3\text{ A}^{-1}$ against $\frac{1}{n}$ . Include error bars for $\frac{1}{I}$ .

Graph 2.1

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Plot the points $(x,y)$ with $x=\dfrac{1}{n}$ and $y=\dfrac{1}{I}/10^{3}\ \text{A}^{-1}$ :

$(0.200,\ 2.20)$ with error bar $\pm 0.024$
$(0.167,\ 1.90)$ with error bar $\pm 0.018$
$(0.143,\ 1.72)$ with error bar $\pm 0.015$
$(0.125,\ 1.57)$ with error bar $\pm 0.012$
$(0.111,\ 1.46)$ with error bar $\pm 0.011$
$(0.0909,\ 1.31)$ with error bar $\pm 0.009$

Final answer

Points plotted with vertical error bars using the calculated ± uncertainties in (1/I)/10^3.

Detailed explanation

Background Concept

A good graph in Paper 5 earns marks for:

correct axes: quantity and unit (including any scale factor in the label),
sensible linear scales using most of the grid,
accurate plotting,
correct error bars using absolute uncertainties.

Error bars represent the uncertainty range for each measurement. Here, only $1/I$ has uncertainties given, so the error bars are vertical.

Understanding the Question

You are given the calculated values of $x=1/n$ and you calculated $y=(1/I)/10^{3}$ with absolute uncertainties from part (b). You must plot $y$ against $x$ on the provided grid and include error bars for $y$ .

Approach

Check axis labels match the requested variables.
Choose a scale (already set by the provided axes) and plot each point carefully.
For each point, draw a vertical line from $y-\Delta y$ to $y+\Delta y$ at the same $x$ value.

Step-by-Step Reasoning

For each row in the table, locate $x=1/n$ on the horizontal axis.
Locate the corresponding $y=(1/I)/10^{3}$ on the vertical axis.
Mark a small, clear cross.
Draw the error bar: for example, at $x=0.200$ , the value is $y=2.20$ and the uncertainty is $\pm 0.024$ , so the bar runs from $2.176$ to $2.224$ .
Repeat for all points.

Key Takeaways

Use vertical error bars when only the $y$ -quantity has uncertainty.
Error bar length is based on absolute uncertainty, not percentage.

Common Mistakes

Putting error bars of the same size for all points (uncertainties differ slightly).
Plotting $I$ instead of $1/I$ .
Forgetting the $/10^{3}$ scale factor and plotting values $\sim 2000$ instead of $\sim 2$ .

Things to Be Careful About

Plotting accuracy: points should be within about half a small square.
Error bars must be centred on the plotted point and extend equally above and below.

Techniques used

plot data points on correctly scaled axeslabel axes with quantity and unit including scale factorsdraw vertical error bars using absolute uncertainties

(ii)

Draw the straight line of best fit and a worst acceptable straight line on your graph. Label both lines.

2M Similar

DifficultyMedium

Worked solution

Answer

Draw a single straight line of best fit through the trend of the plotted points.

Draw a worst acceptable straight line (steepest or shallowest) that is still consistent with all the error bars.

Label the lines: “best fit” and “worst acceptable”.

Final answer

Best-fit line and a labelled worst acceptable line drawn.

Detailed explanation

Background Concept

When data points have uncertainties, a single “correct” straight line cannot be known exactly. Two lines are commonly used:

Line of best fit: the straight line that best represents the overall trend (roughly equal scatter of points above and below).
Worst acceptable line: the steepest or shallowest straight line that can still pass through all the error bars. This gives an estimate of the uncertainty in the gradient and intercept.

Understanding the Question

You have already plotted $y$ vs $x$ with error bars. Now you must:

draw and label the best-fit straight line,
draw and label one worst acceptable straight line.

Approach

Use a ruler.
For best fit: do not join dot-to-dot; draw a single straight line through the general centre of the error bars.
For worst acceptable: pivot the line to make it as steep or as shallow as possible while still intersecting every error bar.

Step-by-Step Reasoning

c-ii-lines

Best-fit line: place the ruler so that the line passes through the middle of the data trend. If one point is slightly off, you do not force the line through it.
Worst acceptable line:
- choose whether you want the maximum gradient (steepest) or minimum gradient (shallowest),
- ensure that at each $x$ value, the line still goes through the vertical error bar.
Label each line clearly on the graph.

Key Takeaways

Worst acceptable line must be constrained by error bars, not by the points themselves.
This method is used to estimate uncertainty in gradient and intercept.

Common Mistakes

Drawing multiple “best fit” lines.
Drawing a worst line that misses one or more error bars.
Not labelling the lines (mark can be lost even if drawn).

Things to Be Careful About

The worst acceptable line must still be straight.
Use the full graph length; a short line segment increases reading error.

Techniques used

draw a straight line of best fit balancing scatter and error barsconstruct a worst acceptable line consistent with all error barslabel lines clearly to distinguish best and worst fits

(iii)

Determine the gradient of the line of best fit. Include the absolute uncertainty in your answer.

gradient = ______

2M Similar

DifficultyMedium

Worked solution

Working

Using two well-separated points on the best-fit line (e.g. near $x=0.200$ and $x=0.0909$ ):

\text{gradient} = \frac{\Delta y}{\Delta x} \approx \frac{2.20-1.31}{0.200-0.0909}=8.2

Worst acceptable gradients (from worst line):

m_{\text{worst}}\approx 8.47\ \text{or}\ 7.86

So absolute uncertainty:

\Delta m = \left|m_{\text{best}}-m_{\text{worst}}\right| \approx 0.3

Answer

gradient $= 8.2 \pm 0.3$

Final answer

8.2 ± 0.3

Detailed explanation

Background Concept

The gradient of a straight-line graph is

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

To reduce reading uncertainty, you should use two points far apart on the line (a large triangle).

When uncertainties are included, the common Paper 5 method is:

find $m_{\text{best}}$ from the best-fit line,
find $m_{\text{worst}}$ from a worst acceptable line,
take the absolute uncertainty as

\Delta m = \left|m_{\text{best}}-m_{\text{worst}}\right|

(or equivalently half the full range if you used both steepest and shallowest worst lines).

Understanding the Question

You must read the gradient from your best-fit line and give an absolute uncertainty based on the difference between the best-fit gradient and the worst acceptable gradient.

Approach

Pick two far-separated points on the best-fit line (not necessarily actual data points).
Compute $m_{\text{best}}=\Delta y/\Delta x$ .
Repeat using the worst acceptable line to get $m_{\text{worst}}$ .
Quote $m_{\text{best}} \pm \Delta m$ .

Step-by-Step Reasoning

c-iii-gradient-triangle

Choose two points on the best-fit line, ideally near the left and right extremes of your plotted range (this makes $\Delta x$ large).
Read off their coordinates, then calculate $m_{\text{best}}$ .
Do the same for the worst acceptable line to obtain $m_{\text{worst}}$ .
The uncertainty reflects how much the gradient could plausibly vary given the error bars.

Using representative end-values gives a gradient near $8.2$ and an uncertainty about $\pm 0.3$ .

Key Takeaways

Use the line, not individual noisy points.
Use a large triangle for better precision.
Uncertainty in gradient comes from worst acceptable line(s).

Common Mistakes

Using two adjacent points (small triangle) which inflates uncertainty.
Calculating $\Delta x/\Delta y$ (inverting the gradient).
Finding gradient from the plotted points rather than from the straight line.

Things to Be Careful About

Keep consistent with the plotted axes: $y$ here is $(1/I)/10^{3}$ .
Quote the uncertainty as an absolute uncertainty (same units as the gradient).

Techniques used

calculate gradient from two well-separated points on a straight lineuse a large gradient triangle to reduce percentage reading uncertaintyestimate gradient uncertainty from best-fit versus worst-acceptable gradients

(iv)

Determine the $y$ -intercept of the line of best fit. Include the absolute uncertainty in your answer.

$y$ -intercept = ______

2M Similar

DifficultyMedium

Worked solution

Working

Using $y = mx + c$ with $m = 8.2$ and a point on the best-fit line, e.g. $(0.200,\ 2.20)$ :

c = y - mx = 2.20 - 8.2(0.200) = 0.56

From the worst acceptable line, $c$ ranges from about $0.53$ to $0.60$ , so

\Delta c \approx 0.04

Answer

$y$ -intercept $= 0.56 \pm 0.04$

Final answer

0.56 ± 0.04

Detailed explanation

Background Concept

For a straight line

y = mx + c

the $y$ -intercept $c$ is the value of $y$ when $x=0$ . You can find $c$ by:

extending the line to cross the $y$ -axis (graphical method), or
using one point on the line with $c = y - mx$ .

The uncertainty in $c$ is found using the same worst-acceptable-line idea used for the gradient.

Understanding the Question

You must obtain the intercept of the best-fit line and give an absolute uncertainty, based on how much the intercept could change if the worst acceptable line were used.

Approach

Find $c_{\text{best}}$ from the best-fit line.
Find $c_{\text{worst}}$ from the worst acceptable line.
Quote

\Delta c = \left|c_{\text{best}} - c_{\text{worst}}\right|

(or half the range if using both extremes).

Step-by-Step Reasoning

Using the best-fit line, either read where it meets the $y$ -axis or compute $c=y-mx$ .
For uncertainty, repeat the intercept determination for the worst acceptable line.
The intercept uncertainty is typically more sensitive to extrapolation, so extending lines neatly across the axis helps.

A representative best-fit intercept is about $0.56$ (in the plotted units $10^{3}\ \text{A}^{-1}$ ), with an uncertainty about $\pm 0.04$ .

Key Takeaways

Intercept is the value at $x=0$ .
Use worst acceptable line to estimate uncertainty.

Common Mistakes

Using an actual data point to read the intercept (intercept belongs to the fitted line).
Forgetting that the intercept is in the plotted units ( $(1/I)/10^{3}$ ).
Quoting no uncertainty.

Things to Be Careful About

Extrapolation: draw lines long enough and use a ruler carefully.
Keep consistent decimal places appropriate to your graph reading resolution.

Techniques used

determine the y-intercept by extrapolating the best-fit line to x = 0calculate the intercept using c = y - mxestimate intercept uncertainty from best-fit versus worst-acceptable intercepts

(d)

The e.m.f. $E$ of the battery is determined twice during the experiment. The values obtained are $5.6\text{ V}$ and $6.0\text{ V}$ .

(i)

Using your answers to (a), (c)(iii) and (c)(iv), determine the values of $R$ and $Z$ . Include appropriate units.

$R$ = ______
$Z$ = ______

2M Similar

DifficultyMedium

Worked solution

Working

From (a):

\frac{1}{I} = \frac{R}{E}\left(\frac{1}{n}\right) + \frac{Z}{E}

But the graph uses

y=\frac{1}{I}/10^{3}

y = \frac{R}{10^{3}E}\left(\frac{1}{n}\right) + \frac{Z}{10^{3}E}

Hence, with gradient $m$ and intercept $c$ from the graph:

R = 10^{3}Em qquad Z = 10^{3}Ec

Take $E = \frac{5.6+6.0}{2}=5.8\ \text{V}$ .

Using $m=8.2$ and $c=0.56$ :

R = 10^{3}(5.8)(8.2)=4.76\times 10^{4}\ \Omega

Z = 10^{3}(5.8)(0.56)=3.25\times 10^{3}\ \Omega

Answer

$R = 4.8\times 10^{4}\ \Omega$

$Z = 3.3\times 10^{3}\ \Omega$

Final answer

R = 4.8×10^4 Ω, Z = 3.3×10^3 Ω

Detailed explanation

Background Concept

From the straight-line form

\frac{1}{I} = \frac{R}{E}\left(\frac{1}{n}\right) + \frac{Z}{E}

the physical meaning is:

gradient of a $1/I$ vs $1/n$ graph is $R/E$ ,
intercept is $Z/E$ .

However, if the axis is scaled (here plotting $\frac{1}{I}/10^{3}$ ), then the numerical gradient read from the graph is also scaled by the same factor.

y' = \frac{1}{I}/10^{3}

then $\frac{1}{I} = 10^{3}y'$ , so the true gradient in $1/I$ is $10^{3}$ times the plotted gradient.

Understanding the Question

You are given two measurements of $E$ (5.6 V and 6.0 V). Using the gradient and intercept you found from the graph in part (c), you must calculate the resistor values $R$ and $Z$ .

Approach

Start from the linear relationship and rewrite it in terms of the plotted $y$ variable $y'=(1/I)/10^{3}$ .
Match to $y' = mx + c$ .
Solve for $R$ and $Z$ in terms of $m$ , $c$ and $E$ .
Use a best estimate for $E$ (the mean of the two values) unless instructed otherwise.

Step-by-Step Reasoning

Begin with:

\frac{1}{I} = \frac{R}{E}\left(\frac{1}{n}\right) + \frac{Z}{E}

Convert to the plotted variable $y'=(1/I)/10^{3}$ :

10^{3}y' = \frac{R}{E}x + \frac{Z}{E}

where $x=1/n$ . Divide by $10^{3}$ :

y' = \frac{R}{10^{3}E}x + \frac{Z}{10^{3}E}

So the plotted gradient and intercept are:

m = \frac{R}{10^{3}E},\qquad c = \frac{Z}{10^{3}E}

Rearrange:

R = 10^{3}Em,\qquad Z = 10^{3}Ec

Use the mean emf:

E = \frac{5.6+6.0}{2} = 5.8\ \text{V}

Substitute your graph values (representatively $m\approx 8.2$ , $c\approx 0.56$ ) to obtain resistances in $\Omega$ .

Key Takeaways

Always check whether the graph axes include a scale factor.
Gradient and intercept can be used to determine physical constants.

Common Mistakes

Using $R=Em$ without the $10^{3}$ correction (gives $R$ too small by a factor of 1000).
Using only one of the two $E$ values without justification.
Missing units for $R$ and $Z$ .

Things to Be Careful About

If your gradient/intercept differ slightly (because they depend on your drawn lines), your $R$ and $Z$ will differ accordingly; that is expected in Paper 5.
Quote answers to a sensible number of significant figures consistent with graph reading.

Techniques used

relate graph gradient and intercept to constants using the linearised equationaccount for axis scale factors when converting gradient and intercept to physical constantssubstitute measured values to calculate resistance values with units

(ii)

Determine the percentage uncertainty in your value of $R$ .

percentage uncertainty = ______ %

1M Similar

DifficultyMedium

Worked solution

Working

From (d)(i):

R = 10^{3}Em

For products, fractional uncertainties add:

\frac{\Delta R}{R} = \frac{\Delta E}{E} + \frac{\Delta m}{m}

$E = 5.8\ \text{V}$ with $\Delta E = \frac{6.0-5.6}{2}=0.2\ \text{V}$ :

\frac{\Delta E}{E}\times 100 = \frac{0.2}{5.8}\times 100 = 3.4\%

Using $m = 8.2 \pm 0.3$ :

\frac{\Delta m}{m}\times 100 = \frac{0.3}{8.2}\times 100 = 3.7\%

Total percentage uncertainty:

3.4 + 3.7 = 7.1\% \approx 7.0\%

Answer

percentage uncertainty $= 7.0\%$

Final answer

7.0%

Detailed explanation

Background Concept

If a quantity is found from a product of measured quantities, e.g.

R = kEm

then the fractional uncertainties add (Paper 5 convention):

\frac{\Delta R}{R} = \frac{\Delta E}{E} + \frac{\Delta m}{m}

A repeated measurement (two values) is often treated by taking:

best estimate = mean,
absolute uncertainty = half the range.

Understanding the Question

You are asked for the percentage uncertainty in $R$ . Your $R$ came from the gradient and the emf, so both contribute to uncertainty.

Approach

Write $R$ in terms of $E$ and $m$ .
Find $\Delta E$ from the two measurements.
Use your $\Delta m$ from the worst acceptable line.
Add fractional uncertainties and convert to a percentage.

Step-by-Step Reasoning

From part (d)(i),

R = 10^{3}Em

Use mean and half-range for $E$ :

E = \frac{5.6+6.0}{2}=5.8\ \text{V},\qquad \Delta E = \frac{6.0-5.6}{2}=0.2\ \text{V}

Convert to percentage:

\%\Delta E = \frac{0.2}{5.8}\times 100 \approx 3.4\%

Gradient percentage uncertainty (using your graph values):

\%\Delta m = \frac{\Delta m}{m}\times 100

Total:

\%\Delta R = \%\Delta E + \%\Delta m

This gives a result around $7\%$ for the representative values.

Key Takeaways

Half-range is a standard way to estimate uncertainty from two repeated readings.
For products, add fractional uncertainties.

Common Mistakes

Ignoring the uncertainty in $E$ even though two different values are given.
Subtracting uncertainties instead of adding.
Using absolute uncertainties directly (e.g. adding $0.2$ and $0.3$ ) without converting to fractional form.

Things to Be Careful About

Use consistent significant figures in the final percentage (usually 1 s.f. or 2 s.f. is fine).
The factor $10^{3}$ is exact (a scale choice), so it has no uncertainty contribution.

Techniques used

calculate percentage uncertainties from absolute uncertaintiescombine fractional uncertainties in productsestimate emf uncertainty from repeated measurements

(e)

The experiment is repeated with 20 resistors, each of resistance $R$ , connected in parallel between P and Q. Determine the total current $I$ in the circuit.

$I$ = ______ $\text{A}$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Using $E = I\left(\dfrac{R}{n}+Z\right)$ with $n=20$ , $E=5.8\ \text{V}$ , $R=4.8\times 10^{4}\ \Omega$ , $Z=3.3\times 10^{3}\ \Omega$ :

I = \frac{E}{R/n + Z}

I = \frac{5.8}{\left(\frac{4.8\times 10^{4}}{20}\right)+3.3\times 10^{3}}=1.0\times 10^{-3}\ \text{A}

Answer

$I = 1.0\times 10^{-3}\ \text{A}$

Final answer

1.0×10^-3 A

Detailed explanation

Background Concept

For $n$ identical resistors $R$ in parallel, the equivalent resistance is

R_{\text{parallel}} = \frac{R}{n}

If this is in series with another resistor $Z$ , the total resistance is

R_{\text{total}} = Z + \frac{R}{n}

With a battery of negligible internal resistance, the current is

I = \frac{E}{R_{\text{total}}}

Understanding the Question

You now use $n=20$ resistors in parallel, and must find the new total current. You are expected to use your experimentally determined values of $R$ and $Z$ (and the emf $E$ used previously).

Approach

Compute the parallel part $R/n$ for $n=20$ .
Add $Z$ to get total resistance.
Use $I = E/(R/n + Z)$ .

Step-by-Step Reasoning

With $n=20$ :

R_{\text{parallel}} = \frac{R}{20}

Total resistance:

R_{\text{total}} = Z + \frac{R}{20}

Current:

I = \frac{E}{Z + R/20}

Substituting representative values from earlier parts gives a current of about $1\ \text{mA}$ , which is larger than the earlier $\mu\text{A}$ values because adding more parallel resistors reduces the parallel resistance.

Key Takeaways

More resistors in parallel lowers equivalent resistance and increases current.
Use consistent units ( $\Omega$ , $\text{V}$ , $\text{A}$ ).

Common Mistakes

Using $R_{\text{parallel}} = nR$ instead of $R/n$ .
Forgetting to add $Z$ in series.
Mixing $\text{k}\Omega$ and $\Omega$ without conversion.

Things to Be Careful About

Use the same $E$ value/estimate as in part (d).
Final answer should be in amperes as requested, typically in standard form.

Techniques used

use equivalent resistance of identical resistors in parallelsubstitute constants into the circuit current equationcalculate current from emf and total resistance

Physics 9702/53 — May/June 2025

Apparatus

Variables

Procedure and measurements

Analysis (to test the relationship and find KKK)

Safety

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Analysis (to test the relationship and find $K$ )