You may not need to use all of the materials provided. In this experiment, you will investigate the relationship between the power dissipated in a filament lamp and the resistance of the lamp. (a) Assemble the circuit of Fig. 1.1. (b) Set the power supply voltage to $12\ \text{V}$ and close the switch so that the lamp lights. Record the voltmeter reading $V$ and the ammeter reading $I$. $V$ = \_\_\_\_\_\_ $\text{V}$ $I$ = \_\_\_\_\_\_ $\text{A}$ (c) Reduce the power supply…

Physics 9702/35 — May/June 2010

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

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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 the relationship between the power dissipated in a filament lamp and the resistance of the lamp.

(a)

Assemble the circuit of Fig. 1.1.

Fig. 1.1

DifficultyEasy

Worked solution

Answer

Circuit assembled as in Fig. 1.1, with the ammeter in series with the lamp and the voltmeter connected in parallel across the lamp.

Final answer

Circuit assembled as in Fig. 1.1.

Detailed explanation

Background Concept

In d.c. circuits:

An ammeter measures current and must be placed in series so the same current flows through it as through the component.
A voltmeter measures potential difference and must be placed in parallel across the component so it measures the p.d. across that component.

Ammeters have very low resistance (so they do not change the circuit current much). Voltmeters have very high resistance (so they draw negligible current).

Understanding the Question

You are asked to build the circuit shown in Fig. 1.1 so you can measure:

the current $I$ through the filament lamp (ammeter reading), and
the potential difference $V$ across the lamp (voltmeter reading).

These two readings are needed later to calculate power $P$ and resistance $R$ .

Approach

Follow the diagram exactly:

Put the supply, switch, and ammeter all in one series loop with the lamp.
Connect the voltmeter directly across the two terminals of the lamp.
Check that connections are secure and polarities are sensible (meter reads positive).

Step-by-Step Reasoning

diagram-a1

Connect the power supply to the switch.
From the switch, connect to the ammeter, then to the lamp, and back to the power supply: this makes the lamp current pass through the ammeter.
Connect the voltmeter across the lamp terminals (one lead to each side of the lamp). This ensures the voltmeter reads the p.d. across the lamp only.
Before switching on, check for loose leads and correct meter ranges.

Key Takeaways

Ammeter in series, voltmeter in parallel.
Correct placement ensures $I$ is the lamp current and $V$ is the lamp p.d.

Common Mistakes

Putting the ammeter in parallel (can short-circuit and blow a fuse).
Putting the voltmeter in series (gives meaningless readings because of its high resistance).
Measuring $V$ across more than just the lamp (e.g. including the ammeter or leads).

Things to Be Careful About

Ensure the voltmeter is across the lamp, not across the whole supply.
Use appropriate ranges (start high, then reduce for better resolution).
Switch off/open the switch when changing connections.

Techniques used

assemble a series circuit with an ammeterconnect a voltmeter in parallel across a componentcheck correct polarity and secure connections

(b)

Set the power supply voltage to $12\ \text{V}$ and close the switch so that the lamp lights. Record the voltmeter reading $V$ and the ammeter reading $I$ .

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

DifficultyEasy

Worked solution

Answer

(Example readings)

$V = 12.1\ \text{V}$

$I = 0.30\ \text{A}$

Final answer

Example: V = 12.1 V, I = 0.30 A

Detailed explanation

Background Concept

To determine how a lamp behaves electrically, you measure:

Potential difference across it, $V$ (volts), using a voltmeter in parallel.
Current through it, $I$ (amperes), using an ammeter in series.

Good experimental practice is to record values to the resolution of the instrument (e.g. 0.1 V, 0.01 A) and to include units.

Understanding the Question

You are told to set the supply to $12\ \text{V}$ and then record the voltmeter reading $V$ and ammeter reading $I$ while the lamp is lit.

Note: the voltmeter reading across the lamp may not be exactly $12\ \text{V}$ if there are internal resistances or other voltage drops, so you must record the meter readings rather than assume $V = 12\ \text{V}$ .

Approach

Set the supply to approximately $12\ \text{V}$ .
Close the switch and allow the reading to settle.
Read $V$ from the voltmeter and $I$ from the ammeter.
Record each with sensible precision and units.

Step-by-Step Reasoning

Adjust the power supply control until its output is near $12\ \text{V}$ .
Close the switch; the lamp heats up and may cause readings to change briefly before stabilising.
Read the voltmeter across the lamp: record to the smallest division/display.
Read the current on the ammeter: again record to the instrument resolution.

(Your actual values depend on the specific lamp and power supply; the values in the solution are a representative example.)

Key Takeaways

Always record the measured $V$ and $I$ , not the nominal supply setting.
Use correct units and appropriate decimal places.

Common Mistakes

Writing $V = 12\ \text{V}$ without reading the voltmeter.
Omitting units.
Over-precision (e.g. writing $12.137\ \text{V}$ when the meter reads only to $0.1\ \text{V}$ ).

Things to Be Careful About

If using analogue meters, avoid parallax error (eye level with the pointer).
Ensure the lamp is not overheating; keep readings brief and open the switch between adjustments if needed.

Techniques used

set the power supply to a stated voltageread analogue/digital meters to appropriate precisionrecord measured quantities with units

(c)

Reduce the power supply voltage, recording $V$ and $I$ until you have six sets of readings.

Open the switch when you have finished your measurements.

Include in your table of results values of $P$ , $R$ and $R^4$ , where $P$ is the power dissipated in the lamp and $R$ is the resistance of the lamp.

(P = VI \text{ and } R = \frac{V}{I})

DifficultyMedium

Worked solution

Answer

Take six sets of readings of $V$ and $I$ as the supply voltage is reduced.

Calculate for each set:

P = VI

R = \frac{V}{I}

R^4 = \left(\frac{V}{I}\right)^4

Record all values in one table with headings and units (example shown).

$V/\text{V}$	$I/\text{A}$	$P/\text{W}$	$R/\Omega$	$R^4/\Omega^4$
12.1	0.30	3.63	40.3	$2.64\times 10^{6}$
9.3	0.26	2.42	35.8	$1.64\times 10^{6}$
7.0	0.22	1.54	31.8	$1.02\times 10^{6}$
5.1	0.18	0.918	28.3	$6.42\times 10^{5}$
3.5	0.15	0.525	23.3	$2.95\times 10^{5}$
2.3	0.12	0.276	19.2	$1.36\times 10^{5}$

(Open the switch when finished.)

Final answer

See working (table of V, I, and calculated P, R, R^4).

Detailed explanation

Background Concept

The question requires you to investigate how the power dissipated in a filament lamp depends on its resistance.

The key electrical relationships are:

P = VI

and

R = \frac{V}{I}

So if you measure $V$ and $I$ for different operating conditions, you can compute:

$P$ (power dissipated in the lamp), and
$R$ (lamp resistance at that operating point).

You are also asked to calculate $R^4$ because plotting $P$ against $R^4$ may produce a straight-line graph.

Understanding the Question

You must:

Reduce the supply voltage in steps and collect six pairs of readings $(V, I)$ .
Put all raw and calculated values into a single results table.
Include calculated columns for $P$ , $R$ and $R^4$ .

The independent variable you control is effectively the supply setting (which changes $V$ and $I$ ), and you observe how $P$ and $R$ change.

Approach

Choose about six different supply settings spanning a good range (from near maximum brightness down to dim, but still measurable).
For each setting, wait for readings to stabilise (filament temperature affects $R$ ).
Record $V$ and $I$ .
Compute $P$ and $R$ from the measured values, then compute $R^4$ .
Present all values with correct headings, units, and sensible significant figures.

Step-by-Step Reasoning

Collecting data
- Start at the highest setting (around $12\ \text{V}$ ) and record $V$ and $I$ .
- Reduce the supply in steps, recording new $V$ and $I$ each time, until you have six sets.
- Using a wide range matters because it makes any relationship clearer on a graph and reduces percentage uncertainty in the gradient.
Calculating $P$
- Multiply the measured values:

P = VI

Units check: $\text{V} \times \text{A} = \text{W}$ .

Calculating $R$

R = \frac{V}{I}

Units check: $\text{V}/\text{A} = \Omega$ .

Calculating $R^4$

R^4 = (R)^4

This number becomes large quickly, so it is often best written in standard form (e.g. $1.02 \times 10^{6}$ ).

Table quality
- One clear table containing all data is expected.
- Column headings should be of the form quantity/unit (e.g. $V/\text{V}$ ).
- Use consistent decimal places for a column of measured quantities (e.g. all $V$ to 0.1 V if that is your meter resolution).

(The numerical table in the solution is an example format; your values will depend on the lamp and readings obtained.)

Key Takeaways

Measure $V$ and $I$ directly, then compute $P$ , $R$ , and $R^4$ .
Good tables need headings with units, consistent precision, and a sensible range of data.

Common Mistakes

Fewer than six sets of readings.
Missing units in headings, or mixing units within a column.
Inconsistent precision (e.g. $0.2\ \text{A}$ then $0.23\ \text{A}$ without justification).
Calculating $R$ using $I/V$ instead of $V/I$ .
Arithmetic errors when raising $R$ to the fourth power.

Things to Be Careful About

The filament lamp is non-ohmic: $R$ changes with temperature, so allow readings to stabilise.
If $I$ becomes very small at low voltage, the percentage uncertainty in $R = V/I$ becomes large; avoid readings where the ammeter resolution dominates.
When computing $R^4$ , use enough significant figures in $R$ to avoid excessive rounding error, then round $R^4$ sensibly (often 3 s.f.).

Techniques used

collect a suitable range of paired measurementscalculate derived quantities using given formulaerecord results in a table with headings and unitsuse consistent significant figures within each column

(d)

(i)

Plot a graph of $P$ on the $y$ -axis against $R^4$ on the $x$ -axis.

DifficultyMedium-Easy

Worked solution

Answer

Plot $P$ (in $\text{W}$ ) on the $y$ -axis against $R^4$ (in $\Omega^4$ ) on the $x$ -axis, using an appropriate scale and plotting all six points.

Final answer

Graph of P (y) against R^4 (x) with correct labels and plotted points.

Detailed explanation

Background Concept

A graph is used to test whether two variables have a linear relationship. If the suggested relationship is of the form

P = aR^4 + b

then plotting $P$ (vertical axis) against $R^4$ (horizontal axis) should produce a straight line.

Good graphing practice includes:

axes labelled with quantity and unit,
a scale that uses at least half the grid in each direction,
neat, accurate points (small crosses or dots in circles),
avoiding awkward scales (e.g. 3 squares = 1 unit).

Understanding the Question

You have calculated $P$ and $R^4$ for six readings. You must now transfer these pairs $(R^4, P)$ onto a graph with:

$P$ on the $y$ -axis,
$R^4$ on the $x$ -axis.

Approach

Decide sensible axis ranges that include all your data.
Label axes: $P/\text{W}$ and $R^4/\Omega^4$ .
Because $R^4$ may be large, use a scaled axis such as $R^4/(10^6\ \Omega^4)$ if helpful (but label it clearly).
Plot all six points accurately.

Step-by-Step Reasoning

diagram-d1

From your table, take each $R^4$ value as an $x$ -coordinate and the corresponding $P$ as a $y$ -coordinate.
Choose a scale so the smallest and largest values are well separated on the page.
Plot each point carefully; if one point appears anomalous, still plot it (do not omit points unless instructed).

Key Takeaways

To test $P = aR^4 + b$ , the correct graph is $P$ vs $R^4$ .
Clear labels and sensible scales are essential for marks.

Common Mistakes

Swapping axes (plotting $R^4$ on $y$ and $P$ on $x$ ).
Missing units or unclear axis labels.
Using a compressed scale so points bunch together.
Plotting $R$ instead of $R^4$ .

Things to Be Careful About

If you rescale the $x$ -axis (e.g. dividing by $10^6$ ), remember it affects the numerical value of the gradient unless you account for the factor.
Points should be precise; avoid large blobs that hide accuracy.

Techniques used

choose suitable axis scales using most of the 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 (balanced so there are roughly equal numbers of points 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 in data when scatter is present due to random uncertainties. For a linear relationship, you should draw one straight line that best represents all points, not join-the-dots.

A good best-fit line:

is a single straight line,
passes through the middle of the cluster of points,
has roughly equal scatter of points above and below.

Understanding the Question

After plotting $P$ against $R^4$ , you are asked to draw the straight line of best fit. This line will then be used to find the gradient and intercept.

Approach

Use a ruler.
Visually judge the overall straight-line trend.
Do not force the line through every point or through the origin unless the points strongly justify it.

Step-by-Step Reasoning

Look for the general trend of the plotted points.
Place a ruler so the line passes as centrally as possible through the data.
Adjust so the number (and spread) of points above the line is similar to the number (and spread) below.
Draw the line as long as practical across the graph area to allow accurate gradient/intercept readings.

Key Takeaways

Best fit is about the overall trend, not perfect agreement with every point.
A long, well-placed line improves the accuracy of gradient/intercept.

Common Mistakes

Joining points with segments.
Drawing a line that goes through an outlier at the expense of the rest.
Drawing a very short line (makes gradient less accurate).

Things to Be Careful About

If one point is clearly anomalous, the best-fit line can still be drawn to fit the remaining trend, but do not automatically exclude points without good reason.
Keep the line thin and precise; a thick line increases reading uncertainty.

Techniques used

draw a straight line of best fit through plotted pointsbalance points above and below the lineignore minor scatter consistent with experimental uncertainty

(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 (example):

\text{gradient} = \frac{\Delta P}{\Delta (R^4)}

\text{gradient} = \frac{3.63 - 0.276}{(2.64\times 10^{6}) - (1.36\times 10^{5})} = 1.34\times 10^{-6}\ \text{W }\Omega^{-4}

$y$ -intercept (at $R^4 = 0$ ) $\approx 0.09\ \text{W}$ .

Answer

gradient $\approx 1.3\times 10^{-6}\ \text{W }\Omega^{-4}$

$y$ -intercept $\approx 0.09\ \text{W}$

Final answer

gradient ≈ 1.3×10^-6 W Ω^-4, y-intercept ≈ 0.09 W (example)

Detailed explanation

Background Concept

For a straight-line graph, the general form is

y = mx + c

where:

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

On a graph of $P$ (y-axis) against $R^4$ (x-axis):

gradient $m = \Delta P / \Delta (R^4)$ ,
intercept $c$ is the value of $P$ when $R^4 = 0$ .

Units:

$P$ is in $\text{W}$ ,
$R^4$ is in $\Omega^4$ ,
so gradient units are $\text{W}/\Omega^4 = \text{W }\Omega^{-4}$ .

Understanding the Question

You must use your best-fit line (not individual data points) to determine:

the gradient, and
the y-intercept.

These will be used in the next part to identify constants in the proposed equation.

Approach

Pick two points on the drawn best-fit line that are far apart (to reduce percentage reading error).
Read off their coordinates $(R_1^4, P_1)$ and $(R_2^4, P_2)$ .
Compute:

\text{gradient} = \frac{P_2 - P_1}{R_2^4 - R_1^4}

Read the $y$ -intercept by extending the line to $R^4 = 0$ and reading $P$ there.

Step-by-Step Reasoning

diagram-d3

Gradient: draw a large triangle on the best-fit line. The vertical side is $\Delta P$ and the horizontal side is $\Delta(R^4)$ . Then calculate $\Delta P / \Delta(R^4)$ .
Using points far apart makes $\Delta P$ and $\Delta(R^4)$ large, so any small reading uncertainty has a smaller percentage effect.
Intercept: where the best-fit line crosses the $P$ -axis (at $R^4=0$ ) is the y-intercept. If the line does not reach the axis within the graph, extend it with a ruler.

The numerical values depend on the graph you have drawn. The worked numbers in the solution are an example to show method and correct units.

Key Takeaways

Always use the best-fit line, not point-to-point calculations.
Gradient is $\Delta y/\Delta x$ , with correct units.
The intercept is read at $x=0$ .

Common Mistakes

Using $\Delta x/\Delta y$ instead of $\Delta y/\Delta x$ .
Choosing two data points instead of two points on the best-fit line.
Using two points too close together (large percentage uncertainty).
Forgetting units, especially the unusual $\Omega^{-4}$ .

Things to Be Careful About

If you scaled the $x$ -axis (e.g. plotted $R^4/(10^6\ \Omega^4)$ ), then the gradient you calculate from the graph must be converted back by the scale factor.
When reading coordinates, use the correct number of significant figures consistent with graph-reading precision.

Techniques used

determine the gradient using a large triangle on the best-fit linecalculate gradient as \Delta y / \Delta x with unitsread the y-intercept from the graph at x = 0

(e)

It is suggested that the relationship between $P$ and $R$ is

P = aR^4 + b

where $a$ and $b$ are constants.

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

$a$ = ______
$b$ = ______

DifficultyMedium-Easy

Worked solution

Working

Given

P = aR^4 + b

Comparing with $y = mx + c$ for a graph of $P$ (y-axis) against $R^4$ (x-axis):

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

Units:

[a] = \frac{\text{W}}{\Omega^4} = \text{W }\Omega^{-4},\quad [b] = \text{W}

Answer

$a = 1.3\times 10^{-6}\ \text{W }\Omega^{-4}$

$b = 0.09\ \text{W}$

Final answer

a = gradient (W Ω^-4), b = y-intercept (W)

Detailed explanation

Background Concept

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

y = mx + c

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

gradient $m$ ,
intercept $c$ .

Here the suggested model is

P = aR^4 + b

This is already linear in $R^4$ , so if you plot $P$ against $R^4$ :

$a$ corresponds to the gradient,
$b$ corresponds to the $y$ -intercept.

Understanding the Question

You have already found the gradient and y-intercept from your graph in part (d)(iii). This part asks you to convert those graphical values into the constants $a$ and $b$ in the proposed equation, and to state the units.

Approach

Identify $y$ and $x$ for the graph: $y=P$ , $x=R^4$ .
Compare $P = aR^4 + b$ with $y = mx + c$ .
Set $a=m$ and $b=c$ .
Determine units from the axes.

Step-by-Step Reasoning

Since the graph is $P$ (y-axis) against $R^4$ (x-axis), the straight-line equation is:

P = (\text{gradient})\,(R^4) + (y\text{-intercept})

Comparing with the model $P = aR^4 + b$ gives:

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

Units:
- $P$ is in watts (W).
- $R^4$ is in $\Omega^4$ .
- Therefore

[a] = \frac{\text{W}}{\Omega^4} = \text{W }\Omega^{-4}

and

[b] = \text{W}

Substitute your measured gradient and intercept values to obtain numerical values of $a$ and $b$ .

Key Takeaways

When you plot the variables in the correct linear form, constants come directly from gradient and intercept.
Units of gradient come from (units of $y$ )/(units of $x$ ).

Common Mistakes

Swapping $a$ and $b$ .
Giving $a$ the wrong unit (e.g. $\text{W}\Omega^4$ instead of $\text{W }\Omega^{-4}$ ).
Using values from two data points rather than the gradient/intercept from the best-fit line.

Things to Be Careful About

If your $x$ -axis was scaled (e.g. $R^4/(10^6\ \Omega^4)$ ), convert the gradient back to per $\Omega^4$ before quoting $a$ .
Quote $a$ and $b$ to a sensible number of significant figures consistent with the graph-reading uncertainty.

Techniques used

match a straight-line graph to y = mx + cidentify constants from gradient and interceptassign units to constants from the graph axes

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