Launch practical
About 60 minutes
- Set up the bench with the 9 items of apparatus.
- Take the readings yourself, into the results table you draw first.
- Process your results and plot the graph.
- Answer exam questions on your own results, which you mark against the mark scheme.
Runs in your browser: a 3D bench with 2D close-ups of each instrument for exact readings. Nothing to install and no account.
What you measure, and how
One run takes 46 readings. You take every one yourself, the way you would in class.
- On the 3D bench (Fig. 1) you set out the apparatus and read each instrument where it stands.
- In the 2D close-up (Fig. 2) the same instrument is drawn large, from the same run, for an exact reading. It is also how the practical runs on a device that cannot show 3D.
- In your table you write what you read, to the instrument’s resolution, and the notebook checks it against the instrument.
How to read it. Ammeter, 200 mA range: it reads to 0.1 mA. Record every digit, and the minus sign if there is one.
What it shows
V = I R
Resistance = potential difference ÷ current, at any point on an I–V graph. A resistor at constant temperature gives a straight line through the origin. A filament lamp’s resistance rises as it heats, so its graph flattens off. A diode lets current flow one way only.
The graph to expect. For the resistor the points lie on a straight line through the origin, in both directions: the current is directly proportional to the pd. The lamp gives an S-shaped curve and the diode a curve in one direction only.
On the AQA 2027 Physics Equations Sheet. Resistance at a point is V ÷ I there, not the gradient of the curve.
Apparatus
| Item | What it is for | Resolution |
|---|---|---|
| Low-voltage power pack (0–12 V DC) | Drives the current; its dial sets the pd. | — |
| Digital ammeter (multimeter) | Measures the current through the component, in series: 200 mA range (0.1 mA) or 10 A range (0.01 A). | 0.1 mA or 0.01 A |
| Digital voltmeter (multimeter on the 20 V range) | Measures the potential difference across the component, in parallel. | 0.01 V |
| Component holder with two 4 mm sockets | Holds each component in turn. | — |
| 100 Ω resistor (1 W) | The first component: an ohmic conductor at constant temperature. | — |
| Filament lamp, 12 V 24 W | The second component. | — |
| Silicon diode with protective resistor P (10 Ω) | The third component; P limits its current. | — |
| 10 Ω rheostat (5 A) | Optional: varies the lamp’s pd with the pack on 12 V. | — |
| Six 4 mm leads | Connect the circuit. | — |
Method
In our own words, as the Lab runs it. Your school's method sheet may differ in detail.
Which version are you doing?
Boards set this practical differently. Your course setting picks one; you can change it before you start.
Why: The components, the bench and the exam questions follow your board’s version.
Connect the supply + to the ammeter (lead 1)
Start at the red + socket of the power pack. Plug the other end into the ammeter’s mA socket: the resistor’s current is under 100 mA, so the 200 mA range reads it to 0.1 mA.
Why: Starting at the positive terminal and following the loop round makes the circuit easy to check against the diagram.
Connect the ammeter to the component (lead 2)
From the ammeter’s COM socket to the left-hand (+) socket of the component holder (the resistor is in it).
Why: The ammeter is in series with the component, so the same current flows through both.
Connect the component back to supply − (lead 3)
From the right-hand (−) socket of the holder back to the black − socket of the power pack. This lead is your switch.
Why: The main loop is now complete. Unplugging this lead between readings switches the current off, so the resistor stays at a constant temperature.
Choose where to connect the voltmeter
The main loop is built. Decide how to connect the voltmeter.
Do: Across the component, in parallel with it.
Why: An ammeter goes in series, in the loop. A voltmeter compares the potential at two points, so it goes across the component.
Connect voltmeter V to the component (lead 4)
From the voltmeter’s V socket to the left-hand (+) socket of the holder, the end the ammeter is joined to.
Why: The voltmeter goes across the component, in parallel, so it measures the potential difference across the component only.
Connect voltmeter COM to the other end (lead 5)
From the voltmeter’s COM socket to the right-hand (−) socket of the holder.
Why: Now the voltmeter is connected across the component, not round the loop.
Choose how to change the lamp’s pd
For the resistor and the diode you turn the power pack’s dial. For the lamp you can do the same, or keep the pack on 12 V and add a 10 Ω rheostat in series (the handbook’s circuit).
Do: Turn the power pack’s dial: 1, 2, 4, 6, 9 and 12 V.
Why: You need readings right from a small pd up to 12 V, so you can see where the lamp’s curve bends.
Choose the ammeter range for the diode
The diode with its protective resistor P takes at most about 150 mA from 2.5 V. Choose the ammeter’s range for it.
Do: 200 mA range (reads to 0.1 mA).
Why: The range sets the resolution: 0.1 mA on the 200 mA range, 0.01 A (10 mA) on the 10 A range.
Take the readings
For each component in turn, the resistor (2–8 V), the lamp (about 1–12 V) and the diode (0.4–2.5 V on the pack): set the pack, read the voltmeter and the ammeter, and switch off between readings. Then reverse the component in the circuit and take the readings again, for negative values. Plot current against potential difference for each.
Repeats: AQA’s handbook takes "several pairs of readings" in each direction rather than repeats. Taking more readings where a curve bends (the diode between 0.5 and 0.8 V, the lamp below 3 V) is worth more than repeating the straight parts.
Make safe and clear upSafety
Turn the power pack down to 0 V and switch it off. Let the lamp cool before you touch it or take it out. Unplug the leads and coil them. Turn both meters to OFF and return the components.
While you take readings
Reverse the supply for negative readings
Switch off and swap the two leads at the power pack, so the current goes through the component the other way. Then read both meters as before: they show negative values.
Why: The I–V characteristic has to show what the component does with the pd the other way round. Swapping the supply reverses the pd across the component itself.
Hazards and control measures
| Hazard | What could happen | Control measure |
|---|---|---|
| The filament lamp gets very hot (24 W) | Burns to the fingers | Do not touch the lamp while it is on, and let it cool before taking it out of the holder |
| A diode connected without its protective resistor | A very large current: the diode overheats (burns) and the meter’s fuse blows | Keep the protective resistor P in series with the diode and keep the supply at 2.5 V or below |
| The resistor run above its power rating | It gets hot enough to burn and can smoke | Keep its pd at 8 V or below and switch off between readings |
| A damaged mains lead on the power pack | Electric shock | Check the mains lead and plug before use and do not use damaged kit |
Variables
- Independent: you change
- Potential difference across the component in V
- Dependent: you measure
- Current through the component in A
- Control: you keep the same
- The component (one at a time). Change only the pd while testing each component.
- Temperature of the resistor. Low pd (8 V or below) and switch off between readings.
- The meters and their ranges. Use the same meters; change range only between components.
Results from a specimen run
| Power pack | Potential difference in V | Current in mA |
|---|---|---|
| 2 V | 2.03 | 20.3 |
| 4 V | 4.05 | 40.6 |
| 6 V | 6.07 | 60.9 |
| 8 V | 8.10 | 81.2 |
| 4 V, reversed | -4.04 | -40.6 |
| 8 V, reversed | -8.10 | -81.3 |
| Power pack | Potential difference in V | Current in A |
|---|---|---|
| 1 V | 0.76 | 0.40 |
| 2 V | 1.64 | 0.62 |
| 4 V | 3.48 | 0.96 |
| 6 V | 5.35 | 1.22 |
| 9 V | 8.19 | 1.56 |
| 12 V | 11.08 | 1.86 |
| 2 V, reversed | -1.64 | -0.62 |
| 6 V, reversed | -5.35 | -1.22 |
| 12 V, reversed | -11.08 | -1.86 |
| Power pack | Potential difference in V | Current in mA |
|---|---|---|
| 0.4 V | 0.41 | 0.0 |
| 0.6 V | 0.61 | 0.7 |
| 0.8 V | 0.72 | 9.2 |
| 1 V | 0.75 | 23.4 |
| 1.5 V | 0.80 | 63.8 |
| 2.5 V | 0.84 | 149.1 |
| 1 V, reversed | -1.03 | 0.0 |
| 2.5 V, reversed | -2.57 | 0.0 |
Processing the data
- Shape of the graph
- A straight line through the origin, the same in both directions: the current is directly proportional to the pd.
- Gradient (current per volt)
- 10.0 mA/V
- Resistance
- 99.7 Ω
- Shape of the graph
- A curve through the origin that gets less steep as the pd increases, and the same shape reversed.
- The lamp at its lowest pd
- 1.90 Ω
- The lamp at its highest pd
- 5.96 Ω
- Why the lamp’s resistance changes
- The current heats the filament. In hotter metal the ions vibrate more, so the electrons collide with them more: the resistance increases.
- What your diode graph shows
- No current in reverse. Forwards, almost none until about 0.6 V, then it rises very steeply.
- Where the diode switches on
- About 0.6–0.8 V
- Inside the filament
- Mechanism explained on the first choice
| Quantity | Specimen run | Accepted |
|---|---|---|
| Resistor (marked 100 Ω, ±5 %) | 99.7 Ω | 100 Ω |
| Lamp, hot (12 V 24 W rating: 12 ÷ 2) | 5.96 Ω | 6.00 Ω |
Across 120 simulated careful groups, resistance of the resistor (R) came out between 95.5 and 104 Ω (median 100). That is the simulation's model, not a promise of what a real class gets.
Common mistakes and the fix
Mistakes to check in this practical, and what to do instead. In a write-up or an exam answer, the mark goes to naming the step and saying what it does to the readings.
Method: In the loop, in series with the component
A voltmeter has a very high resistance (about 10 MΩ). In series it lets almost no current through: the ammeter reads 0.0 mA and the voltmeter reads the whole supply pd. You would have to move it before taking any readings.
The fix: Choose where to connect the voltmeter. An ammeter goes in series, in the loop. A voltmeter compares the potential at two points, so it goes across the component.
Method: Pack on 12 V with a 10 Ω rheostat in series
The lamp’s own resistance is small (under 6 Ω), so even the whole 10 Ω of the rheostat only brings it down to about 3 V. There are no readings below that: the start of the curve has to be guessed.
The fix: Choose how to change the lamp’s pd. You need readings right from a small pd up to 12 V, so you can see where the lamp’s curve bends.
Method: 10 A range (reads to 0.01 A)
Every current under 5 mA reads 0.00 A, so the point where the diode switches on cannot be seen, and the larger currents have only one or two significant figures.
The fix: Choose the ammeter range for the diode. The range sets the resolution: 0.1 mA on the 200 mA range, 0.01 A (10 mA) on the 10 A range.
Potential difference, resistor, 2 V: Typed 20.3 V
The fix: That is the current from the ammeter. The potential difference is on the voltmeter, the meter connected across the component: read its display.
Potential difference, resistor, 2 V: Typed -2.03 V
The fix: There is no minus sign on the display for this reading. Record the number as it shows.
Exam answer: Leaving out: potential difference = current × resistance, V = I R (or any correct rearrangement)
The fix: Potential difference = current × resistance, V = I R.
Exam answer: Leaving out: The voltmeter has a very high resistance
The fix: What is special about a voltmeter’s resistance?
Worked exam-style questions
Written for Study Hack in the style of exam questions, on the specimen run above. They are not taken from any exam board's papers. Try each one before you open the worked answer.
- Question 1V = I R · Knowledge[1 mark]
Write down the equation that links current (I), potential difference (V) and resistance (R).
Worked answer and how the marks are given
A full-mark answer
potential difference = current × resistance
V = I RMark scheme
- potential difference = current × resistance, V = I R (or any correct rearrangement)
Where marks are usually lost
- Missing: potential difference = current × resistance, V = I R (or any correct rearrangement). Potential difference = current × resistance, V = I R.
- Question 2Resistance from V and I (in mA) · Applying knowledge[2 marks]
With the power pack set to 6 V, your voltmeter read 6.07 V and your ammeter read 60.9 mA.
Calculate the resistance of the resistor.
Resistance = Ω
Worked answer and how the marks are given
A full-mark answer
I = 60.9 mA = 0.0609 A
R = V ÷ I = 6.07 ÷ 0.0609
R = 99.7 ΩMark scheme
- Current = 0.0609 A and R = V ÷ I (method)
- R = 99.7 Ω
- Allow answers from 98.67 to 100.7 Ω (the marker's tolerance; correct rounding of the student's own working is also accepted).
- Allow the unit written as ohm, ohms, Ω, Ω.
- A correct answer with no working scores full marks.
Where marks are usually lost
- Answer 0.0997 Ω (no marks). The current must be in amps: 60.9 mA = 0.0609 A, so R = 6.07 ÷ 0.0609 = 99.7 Ω.
- Answer 0.01 Ω (no marks). Upside down: R = V ÷ I = 6.07 ÷ 0.0609 = 99.7 Ω.
- Answer 10 Ω (no marks). Upside down, and the current must be in amps: R = 6.07 ÷ 0.0609 = 99.7 Ω.
- Question 3Describing the method · Knowledge[6 marks]
A student wants to find the I–V characteristic of a filament lamp, for positive and negative potential differences.
Describe a method the student could use. Your answer should include a hazard and how to reduce its risk.
Worked answer and how the marks are given
A full-mark answer
1. Connect the lamp in series with an ammeter and a variable power supply, with a voltmeter in parallel across the lamp.
2. Set the supply to 1 V, switch on and record the current from the ammeter and the potential difference from the voltmeter.
3. Increase the potential difference in steps (1, 2, 4, 6, 9, 12 V) and record the current and potential difference each time, waiting for the readings to settle.
4. Reverse the connections at the power supply and repeat the readings to get negative values.
5. Plot a graph of current against potential difference, with the origin in the middle, and draw a smooth curve.
Hazard: the lamp gets hot and can burn you, so do not touch it while it is on and let it cool before removing it.Mark scheme
Indicative content:
- Ammeter in series with the lamp; voltmeter in parallel (across the lamp) Do not accept “voltmeter in series”.
- Vary the pd using a variable power supply or a variable resistor
- Measure the current with the ammeter
- Measure the potential difference with the voltmeter
- A suitable range and interval of readings (e.g. 0 to 12 V in 1 or 2 V steps)
- Reverse the connections to the power supply to get negative values Do not accept “swap… / revers… / switch… meter… / ammeter / voltmeter lead… / connection…”.
- Plot a graph of current against potential difference
- Hazard: the lamp gets hot and can burn; do not touch it while on / let it cool
Levels of response Level Marks What the answer does 3 5–6 A coherent method that would give valid results: all the important steps are there, in a sensible order, including how negative values are obtained. 2 3–4 A method that might not give valid results: most of the important steps are there, but some are missing or out of order. 1 1–2 A method that would not give valid results: some relevant steps are given, but they are not linked into a workable method. In Study Hack a six-mark answer is not given an automatic level: you tick the points your answer makes, then choose the level that fits it.
- No relevant content: 0 marks.
Evaluate
| Source of error | Type | Effect | Improvement |
|---|---|---|---|
| The ammeter’s resolution at small currents | Random | Where the current is only a few milliamps (the diode just switching on, the resistor at 2 V) the last digit is a big fraction of the reading, so those points scatter. | Use the most sensitive range the current fits on (200 mA reads to 0.1 mA), and take more readings where the diode’s curve bends. |
| Reading the lamp before the filament had settled | Random | Straight after the pd goes up the filament is still cooler than it will be, so its resistance is lower and the current reads too high: one point sits above the curve. | Wait a few seconds for the ammeter reading to stop changing before you read it. |
| The meters’ calibration | Systematic | Each meter reads up to about 1 % high or low on every reading, so every resistance is out by the same fraction. | Check the meters against a calibrated meter, or use the same meters throughout so comparisons are fair. |
| A rheostat in series cannot take the lamp below about 3 V | Systematic | With the pack on 12 V and a 10 Ω rheostat, the lamp always has at least about 3 V across it, so the steep start of its curve is missing. | Vary the pd with the power pack’s dial, or wire the rheostat as a potential divider, so readings can start near 0 V. |
Repeatable. If you set the same pd again with the same component and meters, you read almost the same current.
Reproducible. Another group, with a different power pack and meters and their own components of the same type, gets graphs of the same shapes.
Other exam boards
How this practical lines up with other GCSE boards' required or core practicals. Codes only; check your own specification for the exact activity.
| Board | Practical | Match | What is different |
|---|---|---|---|
| AQA | Required practical 4 | This page | — |
| Edexcel | 10.17 | Similar, with differences | Part a is resistor and lamp I-V. The board combines it with series and parallel circuits in one practical. |
Print it
- Blank write-up: the nine-step lab book with the correct table headings, ready to print for a real lesson.
- Specimen lab book: this page's run written up in full, with its exam answers ticked against the mark scheme.
- Printable method card: the practical on one index card. Practise with the cards in Study Hack.
- Practical workbook: a full simulated data set to process on paper, with exam questions and answers.
- Teacher and technician pack: apparatus lists, technician notes and marking points.