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Specimen lab book: I–V characteristics
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Study Hack Lab
RP 4 · Physics
I–V characteristics
Lab book
Name
Specimen
Date
2 November 2026
Course
GCSE Physics
Board and specification
AQA GCSE Physics (8463), 4.2.1.4AQA GCSE Combined Science: Trilogy (8464), 6.2.1.4Apparatus and techniques: AT 6, AT 7
Run
C29HO
Status
Finished: every stage done
Exam questions
30/30 marks
Aim
Measure how the current through a resistor, a filament lamp and a diode depends on the potential difference across it, in both directions, and plot their I–V characteristics.
Contents
Plan and risk assessment
Apparatus
Method
Results
Graph
Analysis
Conclusion and evaluation
Make safe and clear up
Exam questions on your data
Simulated data produced by Study Hack Lab from the student's own run (seed 20261004). Every reading, value and mark here is from that run; nothing has been retyped. Layout v1, practical v2. Original material; not affiliated with AQA, which has not reviewed it.
1 Plan and risk assessment
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.
V = I R
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); Temperature of the resistor; The meters and their ranges
Prediction
What will a graph of current against potential difference look like for the resistor?
A straight line through the origin: double the pd, double the current.Right first time
At a constant temperature a resistor’s resistance does not change, so the current is directly proportional to the potential difference: a straight line through the origin, in both directions.
And for the filament lamp?
A curve that gets less steep as the pd rises, the same shape when reversed.Right first time
As the current rises the filament gets hotter and its resistance increases, so each extra volt adds less current. It does the same whichever way the current flows.
And for the diode?
No current in reverse; forwards, almost none until about 0.6 V, then a steep rise.Right first time
A diode lets current flow in one direction only. In reverse its resistance is very high. Forwards it needs about 0.6–0.7 V before the current starts to rise steeply.
Risk assessment
Hazard
Harm it could cause
Control you chose
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 holderRight first time
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 belowRight first time
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 readingsRight first time
A damaged mains lead on the power pack
Electric shock
Check the mains lead and plug before use and do not use damaged kitRight first time
2 Apparatus
✓Low-voltage power pack (0–12 V DC) Drives the current; its dial sets the pd.
✓Digital ammeter (multimeter) (resolution 0.1 mA or 0.01 A) Measures the current through the component, in series: 200 mA range (0.1 mA) or 10 A range (0.01 A).
✓Digital voltmeter (multimeter on the 20 V range) (resolution 0.01 V) Measures the potential difference across the component, in parallel.
✓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.
3 Method: your choices
Which version are you doing?: AQA required practical 4: resistor, filament lamp and diodeI–V characteristics of all three, forwards and reversed.
Choose where to connect the voltmeter: Across the component, in parallel with itThe voltmeter measures the potential difference across the component only, and the ammeter the current through it.
Choose how to change the lamp’s pd: Turn the power pack’s dial: 1, 2, 4, 6, 9 and 12 VReadings from about 0.8 V to 11 V: the steep start of the lamp’s curve is measured.
Choose the ammeter range for the diode: 200 mA range (reads to 0.1 mA)Every reading fits on the range, and the small currents where the diode starts to conduct (about 0.6 V) show up.
Reverse the supply for negative readings: DoneThe 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.
4 Results
46 of 46 readings taken.
Resistor (100 Ω) · Reversed readings are negative: write the minus sign.
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
Filament lamp (12 V 24 W)
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
Diode (with protective resistor P)
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
Anomaly: No anomalous reading.
5 Graph
The resistor: current against potential differenceThe filament lampThe diode
6 Analysis
Shape of the graph
What does your graph for the resistor show?
A straight line through the origin, the same in both directions: the current is directly proportional to the pd.Right first time
Gradient (current per volt)
gradient = change in y ÷ change in x
10.0 mA/VRight first time
Resistance
99.7 ΩRight first time
Shape of the graph
What shape is the lamp’s graph?
A curve through the origin that gets less steep as the pd increases, and the same shape reversed.Right first time
The lamp at its lowest pd
resistance = potential difference ÷ current (V = I R)
Calculate the resistance of the lamp at 0.76 V.
1.90 ΩRight first time
The lamp at its highest pd
resistance = potential difference ÷ current (V = I R)
Calculate the resistance of the lamp at 11.08 V.
5.96 ΩRight first time
Why the lamp’s resistance changes
Your lamp’s resistance went from 1.9 Ω at the lowest pd to 5.96 Ω at the highest. Why?
The current heats the filament. In hotter metal the ions vibrate more, so the electrons collide with them more: the resistance increases.Right first time
What your diode graph shows
Look at your diode graph, including the reversed readings. What does it show?
No current in reverse. Forwards, almost none until about 0.6 V, then it rises very steeply.Right first time
Where the diode switches on
At about what potential difference does the current through your diode start to rise?
About 0.6–0.8 VRight first time
Inside the filament
Mechanism explained on the first choiceRight first time
7 Conclusion and evaluation
Quantity
Your value
Accepted value
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
Conclusion
What do your three I–V graphs show?
The resistor has a constant resistance (I ∝ V); the lamp’s resistance increases as it gets hotter; the diode lets current through in one direction only, above about 0.6 V.Right first time
Your resistor gave a straight line through the origin (99.7 Ω); the lamp’s curve flattens off as the filament heats; the diode only conducts forwards.
Sources of error
The ammeter’s resolution at small currents: systematic or random?
RandomRight first time
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.
Reading the lamp before the filament had settled: systematic or random?
RandomRight first time
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.
The meters’ calibration: systematic or random?
SystematicRight first time
Each meter reads up to about 1 % high or low on every reading, so every resistance is out by the same fraction.
A rheostat in series cannot take the lamp below about 3 V: systematic or random?
SystematicRight first time
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.
Improvement
Which change would most improve your results?
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.Right first time
Repeatable or reproducible
You set 4 V across the resistor again a few minutes later and read the same current. This shows your measurement is…
RepeatableRight first time
Repeatable: the same person, method and equipment give similar results.
Another group, with their own lamp of the same type and different meters, gets the same S-shaped curve. This shows the result is…
ReproducibleRight first time
Reproducible: a different person or different equipment gives similar results.
8 Make safe and clear up
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
9 Exam questions on your data
On your own readings, self-marked against the mark scheme, with error carried forward where it applies. Total 30/30.
01.1[1 mark]
Write down the equation that links current (I), potential difference (V) and resistance (R).
Your answer
potential difference = current × resistance
1/1 self-marked
potential difference = current × resistance, V = I R (or any correct rearrangement)
Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.
Model answer: potential difference = current × resistance
V = I R
01.2[2 marks]
A student set up a circuit with a resistor, an ammeter and a voltmeter all in series with the power supply.
The ammeter read 0.0 mA at every setting of the power supply.
Explain why the ammeter read zero.
Your answer
The voltmeter has a very high resistance so almost no current can flow round the circuit.
2/2 self-marked
The voltmeter has a very high resistance
So (almost) no current can flow in the circuit
Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.
Model answer: The voltmeter has a very high resistance, so (almost) no current can flow round the circuit.
01.3[2 marks]
To get negative values of current and potential difference, one student swapped the connections at the power supply.
Another student swapped the connections at the ammeter and at the voltmeter instead.
Explain why only the first student’s method gives a valid I–V characteristic.
Your answer
Swapping the supply reverses the potential difference across the component so the current flows the other way through it. Swapping the meter leads only changes the sign of the readings; the current through the component is still in the same direction.
2/2 self-marked
Swapping the supply reverses the pd across (and current through) the component
Swapping the meter leads only changes the sign of the readings; the current through the component is still in the same direction
Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.
Model answer: Swapping the supply connections reverses the potential difference across the component, so the current through it flows the other way. Swapping the meter leads only makes the readings negative: the current through the component is still in the same direction.
01.4[2 marks]
The student then tests a 12 V 24 W filament lamp.
Give one hazard when testing the lamp and how to reduce the risk of harm.
Your answer
The lamp gets very hot and could burn your skin, so do not touch it and let it cool before removing it.
2/2 self-marked
The lamp gets hot, which could burn (the skin)
Do not touch the lamp / let it cool before handling / switch off between readings
Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.
Model answer: The lamp gets very hot and could burn your skin, so do not touch it while it is on and let it cool before removing it.
02.1[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.
Your answer
99.7
Working: I = 60.9 mA = 0.0609 A
R = V ÷ I = 6.07 ÷ 0.0609
R = 99.7 Ω
2/2
Current = 0.0609 A and R = V ÷ I
R = 99.7 Ω
Full marks. Your answer meets all the points for this question.
Model answer: I = 60.9 mA = 0.0609 A
R = V ÷ I = 6.07 ÷ 0.0609
R = 99.7 Ω
02.2[1 mark]
The I–V graph for the resistor is a straight line through the origin.
What does this show?
Tick (✓) one box.
✓ The current is directly proportional to the potential difference. (your choice)
The resistance increases as the current increases.
Current only flows in one direction.
The resistor has no resistance.
Your answer
The current is directly proportional to the potential difference.
1/1
The current is directly proportional to the potential difference.
Full marks. That is exactly what the mark scheme is looking for.
Model answer: The current is directly proportional to the potential difference (so the resistance is constant).
03.1[3 marks]
Your highest forward reading for the lamp was 11.08 V and 1.86 A.
Calculate the resistance of the lamp at this potential difference.
Your answer
5.96
Working: R = V ÷ I
R = 11.08 ÷ 1.86
R = 5.96 Ω
3/3
Uses the values at this point: 11.08 V and 1.86 A
R = 11.08 ÷ 1.86
R = 5.96 Ω
Full marks. Clear and precise.
Model answer: R = V ÷ I
R = 11.08 ÷ 1.86
R = 5.96 Ω
03.2[3 marks]
The I–V graph for the filament lamp is a curve that gets less steep as the potential difference increases.
Explain why.
Your answer
As the current increases the filament gets hotter. Its ions vibrate more so the electrons collide with them more often. The resistance of the filament increases.
3/3 self-marked
As the current (pd) increases the filament gets hotter
The ions (atoms) vibrate more, so electrons collide with them more
So the resistance of the filament increases
Self-marked: 3 of 3 marks. This is your own judgement against the mark scheme.
Model answer: As the current increases, the filament gets hotter. Its ions vibrate more, so the electrons collide with them more often. The resistance of the filament increases, so each extra volt gives a smaller increase in current.
03.3[1 mark]
Another student tested the same type of lamp. They read one current straight after turning the pd up, before the reading had settled.
Table 1 shows the student’s results.
Which result is anomalous?
Select one row.
Table 1
Potential difference in V
Current in A
0.79
0.45
1.68
0.69
3.54
1.46
5.43
1.30
8.31
1.64
11.20
1.94
Your answer
the reading at 3.54 V
1/1
The reading at 3.54 V: its current is too big to fit the smooth curve through the others
Full marks. Well done. Work out V ÷ I for each row: it should rise steadily. At 3.54 V it drops.
Model answer: The reading at 3.54 V. Its current is too big for the pattern of the others (its resistance is lower than the reading before it).
03.4[2 marks]
A student varied the pd across the lamp with a 10 Ω variable resistor in series with it and a 12 V supply.
Their lowest reading was 2.7 V, so they had no readings between 0 V and 2.7 V.
Explain why, and suggest how the student could get readings at lower potential differences.
Your answer
The variable resistor maximum resistance is not big enough compared with the lamp. Use a variable power supply and turn it down.
2/2 self-marked
The variable resistor’s maximum resistance (10 Ω) is not big enough compared with the lamp
Use a variable power supply (turn the supply down) / a potential divider / a bigger variable resistor
Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.
Model answer: The variable resistor’s maximum resistance (10 Ω) is not large compared with the lamp, so the lamp always has a few volts across it. Use a variable power supply (turned down) or a potential divider instead.
04.1[2 marks]
With the supply reversed, the ammeter in the diode circuit read 0.0 mA at every setting.
Explain what this shows about the diode.
Your answer
There is no current in the reverse direction because the diode has a very high resistance in reverse.
2/2 self-marked
There is no current through the diode in the reverse direction (current flows in one direction only)
The diode has a very high resistance in the reverse direction
Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.
Model answer: There is no current through the diode in the reverse direction: current flows through it in one direction only, because its resistance in reverse is very high.
04.2[1 mark]
The diode is connected in series with a 10 Ω protective resistor.
Why is the protective resistor needed?
Your answer
To limit the current so the diode is not damaged.
1/1 self-marked
To limit the current (so the diode / ammeter is not damaged)
Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.
Model answer: To limit the current, so that the diode (and the ammeter) is not damaged by a very large current once it conducts.
04.3[2 marks]
Another student’s graph for the diode shows the same steep curve for negative potential differences as for positive ones, for example −0.76 V and −60.0 mA.
Suggest what the student did wrong, and why it gave this graph.
Your answer
They swapped the leads at the ammeter and voltmeter instead of at the power supply. The current through the diode was still in the forward direction; only the signs of the readings changed.
2/2 self-marked
They swapped the connections at the meters (not at the power supply)
So the current through the diode was still in the forward direction; only the signs of the readings changed
Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.
Model answer: They swapped the leads at the ammeter and voltmeter instead of at the power supply. The current through the diode was still in the forward direction; only the signs of the readings changed.
05[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.
Your 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.
6/6 self-marked
Ammeter in series with the lamp; voltmeter in parallel (across the lamp)
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
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
Self-marked: Level 3, 6 of 6 marks. You ticked 8 of 8 indicative points. This is your own judgement against the mark scheme.
Model 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.
Unseen data: the same skills, new apparatus
01[2 marks]
Use this new pair from a separate constant-temperature resistor investigation.
Meter readings
Potential difference in V
Current in mA
6.08
61.2
Calculate the resistance. Use potential difference = current × resistance.
Your answer
99.3
Working: 61.2 / 1000 = 0.061200000000000004 A
6.08 / 0.061200000000000004
2/2
Current = 0.061200000000000004 A.
Resistance = 99.3 Ω.
Full marks. Well done.
Model answer: 61.2 mA = 0.061200000000000004 A.
R = V ÷ I = 6.08 ÷ 0.061200000000000004 = 99.3 Ω.
02[2 marks]
A separate simulated filament-lamp investigation gives these two settled readings.
Lamp readings
Potential difference in V
Current in A
0.82
0.46
11.46
1.99
Which two statements explain the behaviour of the filament lamp? Tick (✓) two boxes.
✓ The V/I ratio is greater at the higher potential difference, so resistance has increased. (your choice)
✓ The hotter filament has more lattice vibration, causing more collisions with moving electrons. (your choice)
Any increase in current means it is directly proportional to potential difference.
More current has been used up by the filament.
Resistance at a point is always the tangent gradient of a current-against-pd graph.
Your answer
The V/I ratio is greater at the higher potential difference, so resistance has increased.; The hotter filament has more lattice vibration, causing more collisions with moving electrons.
2/2
The V/I ratio is greater at the higher potential difference, so resistance has increased.
The hotter filament has more lattice vibration, causing more collisions with moving electrons.
Full marks. Your answer meets all the points for this question.
Model answer: V/I increases between these readings. The filament heats up; increased lattice vibration leads to more collisions and greater resistance. Current rises but not in direct proportion to potential difference.
03[2 marks]
Another student swaps the leads at both meters, leaving the supply connected to the diode the same way. Their notebook contains this pair.
Meter readings
Connections
Recorded pd in V
Recorded current in mA
Original
0.73
26.7
Both meters swapped
-0.73
-26.6
Which two actions would produce a valid investigation of the diode in both directions? Tick (✓) two boxes.
✓ Switch off and reverse the supply connections to apply reverse potential difference to the diode. (your choice)
✓ Keep the ammeter in series, the voltmeter across the diode, and the protective resistor in series; record several measured pairs in each direction. (your choice)
Keep the same supply polarity and change the signs of more readings in the table.
Remove the protective resistor to force a large reverse current.
Put the voltmeter in series so it measures the current through the diode.
Your answer
Switch off and reverse the supply connections to apply reverse potential difference to the diode.; Keep the ammeter in series, the voltmeter across the diode, and the protective resistor in series; record several measured pairs in each direction.
2/2
Switch off and reverse the supply connections to apply reverse potential difference to the diode.
Keep the ammeter in series, the voltmeter across the diode, and the protective resistor in series; record several measured pairs in each direction.
Full marks. Well done.
Model answer: Reverse the supply connections, not just the meters. Measure current in series and potential difference across the diode, retaining the protective resistor. Collect a range of measured pairs in both directions within safe component ratings.