Science required practicals
How do you do the I-V characteristics required practical?
Set up a circuit with the component, an ammeter in series and a voltmeter across the component. Change the potential difference in steps with a variable resistor, record the current each time, then reverse the component to get negative values. Plot current against p.d. A fixed resistor gives a straight line through the origin. A filament lamp gives a curve, because its resistance rises as it heats up, and a diode lets current flow one way only.
Last updated 23 September 2026 · Written and fact-checked by the GCSEwiz team
What does the I-V characteristics practical investigate?
It shows how the current through a component changes as you change the potential difference (p.d.) across it. The graph of current against p.d. is the component's I-V characteristic, and its shape tells you whether the resistance stays the same or changes. You test a fixed resistor at constant temperature, a filament lamp and a diode.
It is a required practical on AQA GCSE Physics and on AQA GCSE Combined Science: Trilogy.
What equipment do you need?
You need the three components and a circuit that lets you vary the p.d. while you measure the current and the p.d.
- A low-voltage power supply or battery pack, and a switch
- A variable resistor, to change the p.d. across the component
- An ammeter and a voltmeter (or two multimeters)
- A fixed resistor, a filament lamp in a holder and a diode
- A protective resistor to go in series with the diode
- Connecting leads
What is the method?
You take a set of current and p.d. readings for each component in turn, first one way round and then the other.
- Connect the power supply, switch, variable resistor, ammeter and fixed resistor in series.
- Connect the voltmeter across the fixed resistor only.
- Close the switch and set the variable resistor to give a small p.d. Record the p.d. and the current, then open the switch.
- Change the variable resistor in steps, recording a reading each time, until you have a good spread of values.
- Turn the component round in the circuit (or swap the power supply leads) and repeat. Record these readings as negative values.
- Repeat the whole process with the filament lamp, then with the diode, keeping its protective resistor in series and the voltmeter across the diode alone.
What are the variables?
You set the p.d. across the component and measure the current that results. For the lamp you cannot keep the temperature steady, and you should not try: the filament heating up is the effect you are looking for.
| Variable | In this practical |
|---|---|
| Independent variable | The potential difference across the component |
| Dependent variable | The current through the component |
| Control variable | The component: one resistor, lamp or diode for its whole set of readings |
| Control variable | The temperature of the fixed resistor, kept steady with small currents and the switch opened between readings |
How do you work out and present the results?
Find the resistance at any point with R = V ÷ I, using the p.d. and the current at that point. Don't use the gradient: on an I-V graph a straight line's gradient is 1 ÷ R, and on a curve the gradient does not give the resistance at all.
Worked example: a filament lamp carries 0.20 A at 2.0 V, so R = 2.0 ÷ 0.20 = 10 Ω. At 6.0 V it carries 0.30 A, so R = 6.0 ÷ 0.30 = 20 Ω. The p.d. tripled but the current only went up by half, because the resistance doubled as the filament got hotter.
Plot current on the y-axis against p.d. on the x-axis, with negative values too, so the origin sits in the middle of the graph. The three shapes to know:
- Fixed resistor: a straight line through the origin. Current is directly proportional to p.d., so the resistance stays the same. A steeper line means a smaller resistance.
- Filament lamp: a curve through the origin that gets less steep as the p.d. goes up, in both directions. The filament gets hotter and its resistance increases.
- Diode: almost no current in the reverse direction, where its resistance is very high. In the forward direction the current stays near zero until the p.d. passes a small value, then rises steeply.
How do you make it accurate and safe?
Keep the currents small and open the switch between readings, so the fixed resistor does not warm up and change its resistance. Take readings close together where a graph bends, such as the point where the diode starts to conduct, so the shape comes out clearly. Digital multimeters save you judging a needle between two marks.
On safety: the lamp gets hot, so don't touch the bulb while it is on, and resistors can warm up too. Always keep the protective resistor in series with the diode, or a large current can damage it.
What do exam questions ask about it?
Questions often show you a graph and ask you to name the component and explain its shape. Others want the resistance at one point on it. You may also be asked to draw the circuit with the right symbols, or to explain why the lamp's resistance changes.
Here is one we wrote in the exam style: "The current through a component is 0.25 A when the p.d. across it is 3.0 V, and 0.40 A when the p.d. is 6.0 V. Calculate the resistance at each p.d. and name the component. (4 marks)" The resistances are 12 Ω and 15 Ω. Resistance rising with p.d. points to a filament lamp.
Common mistakes
- Putting the voltmeter across the whole circuit. It goes across the component only.
- Working out resistance from the gradient. Use R = V ÷ I with one pair of readings.
- Leaving out the negative half. Reverse the component, or you will miss the diode blocking current in the reverse direction.
- Joining the lamp's points with straight lines. Draw one smooth curve through the origin.
- Forgetting the protective resistor in the diode circuit. Without it, a large current can flow and damage the diode.