AQA GCSE Combined Science Foundation (8464), Foundation tier · Physics › Electricity › Current, potential difference and resistance
Practise Resistors. 9 exam-style questions on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
How the resistance of ohmic conductors, filament lamps, diodes, thermistors and LDRs behaves, their I–V characteristics, and how thermistors and LDRs are used in circuits. Expect graph sketching and reading, 'explain the shape' questions and the I–V characteristics required practical.
Key facts
Ohmic conductor (constant temperature): I ∝ V, constant resistance, straight line through the origin
Filament lamp: resistance increases as the filament's temperature increases
Diode: current in one direction only; very high resistance in reverse
Thermistor: temperature up → resistance down
LDR: light intensity up → resistance down
Resistance at a point on an I–V graph: R = V ÷ I
I–V graph: current on the y-axis, pd on the x-axis
Notes
Ohmic conductors
For an ohmic conductor (such as a fixed resistor) at constant temperature, the current is directly proportional to the potential difference.
Its resistance stays the same as the current changes, so its I–V graph is a straight line through the origin.
For lamps, diodes, thermistors and LDRs the resistance is not constant: it changes as the current through them changes.
I–V characteristics (current on the y-axis, pd on the x-axis)
diagram
Resistor at constant temperature: a straight line through the origin, the same shape for negative values.
Filament lamp: a curve through the origin that gets less steep as the pd increases (in both directions). As the current increases the filament gets hotter, and its resistance increases as its temperature increases.
Diode: current flows in one direction only. In the forward direction the current stays almost zero until the pd reaches a small value, then rises steeply. In the reverse direction the diode has a very high resistance, so the current is (almost) zero.
I–V characteristics: current (I) up, potential difference (V) across.
For a straight line through the origin, a steeper line means a smaller resistance. For a curve, work out R = V ÷ I at the point you need.
Thermistors and LDRs
diagram
Thermistor: resistance decreases as temperature increases. Used as a temperature sensor, e.g. in a thermostat that switches a heating system on and off.
LDR: resistance decreases as light intensity increases. Used as a light sensor, e.g. in lights that switch on automatically when it gets dark.
Thermistor and LDR: resistance falls as temperature or light intensity rises.
Required practical: I–V characteristics
diagram
Connect the component in series with an ammeter, a variable resistor and the supply, with a voltmeter in parallel across the component.
Change the variable resistor to get a range of readings; reverse the component for negative values.
Adjust the variable resistor to get a range of pd values, and record V and I each time.
Reverse the connections to the component to get negative values of V and I.
Plot I against V. Do this for a resistor at constant temperature, a filament lamp and a diode.
How to answer each type of question
Sketch or identify an I–V graph
1 to 3 marksGrade 5
Label the axes: current (y) and potential difference (x), crossing at the origin.
Resistor: straight line through the origin. Lamp: S-shaped curve that flattens at both ends. Diode: current only for positive pd, rising steeply after a small pd.
Show negative values if the question asks for them.
Example. Describe the shape of the I–V graph for a filament lamp, for positive and negative values of potential difference.
Show the model answerHide the model answer
The curve passes through the origin (1). The curve gets less steep (flattens) as the pd increases (1). The graph has the same shape for negative values, as if rotated half a turn about the origin (1).
Don’t lose marks
Using the gradient of a curved I–V graph as the resistance. Use R = V ÷ I at the point.
Saying a thermistor's resistance increases with temperature (it decreases).
Saying the lamp's resistance changes 'because the current changes' without mentioning the temperature of the filament.
Drawing a diode graph with current flowing in both directions.
Swapping the axes: on an I–V graph, current goes on the y-axis.
More tips
Memory tricks
Both sensors go DOWN: more light (LDR) or more heat (thermistor) means less resistance.
Lamp and thermistor are opposites: a hotter filament has MORE resistance, a hotter thermistor has LESS. Learn them as a pair.
A diode is a one-way valve for current.
Straight line through the origin = ohmic = constant resistance.
Exam technique
In 'explain' questions link cause and effect with 'so': the current increases, so the filament gets hotter, so its resistance increases.
When sketching, label the axes, draw through the origin and show both quadrants if negative values are asked for.
For sensor circuits, go step by step: condition changes → resistance → current → pd across each component.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade 3
Recall that an LDR's resistance falls in lightThe resistance of an LDR decreases as light intensity increases.
Grade 4
Recall that a thermistor's resistance falls when hotThe resistance of a thermistor decreases as temperature increases.
Grade 5
Identify an ohmic conductor from its graphAt constant temperature, current is directly proportional to pd, so the I–V graph is a straight line through the origin.
Grade 5
Describe how a diode controls currentCurrent flows in one direction only; the diode has a very high resistance in the reverse direction.
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy6 marks
A student investigates how the current in a filament lamp depends on the potential difference across it, using a lamp, a variable resistor, a battery, an ammeter and a voltmeter.
(a) Write a hypothesis for this investigation.[1]
(b) Identify the independent variable and the dependent variable, and state one variable that must be controlled.[3]
(c) Give one hazard in this investigation and a suitable control measure.[2]
Show the answer and mark scheme
(a)Answer: As the potential difference across the lamp increases, the current increases, but not in direct proportion, because the resistance of the filament increases as it gets hotter.
as potential difference increases, current increases, but resistance also increases (so current is not directly proportional to potential difference), because the filament gets hotter
(b)Answer: Independent: the potential difference across the lamp (set using the variable resistor). Dependent: the current in the lamp. Control: the same lamp must be used throughout.
independent variable: the potential difference across the lamp (set by the variable resistor)
dependent variable: the current in the lamp
a control variable, e.g. using the same lamp throughout, or allowing the lamp to return to the same starting temperature before each reading
(c)Answer: A high current can make the lamp and connecting wires hot enough to burn skin; switch off between readings and avoid touching the lamp while it is lit or straight after.
a hazard, e.g. the lamp or wires become hot and could burn skin, or a short circuit could damage the power supply
a control measure that matches the hazard, e.g. switch off between readings, avoid touching the lamp while lit, or check connections before switching on
Question 2Medium4 marks
A student investigates the current–potential difference characteristic of a diode, using a variable power supply, an ammeter, a voltmeter and a protective resistor connected in series with the diode.
(a) Explain why a protective resistor is included in series with the diode.[2]
(b) Describe how the student obtains readings for negative values of potential difference across the diode.[2]
Show the answer and mark scheme
(a)Answer: Once a diode conducts, its resistance is very low, so without a protective resistor a small increase in potential difference could cause a very large, damaging current; the protective resistor limits the current to a safe value.
once the diode conducts (in its forward direction) its resistance is very low (and changes rapidly), so a small increase in potential difference could cause a very large current
the protective resistor limits the current to a safe value, so the diode is not damaged
(b)Answer: Reverse the connections to the power supply (swap the two connecting leads), so the potential difference across the diode is in the opposite direction, and take current and potential difference readings again.
reverse the connections to the power supply / swap the two connecting leads to the diode
take current and potential difference readings again, with the potential difference now in the opposite (reverse) direction