Practise Resistors. 19 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.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
3
Recall that an LDR's resistance falls in lightThe resistance of an LDR decreases as light intensity increases.
4
Recall that a thermistor's resistance falls when hotThe resistance of a thermistor decreases as temperature increases.
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.
5
Describe how a diode controls currentCurrent flows in one direction only; the diode has a very high resistance in the reverse direction.
6
Explain the filament lamp I–V graphThe line curves because the filament gets hotter as the current increases, so its resistance increases.
6
Describe the I–V characteristics practicalVary the pd with a variable resistor, record V and I, then reverse the connections for negative values.
7
Find resistance at a point on I–V graphRead V and I at that point and use R = V ÷ I, not the gradient.
7
Explain thermistor and LDR sensing circuitsLink a change in temperature or light to a change in resistance, current and the pd across each component.
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)
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.
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
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.
In series with a fixed resistor: if the sensor's resistance falls, the total resistance falls, so the current rises. The sensor takes a smaller share of the supply pd, and the pd across the fixed resistor rises. grade 7+
Required practical: I–V characteristics
Connect the component in series with an ammeter, a variable resistor and the supply, with a voltmeter in parallel across the component.
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.
Cheatsheet
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
How to answer each type of question
Sketch or identify an I–V graph
1 to 3 marks5
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 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).
Explain the shape of the filament lamp graph
3 marks6
Say that as the pd increases, the current increases.
Say that the filament gets hotter.
Say that its resistance increases as its temperature increases, so the current increases by less for each extra volt.
Example. Explain why the I–V graph for a filament lamp is not a straight line.
Show the model answer
As the pd increases, the current through the filament increases (1). The filament gets hotter / its temperature increases (1). The resistance of the filament increases as its temperature increases, so the current does not increase in proportion to the pd (1).
Calculate resistance from graph readings
2 to 3 marks7
Read a pair of values (V and I) from the graph at the point asked about.
Use R = V ÷ I for that point. Do not use the gradient of a curve.
Compare values to describe how the resistance changes.
Example. At a potential difference of 6.0 V the current through a filament lamp is 0.25 A. At 12 V the current is 0.40 A. (a) Calculate the resistance of the lamp at 6.0 V and at 12 V. (b) Describe how the resistance of the lamp changes as the potential difference increases.
Show the model answer
(a) R = 6.0 ÷ 0.25 = 24 Ω (1) R = 12 ÷ 0.40 = 30 Ω (1) (b) The resistance increases as the pd increases (1)
Explain how a sensing circuit works
3 to 4 marks7
Start with what happens to the sensor's resistance.
Then the total resistance and the current.
Then the pd across each component (the pd is shared in proportion to resistance).
End with what the meter reads or what the circuit does.
Example. A thermistor and a fixed resistor are connected in series to a 6.0 V battery. A voltmeter is connected across the fixed resistor. Explain what happens to the reading on the voltmeter when the temperature of the thermistor increases.
Show the model answer
The resistance of the thermistor decreases (1). The total resistance decreases, so the current in the circuit increases (1). The pd across the fixed resistor increases, because V = I R and its resistance does not change / the thermistor now takes a smaller share of the 6.0 V (1). So the voltmeter reading increases (1).
Shortcuts and 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.
Where marks are lost
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.
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.
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
Question 3Hard6 marks
A student wants to plot a current–potential difference graph for a filament lamp, obtaining data for both positive and negative values of potential difference.
Describe a method the student could use, including how both positive and negative values of potential difference and current are obtained.[6]
Show the answer and mark scheme
Answer: Connect the lamp in series with a variable resistor (or variable power supply), an ammeter, with a voltmeter across the lamp; take readings over a range of positive potential differences; reverse the connections to obtain negative values; repeat and plot a graph.
connect the lamp in series with a variable resistor (or use a variable power supply) and an ammeter, with a voltmeter connected in parallel across the lamp
adjust the variable resistor (or power supply) to obtain a range of potential difference values, e.g. 0 V up to the maximum safe value, recording the current at each
repeat each reading and calculate a mean, to check the results are repeatable
reverse the connections to the power supply (swap the two leads) so that current now flows through the lamp in the opposite direction
take current and potential difference readings again with the connections reversed, to obtain negative values
take enough readings to plot a smooth curve, including some closely spaced near 0 V
plot a graph of current (y-axis) against potential difference (x-axis), with axes through the origin, and draw a smooth curve through the points
Marked with levels of response: the full level descriptors are in the app.