Practise Current, resistance and potential difference. 20 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
How the current through a component depends on the potential difference across it and its resistance, and the equation V = I R. This is one of the most used equations in Paper 1. The required practical on the resistance of a wire and of resistors in series and parallel is also assessed here.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
3
State the unit of resistanceResistance is measured in ohms (Ω).
4
Describe how resistance affects currentFor a given potential difference, the greater the resistance, the smaller the current.
4
Calculate pd using V = I RMultiply the current in amps by the resistance in ohms to get the pd in volts.
5
Rearrange V = I R for I or RUse I = V ÷ R and R = V ÷ I, converting kΩ and mA first.
6
Describe the resistance of a wire practicalMeasure V and I for different lengths of wire and calculate R = V ÷ I each time.
7
Analyse resistance against length resultsA straight line through the origin shows resistance is directly proportional to length.
Notes
Current, potential difference and resistance
The current through a component depends on the potential difference (pd) across it and on its resistance.
For a given pd, the greater the resistance, the smaller the current.
For a given resistance, the greater the pd, the greater the current.
Potential difference is measured in volts (V) with a voltmeter. Resistance is measured in ohms (Ω).
Exam questions say 'potential difference', but you get credit for 'voltage' too.
The equation V = I R
potential difference = current × resistance: V = I R (V in volts, I in amps, R in ohms).
Rearranged: I = V ÷ R and R = V ÷ I.
Convert first: kΩ × 1000 = Ω and mA ÷ 1000 = A.
Required practical: resistance of a wire
Fix a wire along a metre ruler. Connect it in series with an ammeter and a power supply, using crocodile clips, and connect a voltmeter in parallel across the length being tested.
Measure the current and the pd for a range of lengths (e.g. 10 cm to 100 cm) and calculate R = V ÷ I for each length.
Keep the same wire (material and thickness) throughout, and keep the temperature constant: use a low current and switch off between readings, because a hotter wire has a higher resistance.
Plot resistance (y-axis) against length (x-axis). For a uniform wire you get a straight line through the origin: resistance is directly proportional to length.
The same circuit is used to find the resistance of resistors joined in series and in parallel: measure V and I for each combination and calculate R = V ÷ I.
Cheatsheet
V = I R: pd (V) = current (A) × resistance (Ω)
I = V ÷ R and R = V ÷ I
Greater resistance → smaller current (for the same pd)
1 kΩ = 1000 Ω; 1 mA = 0.001 A
Resistance of a wire ∝ length (at constant temperature)
Ammeter in series with the wire; voltmeter in parallel across it
How to answer each type of question
Calculate using V = I R
2 marks4
Write V = I R (or the rearranged form you need).
Substitute values in V, A and Ω.
Give the answer with its unit.
Example. The current through a resistor is 0.30 A. The resistance of the resistor is 40 Ω. Calculate the potential difference across the resistor.
Show the model answer
V = 0.30 × 40 (1) V = 12 V (1)
Rearrange V = I R, with unit conversions
3 marks6
Convert kΩ to Ω and mA to A.
Rearrange to I = V ÷ R or R = V ÷ I.
Calculate, then convert to the unit the question asks for.
Example. A potential difference of 6.0 V is applied across a resistor of resistance 2.4 kΩ. Calculate the current through the resistor. Give your answer in milliamps.
Show the model answer
R = 2.4 × 1000 = 2400 Ω (1) I = V ÷ R = 6.0 ÷ 2400 = 0.0025 A (1) I = 2.5 mA (1)
Describe a method: the resistance of a wire
4 to 6 marks6
Describe the circuit: wire, ammeter and power supply in series; voltmeter across the wire.
Say how you measure and change the length (metre ruler, crocodile clips).
Say what you record and how you calculate resistance (R = V ÷ I).
Give a range of lengths and a control (same wire, low current so the temperature stays constant).
Example. A student investigates how the resistance of a wire depends on its length. She has a power supply, an ammeter, a voltmeter, a metre ruler, two crocodile clips and a long piece of wire. Describe a method she could use. (4 marks)
Show the model answer
Connect the wire in series with the ammeter and the power supply, with the voltmeter connected in parallel across the length of wire being tested (1). Use the metre ruler to set the length of wire between the crocodile clips, e.g. 20 cm (1). Record the ammeter and voltmeter readings and calculate the resistance using R = V ÷ I (1). Repeat for at least five different lengths, e.g. 20, 40, 60, 80 and 100 cm, keeping the current low and switching off between readings so the temperature of the wire stays constant (1).
Describe a relationship and make a prediction
2 to 3 marks7
Straight line through the origin: say 'directly proportional'.
Straight line not through the origin: say 'as length increases, resistance increases' (a linear relationship).
For a prediction, scale up in proportion or read the value from the line.
Example. A graph of resistance against length for a wire is a straight line through the origin. A 50 cm length of the wire has a resistance of 1.6 Ω. (a) Describe the relationship between resistance and length. (b) Calculate the resistance of 80 cm of the same wire.
Show the model answer
(a) Resistance is directly proportional to length (1) (b) R = 1.6 × (80 ÷ 50) (1) R = 2.56 Ω, so about 2.6 Ω (1)
Shortcuts and memory tricks
Formula triangle: V on top, I and R underneath.
Sense check: 12 V across 4 Ω gives 3 A, not 48 A. More resistance should always mean less current.
Directly proportional = straight line through the origin: double the length, double the resistance.
Where marks are lost
Multiplying when you should divide: I = V ÷ R, not V × R.
Forgetting to convert kΩ to Ω or mA to A.
Saying 'directly proportional' when the line does not go through the origin.
Letting the wire heat up, which increases its resistance and spoils the results.
Connecting the voltmeter across the power supply instead of across the wire being tested.
Exam technique
V = I R must be learned: it is not on the equation sheet.
In method questions, name the equipment, say what you measure and how you calculate resistance, give the range of values and say how you keep the test fair.
Show R = V ÷ I working for each value. The method mark can be awarded even if you slip on the arithmetic.
If a graph line misses the origin, suggest a systematic error, such as the crocodile clip not being exactly at the zero mark on the ruler.
Required practical:Resistance (method, variables and exam tips)
Quick recall
Cover the answers and test yourself. The app has these as flashcards that come back just before you'd forget them.
Write down the equation that links current (I), potential difference (V) and resistance (R).
V = I R
A current of 0.20 A flows through a component of resistance 25 Ω. Calculate the potential difference across the component. Use the equation: potential difference = current × resistance
5.0 V
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy4 marks
A resistor is connected to a battery.
(a) What is the unit of resistance? Tick (✓) one box.[1]
ampere
ohm
volt
watt
(b) The current in the resistor is 0.50 A. The resistance of the resistor is 12 Ω. Calculate the potential difference across the resistor. Use the equation: potential difference = current × resistance[2]
(c) The resistor is replaced with a resistor that has a greater resistance. The potential difference across it stays the same. What happens to the current in the resistor? Tick (✓) one box.[1]
It decreases.
It increases.
It stays the same.
Show the answer and mark scheme
(a)Answer: ohm
(b)Answer: 6.0 V
V = 0.50 × 12
V = 6.0 (V)
(c)Answer: It decreases.
Question 2Medium7 marks
Electrical components have resistance.
(a) Write down the equation that links current (I), potential difference (V) and resistance (R).[1]
(b) The potential difference across the heating element of a kettle is 230 V. The current in the element is 4.6 A. Calculate the resistance of the heating element.[3]
(c) The current in a 150 Ω resistor is 20 mA. Calculate the potential difference across the resistor.[3]
Show the answer and mark scheme
(a)Answer: V = I R
V = I × R / potential difference = current × resistance
(b)Answer: 50 Ω
230 = 4.6 × R
R = 230 ÷ 4.6
R = 50 (Ω)
(c)Answer: 3.0 V
I = 0.020 A
V = 0.020 × 150
V = 3.0 (V)
Question 3Hard7 marks
A heating element is made from a wire that has a resistance of 0.80 Ω for each metre of its length. The element is connected to a 12 V power supply.
(a) The current in the element must be 1.5 A. Calculate the length of wire needed.[3]
(b) The wire is cut into two equal halves. One half is connected to the 12 V supply. Calculate the current in this half of the wire.[2]
(c) The current in the element decreases slightly in the first few seconds after it is switched on. Explain why.[2]
Show the answer and mark scheme
(a)Answer: 10 m
12 = 1.5 × R
R = 8.0 (Ω)
length = 8.0 ÷ 0.80 = 10 (m)
(b)Answer: 3.0 A
R = 4.0 (Ω)
I = 12 ÷ 4.0 = 3.0 (A)
(c)Answer: The wire gets hotter, so its resistance increases; with the same potential difference, the current decreases.
the wire gets hotter / temperature increases, so its resistance increases
(so) for the same potential difference the current is smaller