AQA AQA 2019-style practice: A resistor has a current of 0.40\,A through it when the potential difference across it is 6.0\,V. Calculate the resistance of the resistor, then calculate the charge that flows through it in 2.0 minutes.

First use V = IR rearranged to R = V / I = 6.0 / 0.40 = 15\, (2 marks: one for rearrangement, one for the value). Then find the charge using Q = It. Convert the time to seconds: 2.0 minutes is 120\,s (this conversion is often where marks are lost). So Q = 0.40 × 120 = 48\,C (2 marks). Markers reward the correct rearrangement, the time conversion to seconds, and correct units ( and C). This is a typical structured calculation on AQA Physics Paper 1.

EnglandPhysicsSyllabus dot point

What are current, potential difference and resistance, and how are they linked?

Current, potential difference and resistance: the meaning of each quantity, the charge equation, Ohm's law and the I-V characteristics of resistors, lamps and diodes.

A focused answer to AQA GCSE Physics 4.2.1, covering current as the rate of flow of charge, potential difference, resistance, Ohm's law and the current-voltage characteristics of ohmic resistors, filament lamps and diodes.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Current
Potential difference
Resistance and Ohm's law
I-V characteristics
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What this dot point is asking

AQA wants you to define current, potential difference and resistance, use the charge equation and Ohm's law, and describe the current-voltage (I-V) characteristics of an ohmic resistor, a filament lamp and a diode. This is topic 4.2.1 of the AQA GCSE Physics (8463) specification and the I-V investigation is a named required practical.

Current

Potential difference

Resistance and Ohm's law

Resistance opposes the flow of charge and is measured in ohms ( $\Omega$ ).

I-V characteristics

The I-V (current against potential difference) characteristic of a component is found in a required practical by varying the potential difference, recording the matching current, and plotting current on the vertical axis against potential difference on the horizontal axis. The shape of the graph reveals how the resistance behaves.

Ohmic resistor (fixed resistor at constant temperature): straight line through the origin; constant resistance. Because $R = V/I$ stays the same at every point, the gradient is constant. Keeping the temperature constant (for example by using short bursts of current) is essential, which is why the practical is done quickly.
Filament lamp: as current increases the filament heats up, so its resistance increases; the I-V graph is an S-shaped curve that gets less steep at higher voltages. The metal ions vibrate more when hot, obstructing the electrons.
Diode: current flows in one direction only (the forward direction). The resistance is very high in the reverse direction, so almost no current flows that way, and even in the forward direction very little current flows until the voltage passes a small threshold.

The resistance of two further components changes with their surroundings rather than with their own temperature. A thermistor's resistance falls as temperature rises, which makes it useful in temperature sensors and thermostats. A light-dependent resistor (LDR) has a resistance that falls as light intensity rises, which makes it useful in light-sensing circuits such as automatic street lighting.

Worked example: charge through a wire

A current of $I = 0.5\,A$ flows through a wire for $t = 120\,s$ . Calculate the total charge that passes.

Step 1: Identify the relevant equation and the known quantities

Current is defined as the rate of flow of charge, so the charge $Q$ that flows in a time $t$ is given by:

Q = It

We know $I = 0.5\,A$ and $t = 120\,s$ . The time is already in seconds, so no conversion is needed.

Step 2: Substitute and calculate

Replacing the symbols with the values:

Q = 0.5 \times 120 = 60\,\text{C}

Step 3: State the answer with units

The unit of charge is the coulomb (C). A current of $0.5\,A$ delivers $0.5$ coulombs every second, so over $120$ seconds it delivers $60$ coulombs in total.

Final answer: $Q = 60\,\text{C}$ .

Try this

Q1. Define electric current. [1 mark]

Cue. The rate of flow of electric charge.

Q2. A current of $2\,A$ flows for $30\,s$ . Calculate the charge that flows. [2 marks]

Cue. $Q = It = 2 \times 30 = 60\,C$ .

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20194 marksA resistor has a current of

0.40\,\text{A}

through it when the potential difference across it is

6.0\,\text{V}

. Calculate the resistance of the resistor, then calculate the charge that flows through it in

2.0

minutes.

Show worked answer →

First use $V = IR$ rearranged to $R = V / I = 6.0 / 0.40 = 15\,\Omega$ (2 marks: one for rearrangement, one for the value). Then find the charge using $Q = It$ . Convert the time to seconds: $2.0$ minutes is $120\,\text{s}$ (this conversion is often where marks are lost). So $Q = 0.40 \times 120 = 48\,\text{C}$ (2 marks). Markers reward the correct rearrangement, the time conversion to seconds, and correct units ( $\Omega$ and $C$ ). This is a typical structured calculation on AQA Physics Paper 1.

AQA 20214 marksExplain, in terms of the resistance of the filament, why the current-potential difference graph for a filament lamp is a curve rather than a straight line, and contrast this with the graph for a fixed resistor at constant temperature.

Show worked answer →

As the potential difference across a filament lamp increases, a larger current flows and transfers more energy to the filament, so its temperature rises (1 mark). The hotter filament has a higher resistance because the metal ions vibrate more and obstruct the flow of electrons (1 mark). A higher resistance means the current increases less steeply than the potential difference, so the graph curves and becomes less steep at higher voltages, giving the characteristic S-shape (1 mark). A fixed resistor kept at constant temperature has a constant resistance, so current is directly proportional to potential difference and its graph is a straight line through the origin (1 mark). Markers reward the temperature-resistance link and the explicit contrast.

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Sources & how we know this

AQA GCSE Physics (8463) specification — AQA (2016)