AQA AQA 2018-style practice: A 3\, resistor and a 9\, resistor are connected in series across a 6.0\,V battery. Calculate the total resistance, the current in the circuit, and the potential difference across the 9\, resistor.

In series the resistances add, so the total resistance is R_total = 3 + 9 = 12\, (1 mark). The current from the battery is I = V / R = 6.0 / 12 = 0.50\,A, and this current is the same through both resistors because they are in series (2 marks). The potential difference across the 9\, resistor is V = IR = 0.50 × 9 = 4.5\,V (2 marks). As a check, the 3\, resistor would have 0.50 × 3 = 1.5\,V across it, and 4.5 + 1.5 = 6.0\,V equals the supply, confirming the series rule that the potential differences add. Markers reward the total resistance, the shared current, and correct use of V = IR for one resistor.

EnglandPhysicsSyllabus dot point

How do current, potential difference and resistance behave in series and parallel circuits?

Series and parallel circuits: the rules for current, potential difference and resistance in each arrangement, and how to combine resistors.

A focused answer to AQA GCSE Physics 4.2.2, covering the rules for current, potential difference and resistance in series and parallel circuits and how total resistance changes when components are combined.

Generated by Claude Opus 4.88 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
Series circuits
Parallel circuits
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What this dot point is asking

AQA wants you to apply the rules for current, potential difference and resistance in series and parallel circuits, and to explain how adding resistors in series or parallel changes the total resistance. This is topic 4.2.2 of the AQA GCSE Physics (8463) specification.

Series circuits

The behaviour of current, potential difference and resistance is completely different in the two arrangements, and a clear grasp of the rules lets you predict ammeter and voltmeter readings without calculation. The fundamental principles behind the rules are conservation of charge (which fixes how current behaves) and conservation of energy (which fixes how potential difference is shared), so the rules are not arbitrary but follow from these conservation laws.

In a series circuit, all components are connected in a single loop.

Because resistances add, putting two resistors in series gives a larger total resistance than either one alone, so the current from the supply is smaller. The potential difference rule follows from energy conservation: the energy given to each unit of charge by the supply is shared between the components as the charge passes through each one, so the individual potential differences must add up to the supply potential difference. The current rule follows from conservation of charge: since there is only one loop, charge cannot build up anywhere, so the same number of coulombs per second passes every point. Christmas-tree lights wired in series show the downside of this arrangement: if one bulb fails and breaks the loop, the whole string goes out because the single path is broken.

Parallel circuits

In a parallel circuit, components are connected across each other on separate branches.

Adding a resistor in parallel provides another path for the current, so the total current increases and the total resistance falls. Each branch behaves as if it were the only component connected to the supply, which is why every branch gets the full supply potential difference. This is exactly how household wiring works: appliances are connected in parallel so that each one receives the full mains voltage and can be switched on or off independently without affecting the others. The branch-current rule again comes from conservation of charge: the current leaving the supply splits between the branches and recombines, so the branch currents must add up to the total supply current.

Worked example: two resistors in series

A $4\,\Omega$ resistor and a $6\,\Omega$ resistor are connected in series across a $10\,\text{V}$ supply. Find the total resistance and the current in the circuit.

Step 1: Apply the series resistance rule

In a series circuit there is only one path for current to flow, so charges must pass through every component in turn. Each component adds its resistance to the total opposition the current faces:

R_{\text{total}} = 4 + 6 = 10\,\Omega

Step 2: Find the current using Ohm's law

Now apply $V = IR$ rearranged to $I = \dfrac{V}{R}$ , using the supply voltage and the total resistance. In a series circuit this current is identical through every component:

I = \dfrac{V}{R_{\text{total}}} = \dfrac{10}{10} = 1\,\text{A}

Final answer: total resistance $= 10\,\Omega$ ; current $= 1\,\text{A}$ (the same through both resistors).

Try this

Q1. State what happens to the total resistance when a second resistor is added in series. [1 mark]

Cue. The total resistance increases (the resistances add).

Q2. Two $6\,\Omega$ resistors are in series across a $12\,V$ supply. Calculate the current. [3 marks]

Cue. $R_{total} = 6 + 6 = 12\,\Omega$ , so $I = \dfrac{12}{12} = 1\,A$ .

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 20185 marksA

3\,\Omega

resistor and a

9\,\Omega

resistor are connected in series across a

6.0\,\text{V}

battery. Calculate the total resistance, the current in the circuit, and the potential difference across the

9\,\Omega

resistor.

Show worked answer →

In series the resistances add, so the total resistance is $R_\text{total} = 3 + 9 = 12\,\Omega$ (1 mark). The current from the battery is $I = V / R = 6.0 / 12 = 0.50\,\text{A}$ , and this current is the same through both resistors because they are in series (2 marks). The potential difference across the $9\,\Omega$ resistor is $V = IR = 0.50 \times 9 = 4.5\,\text{V}$ (2 marks). As a check, the $3\,\Omega$ resistor would have $0.50 \times 3 = 1.5\,\text{V}$ across it, and $4.5 + 1.5 = 6.0\,\text{V}$ equals the supply, confirming the series rule that the potential differences add. Markers reward the total resistance, the shared current, and correct use of $V = IR$ for one resistor.

AQA 20213 marksExplain why connecting a second identical resistor in parallel with a first resistor decreases the total resistance of the circuit.

Show worked answer →

Adding a resistor in parallel creates a second path (branch) for the current to flow through (1 mark). With two paths instead of one, more current can flow from the supply for the same potential difference (1 mark). Because resistance is potential difference divided by current, a larger total current at the same voltage means a smaller total resistance, which is always less than the smallest single resistor in the combination (1 mark). Markers reward the "extra path" idea, the increased total current, and the conclusion that total resistance falls. A common error is to add parallel resistances as if they were in series.

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Sources & how we know this

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