GCSE Physics resources

AQA GCSE Electricity

Series and Parallel Circuits

Circuit topology chooses the rule. A series route carries every coulomb through each component; parallel branches connect between the same two points.

AQA 8463AQA 8464 GCSE PhysicsCombined Science: Trilogy FoundationHigher

Start here

The revision snapshot

Circuit topology chooses the rule. A series route carries every coulomb through each component; parallel branches connect between the same two points.

By the end, you should be able to:

  • Apply current and p.d. rules in series.
  • Apply junction and p.d. rules in parallel.
  • Explain how adding resistance changes a circuit.

Before you revise

Diagnostic question

Choose an answer from memory. Your result tells you what to watch for in the guide.

4 Ω and 8 Ω are in series across 12 V. Which current passes each resistor?

Read the feedback for every option
  1. This uses only 4 Ω.
  2. This uses only 8 Ω.
  3. Rtotal = 12 Ω and I = 12/12.
  4. This reverses I = V/R.

Core revision guide

Learn the model, then use it

Circuit topology chooses the rule. A series route carries every coulomb through each component; parallel branches connect between the same two points.

Series: add resistance, share potential difference

Scope: 8463 and 8464 · Foundation and Higher

Current is equal throughout a single loop. Rtotal = R₁ + R₂ because each component impedes the same charge flow. The supply potential difference equals the sum across components. For a 12 V supply with 4 Ω and 8 Ω in series, Rtotal = 12 Ω, I = 1 A and the drops are 4 V and 8 V; equal current does not mean equal voltage.

Parallel: equal potential difference, add currents

Scope: 8463 and 8464 · Foundation and Higher

Each branch has the same supply potential difference because it connects to the same terminals. Current entering a junction equals the sum leaving. Adding a parallel resistor offers another path, so total resistance falls below the smallest individual resistance. AQA requires this qualitative explanation; reciprocal-resistance calculation is not required.

Faults and complete loops

Scope: 8463 and 8464 · Foundation and Higher

A break in a series loop stops all its current. Breaking one parallel branch leaves another complete branch operating. Do not infer equal branch currents unless the branches have equal resistance. Identify nodes on a circuit diagram rather than relying on the shapes of wires.

Standard-symbol series and parallel circuits show equal series current and current splitting at a parallel junction.
In series the current is the same throughout; in parallel each branch has the supply potential difference and branch currents add.Open the full-size exam diagram
ResistanceV = IRV, A, Ω

Misconception clinic

Replace the tempting answer

Read each belief, say what is wrong with it, then compare your correction.

Tempting idea: 3 A.

Use instead: This uses only 4 Ω. Rtotal = 12 Ω and I = 12/12.

Tempting idea: It decreases because total resistance rises.

Use instead: Parallel addition lowers total resistance. The same voltage drives current in an extra path.

Retrieval practice

Quick checks

1. An extra resistor is added in parallel across an ideal fixed-voltage supply. What happens to total current?

Read the feedback for every option
  1. Parallel addition lowers total resistance.
  2. The same voltage drives current in an extra path.
  3. Branch currents add, rather than equal the total.
  4. The original branch still has the supply voltage.

Worked examples

See the method being built

Apply series: add resistance, share potential difference in the worked reasoning, then try the changed-context questions below.

Example 1 · Series p.d.

Scope: 8463 and 8464 · Foundation and Higher

Two series lamps have potential differences 2.0 V and 4.0 V. Find the supply p.d.

  1. Use the series rule.
  2. Add the potential differences.
  3. State volts.
Show the answer

Supply p.d. = 6.0 V.

Example 2 · Parallel current

Scope: 8463 and 8464 · Foundation and Higher

A 1.2 A current reaches a junction. One branch carries 0.50 A. Find the other branch current.

  1. Use conservation of charge.
  2. Subtract branch current from total.
  3. State amperes.
Show the answer

Other branch current = 1.2 − 0.50 = 0.70 A.

Example 3 · Choose circuit type

Scope: 8463 and 8464 · Foundation and Higher

A student wants two lamps that work independently. Choose a circuit and explain.

  1. Identify the required behaviour.
  2. Choose series or parallel.
  3. Use a circuit rule as the reason.
Show the answer

Use parallel. Each lamp has the supply potential difference and one branch can remain complete if the other lamp is removed.

Exam precision

Use the physics precisely

Do this: Mark series paths and junctions first. State the relevant conservation rule before calculating each current or potential difference.

Independent practice

Exam-style questions

Write an answer before opening the marking guidance. Then edit the exact phrase or step that would gain the next mark.

1. 3 Ω and 6 Ω are in series across 9 V. Calculate total resistance, current and the voltage across 6 Ω.

[3 marks]

Scope: 8463 and 8464 · Foundation and Higher

Show marking guidance and model answer

Original PhysicsUK mark scheme: Point 1 (1 mark): Rtotal = 3 + 6 = 9 Ω. Point 2 (1 mark): I = 9/9 = 1 A through both resistors. Point 3 (1 mark): V6 = 1 × 6 = 6 V. Award each distinct point once. Accept equivalent scientifically correct wording. No half marks.

  1. Rtotal = 3 + 6 = 9 Ω. (1 mark)
  2. I = 9/9 = 1 A through both resistors. (1 mark)
  3. V6 = 1 × 6 = 6 V. (1 mark)

Model answer: Rtotal = 9 Ω. The series current is 1 A, and the 6 Ω resistor has 6 V across it.

Marking notes: Carry an incorrect total resistance into a correct I = V/R method, and an earlier current into V = IR; award the relevant later method points. Accept equivalent units.

2. Explain why adding a resistor in parallel lowers total resistance at fixed supply voltage.

[3 marks]

Scope: 8463 and 8464 · Foundation and Higher

Show marking guidance and model answer

Original PhysicsUK mark scheme: Point 1 (1 mark): The extra branch provides an additional route for charge. Point 2 (1 mark): The total current increases at the same potential difference. Point 3 (1 mark): Rtotal = Vsupply/Itotal therefore decreases. Award each distinct point once. Accept equivalent scientifically correct wording. No half marks.

  1. The extra branch provides an additional route for charge. (1 mark)
  2. The total current increases at the same potential difference. (1 mark)
  3. Rtotal = Vsupply/Itotal therefore decreases. (1 mark)

Model answer: The additional branch lets more charge pass each second at the same voltage, so total V/I decreases.

Questions pupils ask

Series and Parallel Circuits FAQs

Why can adding a parallel branch increase the supply current?

The new branch provides another charge-flow path at the same potential difference, so the branch currents sum to a larger total.

Useful next steps

Connect this guide to your next task

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