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Series & Parallel Circuits

Key idea: Learn series and parallel circuit rules for current, potential difference and effective resistance, and practise typical O Level calculations using V = IR.

  • SEC G3 Physics 2027
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Learning objectives

  • Recognise and interpret circuit symbols for cells, batteries, switches, lamps, LEDs, resistors, fuses, ammeters and voltmeters
  • Draw circuit diagrams with cells, batteries, switches, lamps, LEDs, fixed and variable resistors, fuses, ammeters and voltmeters
  • Recognise and interpret circuit symbols for d.c. and a.c. supplies, potentiometers, bells, light-dependent resistors and thermistors
  • Draw circuit diagrams with d.c. and a.c. supplies, potentiometers, bells, light-dependent resistors and thermistors
  • Apply the same-current rule in series circuits
  • Apply the potential-difference sum in series circuits
  • Apply current conservation at parallel junctions
  • Apply equal potential difference across parallel branches
  • Calculate effective resistance in series
  • Calculate effective resistance in parallel
  • Solve whole-circuit problems using consistent quantities
  • Describe variable-potential-divider action
  • Describe NTC thermistor action as an input transducer
  • Describe light-dependent resistor (LDR) action as an input transducer
  • Solve simple NTC and LDR potential-divider problems

1. Definition

A. Series circuit

In a series circuit, components are connected end-to-end in a single loop (one path for current).

B. Parallel circuit

In a parallel circuit, components are connected in separate branches between the same two points (multiple paths for current).

What you need for this course

You should be able to:

  • use series/parallel rules for current and p.d.
  • calculate effective resistance in series and parallel
  • solve circuit problems using V = IR

2. Key Ideas

RuleSeriesParallel
Currentsame everywheresplits; Iₜₒₜₐₗ = I₁ + I₂ + …
Potential differenceadds up: Vₜₒₜₐₗ = V₁ + V₂ + …same across each branch
Effective resistanceRₜₒₜₐₗ = R₁ + R₂ + …1/Rₜₒₜₐₗ = 1/R₁ + 1/R₂ + …

Quick checks:

  • In series, adding a resistor makes Rₜₒₜₐₗ bigger.
  • In parallel, adding a resistor makes Rₜₒₜₐₗ smaller than the smallest branch resistor.

3. Detailed Explanations

A. Series: same current, shared voltage

There is only one path, so the same current must pass through every component.

Series and parallel circuit rulesTwo circuit diagrams. The series circuit has two resistors in one loop labelled with the same current I and potential differences V1 and V2 that add to the supply. The parallel circuit has two resistor branches, both across supply V, with currents I1 and I2 that add to the source current I.Series circuitone pathR₁, p.d. V₁R₂, p.d. V₂II is the same; V₁ + V₂ = VParallel circuitseparate branches between the same two pointsR₁R₂I₁I₂V₁ = V₂ = V; I = I₁ + I₂
Scroll diagram horizontally to read all labels.
Series has one path: current is unchanged and component p.d.s add. Parallel branches share the same two endpoints: branch p.d.s are equal and branch currents add at each junction.

In series, the supply p.d. is shared between components (the p.d.s add up to the supply).

B. Parallel: same voltage, split current

Branches are connected across the same two points, so each branch has the same p.d. as the supply.

The current splits at a junction. The total current from the source equals the sum of branch currents:

Iₜₒₜₐₗ = I₁ + I₂ + …

C. Effective resistance (why the formulas make sense)

  • Series: resistors “add obstacles”, so total resistance increases.
  • Parallel: you add more paths for charge, so total resistance decreases.

4. Common Mistakes

  • Saying p.d. is the same in a series circuit (it is shared; p.d.s add up).
  • Saying current is the same in a parallel circuit (it splits; branch currents add up).
  • Using Rₜₒₜₐₗ = R₁ + R₂ for parallel resistors (parallel uses reciprocals).
  • Forgetting units (V, A, Ω).

5. Exam Tips

  1. Label the circuit first: mark what is in series and what is in parallel.
  2. Use the “same / add up” rules from the table in Section 2.
  3. For mixed circuits, simplify step-by-step using effective resistance.
  4. Sanity checks:
    • parallel Rₜₒₜₐₗ must be smaller than the smallest resistor
    • series Rₜₒₜₐₗ must be larger than each individual resistor

6. Worked Examples

Modelled example 1

Series total resistance and current

Core

Problem

Resistors 2 Ω, 3 Ω and 5 Ω are in series across 12 V. Find total resistance and current.
Study the worked solution

  1. Reduce the series chain

    Method

    Add all series resistances.

    Reason

    The same current passes through each component, and their p.d.s add.

    Working

    Rₜₒₜₐₗ = 2 + 3 + 5 = 10 Ω

  2. Find supply current

    Method

    Divide supply p.d. by total resistance.

    Reason

    The supply sees the complete effective resistance.

    Working

    I = 12/10 = 1.2 A

Guided practice 2

Parallel branch currents and total current

About 5 min

Problem

Resistors 4 Ω and 6 Ω are in parallel across 12 V. Find both branch currents and total current.

Use common branch p.d. before adding currents

Hints

Hint 1: same p.d.
Each parallel branch has 12 V.
Hint 2: junction rule
Add the two branch currents for supply current.
View solution step by step

  1. Calculate branch currents

    Method

    Apply I = V/R to each branch.

    Reason

    Parallel components share the same p.d.

    Working

    I₄ = 12/4 = 3.0 A; I₆ = 12/6 = 2.0 A

  2. Add at the junction

    Method

    Add branch currents.

    Reason

    Charge flow into a junction equals charge flow out.

    Working

    Iₜₒₜₐₗ = 3.0 + 2.0 = 5.0 A

Common misconception 3

Effective resistance in parallel

Find and correct the mistake

Learner response

A learner adds 3 Ω + 6 Ω for two parallel resistors. Diagnose the method and find the effective resistance.

Use the reciprocal rule and sanity bound

Unit: ohm

View solution step by step

  1. Add reciprocal resistances

    Method

    Use the parallel relationship.

    Reason

    Parallel branches provide additional paths for current.

    Working

    1/Rₜₒₜₐₗ = 1/3 + 1/6 = 1/2

  2. Invert and check

    Method

    Take the reciprocal to obtain 2.0 Ω.

    Reason

    Parallel effective resistance must be smaller than the smallest branch resistance.

    Working

    Rₜₒₜₐₗ = 2.0 Ω < 3 Ω

Examiner practice 4

Mixed circuit effective resistance

4 marks

Examination question

A 6 Ω and 3 Ω series pair is connected in parallel with 2 Ω. Find total resistance. [4 marks]

Reduce one identifiable group at a time

View solution step by step

  1. Reduce the series branch

    1 mark

    Method

    Add 6 Ω and 3 Ω.

    Reason

    They lie on one unbranched path.

    Working

    Rₛ = 9 Ω

  2. Combine parallel branches

    3 marks

    Method

    Combine 9 Ω in parallel with 2 Ω and invert.

    Reason

    These are the two complete paths between the same junctions.

    Working

    1/Rₜₒₜₐₗ = 1/9 + 1/2 = 11/18; Rₜₒₜₐₗ = 18/11 ≈ 1.64 Ω

Challenge 5

Voltage sharing in series

Minimal support

Component-voltage transfer

Resistors 2 Ω and 4 Ω are in series across 12 V. Find the p.d. across each.

Use common series current before assigning p.d.s

Hints

Hint 1: find total current
First use Rₜₒₜₐₗ = 6 Ω.
Hint 2: reuse the same current
Apply V = IR separately to each resistor.
View solution step by step

  1. Find common series current

    Method

    Add resistances and divide supply p.d. by the total.

    Reason

    One path means the same current through both resistors.

    Working

    I = 12/(2 + 4) = 2.0 A

  2. Find and check component p.d.s

    Method

    Multiply the common current by each resistance.

    Reason

    Series p.d.s must add to the supply.

    Working

    V₂ = (2.0)(2) = 4.0 V; V₄ = (2.0)(4) = 8.0 V; 4.0 + 8.0 = 12

7. Mind Stretchers

Mind stretcher 1: Series vs parallel total currentExtension

Two identical 6 Ω resistors are connected to a 12 V supply.

  1. If they are in series, what is the current from the supply?
  2. If they are in parallel, what is the current from the supply?
Show Answer

Series: Rₜₒₜₐₗ = 6 + 6 = 12 Ω, so I = 12/12 = 1.0 A.

Parallel: 1/Rₜₒₜₐₗ = 1/6 + 1/6 = 2/6 = 1/3, so Rₜₒₜₐₗ = 3 Ω and I = 12/3 = 4.0 A.

Mind stretcher 2: Parallel resistance checkExtension

Two resistors are connected in parallel. One is 10 Ω. The other is 20 Ω. Is the effective resistance closer to 10 Ω or 20 Ω? Explain.

Show Answer

Closer to 10 Ω, because in parallel the effective resistance is always less than the smallest resistor.

8. Practice and next step

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027