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Resistors in Series and Parallel

Key idea: Calculate combined resistance for series and parallel resistor networks and solve one-source circuits using current and potential-difference relationships.

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Analyse series, parallel and potential-divider resistor networks.

1. Series and parallel conditions

Resistors in series and parallelTwo circuit diagrams compare resistors R1 and R2 in series with a common current, and in parallel with a common potential difference and split branch currents.Seriessame current I; p.d.s addR₁R₂IRₑq = R₁ + R₂Parallelsame p.d. V; branch currents addR₁R₂1/Rₑq = 1/R₁ + 1/R₂
Series resistors carry the same current and their p.d.s add. Parallel resistors share the same p.d. and their branch currents add.

Resistors in series

Series components lie on the same unbranched path, so they carry the same current. Their potential differences add:

R_eq = R₁ + R₂ + …

Resistors in parallel

Parallel branches connect between the same two nodes, so they have the same potential difference. Their branch currents add:

1/R_eq = 1/R₁ + 1/R₂ + …

For two resistors only,

R_eq = R₁R₂/(R₁ + R₂).

2. Checks that catch errors

  • A series equivalent resistance must be larger than every individual resistance.
  • A parallel equivalent resistance must be smaller than the smallest branch resistance.
  • The source current equals the sum of currents entering parallel branches.
  • Potential difference is the same across every branch connected to the same two nodes.

3. Reducing a mixed network

  1. Mark the nodes; components are parallel only if both ends share the same pair of nodes.
  2. Combine the innermost clear series or parallel group.
  3. Redraw the simpler circuit if the topology is not obvious.
  4. Find the source current from I = E/Rₜₒₜₐₗ for an ideal source.
  5. Work back through the network to find branch currents and p.d.s.
Geometry is not topology

Two resistors drawn side by side are not necessarily parallel. Check their end nodes, not their visual position.

4. Common Mistakes

  • Adding parallel resistances directly.
  • Assuming series components have the same p.d.; they have the same current.
  • Assuming parallel components have the same current; they have the same p.d.
  • Using the two-resistor product-over-sum shortcut for three or more branches.
  • Forgetting to include source internal resistance when the source is not ideal.

5. Exam Tips

  • Identify series and parallel relationships from shared nodes, not the drawing’s shape.
  • Check that a parallel equivalent is smaller than its smallest branch resistance.
  • After solving, check current conservation at a junction and p.d. addition around each series path.

6. Worked Examples

Modelled example 1

Two resistors in series

Core

Problem

A 4.0 Ω resistor and an 8.0 Ω resistor are connected in series across a 12 V ideal source. Find the current and the p.d. across each resistor.
Study the worked solution

  1. Reduce the series pair

    Method

    Add the two resistances.

    Reason

    Series resistors carry the same current and their p.d.s add.

    Working

    R_eq = 4.0 + 8.0 = 12 Ω

  2. Find the shared current

    Method

    Apply Ohm’s law to the complete network.

    Reason

    The ideal source places 12 V across 12 Ω.

    Working

    I = 12/12 = 1.0 A

  3. Find and check the p.d.s

    Method

    Use V = IR for each resistor.

    Reason

    The same current passes through both components.

    Working

    V_4Ω = 4.0 V and V_8Ω = 8.0 V; their sum is 12 V.

Guided practice 2

Two resistors in parallel

About 5 min

Problem

A 6.0 Ω resistor and a 3.0 Ω resistor are connected in parallel across 12 V. Find the equivalent resistance and both branch currents.

Try this before viewing the solution

Hints

Hint 1: shared quantity
Both branches have the full 12 V p.d.
Hint 2: reduce and check
Use product-over-sum for these two branches, then verify that branch currents add to the source current.
View solution step by step

  1. Reduce the parallel pair

    Method

    Use the two-resistor product-over-sum relation.

    Reason

    Both resistor ends share the same two nodes.

    Working

    R_eq = (6.0)(3.0)/(6.0 + 3.0) = 2.0 Ω

  2. Find branch currents

    Method

    Apply I = V/R separately.

    Reason

    Each branch has 12 V across it.

    Working

    I₆ = 2.0 A and I₃ = 4.0 A.

  3. Check the source current

    Method

    Add the branch currents.

    Reason

    Charge flow is conserved at the junction.

    Working

    I_source = 6.0 A = 12/2.0.

Common misconception 3

Parallel requires the same two nodes

Find and correct the mistake

Learner claim

Two resistors are drawn side by side, so a learner combines them in parallel. Their left ends share a node, but their right ends connect to different nodes. Diagnose the claim.

Try this before viewing the solution

Are the resistors parallel?

View solution step by step

  1. Ignore drawing geometry

    Method

    Label the node at each resistor end.

    Reason

    Wire shape and visual proximity do not determine topology.

    Working

    Resistor 1 connects nodes A–B; resistor 2 connects nodes A–C.

  2. Apply the parallel definition

    Method

    Reject the parallel combination.

    Reason

    Only one end node is shared; the components do not have the same p.d.

    Working

    A–B is not the same node pair as A–C.

Examiner practice 4

Mixed network source current

4 marks

Examination question

A 5.0 Ω resistor is in series with a parallel pair of 12 Ω and 4.0 Ω. The network is connected to an ideal 16 V source. Find the source current. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Reduce the parallel block

    1 mark

    Method

    Combine 12 Ω and 4.0 Ω.

    Reason

    They share the same two nodes.

    Working

    Rₚ = (12)(4.0)/(12 + 4.0) = 3.0 Ω

  2. Find total resistance

    1 mark

    Method

    Add the series resistor.

    Reason

    The reduced block is in series with 5.0 Ω.

    Working

    Rₜₒₜₐₗ = 5.0 + 3.0 = 8.0 Ω

  3. Find source current

    1 mark

    Method

    Apply Ohm’s law to the whole circuit.

    Reason

    The ideal source has no added internal resistance.

    Working

    I_source = 16/8.0 = 2.0 A

  4. Check the reduction

    1 mark

    Method

    Verify the parallel and total bounds.

    Reason

    3.0 Ω is below both branches and the series addition raises the total.

    Working

    3.0 < 4.0 < 12 and 8.0 > 5.0.

Challenge 5

A mixed series–parallel network

Minimal support

Independent transfer

A 4.0 Ω resistor is in series with a parallel pair of 6.0 Ω and 3.0 Ω across a 12 V ideal source. Find the source current and the p.d. across the parallel pair.

Try this before viewing the solution

Hints

Hint 1: work back after reducing
Reduce the parallel block first. After finding source current, treat that block as one series component to find its p.d.
View solution step by step

  1. Reduce the network

    Method

    Replace the parallel pair by 2.0 Ω, then add the series resistor.

    Reason

    The reduced parallel block remains in series with 4.0 Ω.

    Working

    Rₚ = 2.0 Ω, Rₜₒₜₐₗ = 4.0 + 2.0 = 6.0 Ω

  2. Find source current

    Method

    Apply Ohm’s law to the total.

    Reason

    The ideal source places 12 V across the network.

    Working

    I_source = 12/6.0 = 2.0 A

  3. Work back to branch p.d.

    Method

    Multiply series current by the parallel equivalent.

    Reason

    The source current enters the reduced parallel block before dividing.

    Working

    Vₚ = (2.0)(2.0) = 4.0 V

7. Mind Stretchers

Mind stretcher 1: Network sanity checksExtension

Without calculating an exact value, explain why adding another resistor in parallel must reduce the equivalent resistance and increase source current for a fixed ideal-source voltage.

Show Answer

The new branch provides another path for charge flow, so total conductance increases and equivalent resistance decreases. With fixed source p.d., I_source = V/R_eq therefore increases.

Next: Potential Divider Principle

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027