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Effective Resistance Of Resistors

Key idea: Learn how to calculate effective resistance for resistors in series and parallel (including mixed circuits), with step-by-step worked examples for O Level.

  • 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

The effective resistance, R_eff (Ω), of a combination of resistors is the resistance of a single resistor that would draw the same current from the supply for the same potential difference.

For resistors:

  • Series: R_eff = R₁ + R₂ + … + Rₙ
  • Parallel: 1/R_eff = 1/R₁ + 1/R₂ + … + 1/Rₙ

2. Key Ideas

  • In series, the same current flows through every resistor, so resistances add.
  • In parallel, the same p.d. is across each branch and currents add up, so you add reciprocals.
  • Quick checks:
    • series: R_eff is greater than each individual resistor
    • parallel: R_eff is less than the smallest resistor
Useful shortcut (two resistors in parallel)

If there are exactly two resistors in parallel: R_eff = R₁R₂/(R₁ + R₂)

3. Detailed Explanations

A. Series connection

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.
Use the current and p.d. rules shown here to justify the series and parallel effective-resistance formulae.

If resistors are connected end-to-end, they are in series:

R_eff = R₁ + R₂ + R₃ + …

B. Parallel connection

If resistors are connected in separate branches between the same two points, they are in parallel:

1/R_eff = 1/R₁ + 1/R₂ + 1/R₃ + …

C. Why the formulas make sense (optional)

In series, the same current flows and the p.d.s add up: R_eff = Vₜₒₜₐₗ/I = (V₁ + V₂ + …)/I = V₁/I + V₂/I + … = R₁ + R₂ + …

In parallel, the same p.d. is across each branch and currents add up: 1/R_eff = Iₜₒₜₐₗ/V = (I₁ + I₂ + …)/V = I₁/V + I₂/V + … = 1/R₁ + 1/R₂ + …

4. Common Mistakes

  • Parallel reciprocal trap: you find 1/R_eff but forget to invert to get R_eff.
  • Using series addition for a parallel part (or vice versa).
  • Not checking if the answer makes sense using the quick checks in Section 2.

5. Exam Tips

  • Reduce mixed circuits step-by-step (do one clear series/parallel part at a time).
  • Write units with your final answer (Ω).
  • For parallel, do a quick estimate: the answer must be smaller than the smallest resistor.

6. Worked Examples

Modelled example 1

Series resistors

Core

Problem

Resistors 3 Ω, 5 Ω and 12 Ω are in series. Find R_eff.
Study the worked solution

  1. Identify and combine the chain

    Method

    Add all three resistances.

    Reason

    There is one current path, so the resistors are in series.

    Working

    R_eff = 3 + 5 + 12 = 20 Ω

Guided practice 2

Parallel resistors

About 5 min

Problem

Resistors 6 Ω and 3 Ω are in parallel. Find R_eff.

Find the reciprocal total, then invert

Unit: ohm

Hints

Hint 1: set up reciprocals
Start with 1/R_eff = 1/6 + 1/3.
Hint 2: finish the operation
The reciprocal sum is 1/2; invert it.
View solution step by step

  1. Add reciprocals

    Method

    Use the parallel-resistance relationship.

    Reason

    The branches lie between the same two junctions.

    Working

    1/R_eff = 1/6 + 1/3 = 1/2

  2. Invert

    Method

    Take the reciprocal of 1/2.

    Reason

    The calculation so far gives 1/R_eff, not R_eff.

    Working

    R_eff = 2.0 Ω

Common misconception 3

Three resistors in parallel

Find and correct the mistake

Learner response

For 2 Ω, 3 Ω and 6 Ω in parallel, a learner obtains 1 and writes R_eff = 1 Ω. Decide whether the conclusion is valid.

Check what the intermediate value represents

View solution step by step

  1. Interpret the reciprocal sum

    Method

    Add the reciprocal resistances.

    Reason

    The formula first calculates 1/R_eff.

    Working

    1/R_eff = 1/2 + 1/3 + 1/6 = 1

  2. Invert explicitly

    Method

    Take the reciprocal of 1.

    Reason

    Even though the numerical value stays 1, the inversion step and unit must be stated.

    Working

    R_eff = 1/1 = 1.0 Ω

Examiner practice 4

Mixed circuit (reduce step-by-step)

4 marks

Examination question

A 4 Ω resistor is in parallel with a 12 Ω resistor. This combination is in series with 5 Ω. Find the total effective resistance. [4 marks]

Show each network reduction

View solution step by step

  1. Reduce the parallel group

    3 marks

    Method

    Add reciprocals and invert.

    Reason

    The 4 Ω and 12 Ω resistors share both junctions.

    Working

    1/Rₚ = 1/4 + 1/12 = 1/3; Rₚ = 3 Ω

  2. Add the remaining series resistor

    1 mark

    Method

    Add Rₚ and 5 Ω.

    Reason

    The reduced parallel group is in series with the final resistor.

    Working

    R_eff = 3 + 5 = 8 Ω

Challenge 5

Effective resistance and supply current

Minimal support

Whole-circuit transfer

Two resistors 10 Ω and 15 Ω are in parallel across a 12 V supply. Find (i) R_eff and (ii) the total supply current.

Reduce the load before applying I = V/R

Hints

Hint 1: parallel load
Use 1/R_eff = 1/10 + 1/15.
Hint 2: whole circuit
Use the supply p.d. with the effective resistance.
View solution step by step

  1. Find effective resistance

    Method

    Combine the parallel resistors.

    Reason

    The supply sees the two branches as one equivalent load.

    Working

    1/R_eff = 1/10 + 1/15 = 1/6; R_eff = 6.0 Ω

  2. Find supply current

    Method

    Apply I = V/R to the whole circuit.

    Reason

    The 12 V supply is across the complete 6.0 Ω load.

    Working

    Iₜₒₜₐₗ = 12/6.0 = 2.0 A

7. Mind Stretchers

Mind stretcher 1: Why does parallel reduce resistance?Extension

Explain (in words) why adding a resistor in parallel makes the effective resistance smaller.

Show Answer

In parallel you add an extra path for current. For the same supply p.d., the total current increases (currents in branches add up). Since R_eff = V/Iₜₒₜₐₗ, a larger total current means a smaller effective resistance.

Mind stretcher 2: Estimation checkExtension

Two resistors 8 Ω and 12 Ω are in parallel. Is R_eff closer to 8 Ω or 12 Ω? Explain.

Show Answer

Closer to 8 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