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IGCSE · Cambridge (CIE) · 0652

Add resistances in one series path

Calculate a series resistance total and distinguish it from a current shared by all loads.

Section P4.3.2 · Series and parallel circuits

Series and parallel circuits · Cambridge IGCSE Physical Science 0652 · P4.3.2

Cross-referenced from the related Cambridge IGCSE Chemistry 0620 or Physics 0625 teaching to the 0652 syllabus. Exact 0652 Core/Supplement objective coverage and this final-series course’s examiner fit still need specialist review. Written practice is self-checked and does not predict a Cambridge grade or demonstrate supervised laboratory competence.

Before you begin

  • Recognize an unbranched series path.
  • Convert resistance units.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

How many ohms are in 0.5 kΩ?

What you will learn

  • Calculate combined resistance of two or more series resistors.
  • Find a missing positive resistance from a supplied total.
Original one-path resistor arrangement shows 3, 4 and 5 ohms in series, giving 12 ohms. The order does not alter their sum, and a positive added series resistance increases the total.Resistors in one series path3 Ω4 Ω5 ΩRtotal = 3 + 4 + 512 ΩAdd positive series resistanceThe total increases
Original one-path resistor arrangement shows 3, 4 and 5 ohms in series, giving 12 ohms. The order does not alter their sum, and a positive added series resistance increases the total.

Add the actual series values

For resistors in series, the combined resistance is Rtotal = R1 + R2 + … . A 3 Ω resistor and a 7 Ω resistor in the same path give 10 Ω. Each additional positive resistance increases the total. Do not average the values or use the parallel rule.

Convert all values to one unit before adding: 0.5 kΩ is 500 Ω, so it and 250 Ω total 750 Ω. The physical order of the same series resistors does not change their sum in this model, although it changes where individual components appear in the diagram.

Check an unknown value

Subtract known series resistances from a supplied total to find a missing resistance. A 12 Ω total containing 3 Ω and 4 Ω leaves 5 Ω. A negative inferred value would contradict a model made only of the stated ordinary positive resistors; it signals an inconsistent total, units or connection assumption.

The current through each series resistor is the same in a steady path. At a fixed ideal source p.d., adding positive series resistance reduces the common current. These statements concern different quantities: resistance adds, while the current does not add across series loads.

Sources for this lesson

Pause and explain

A 12 Ω series total contains 3 Ω and 4 Ω plus an unknown resistor. What is the unknown?

Put the idea to work

Worked example

Find the equivalent resistance of 3 Ω, 4 Ω and 5 Ω in series.

Show the worked solution
  1. The components share one unbranched path.
  2. Add their resistances in ohms.
  3. 3 + 4 + 5 = 12 Ω.

Answer 12 Ω.

Try a topic test

Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

Find the combined resistance of 3 Ω, 4 Ω and 5 Ω in series.

Give me a hint

Add every component in the one path.

Compare my reasoning
  1. All resistances are already in ohms.
  2. Rtotal = 3 + 4 + 5.
  3. Rtotal = 12 Ω.

12 Ω.

Look for these in your work

  • Use the series rule.
  • Include all three values.
2 · Independent

Find the total of 0.5 kΩ and 250 Ω in series.

Give me a hint

Convert the prefixed resistance before adding.

Compare my reasoning
  1. 0.5 kΩ = 500 Ω.
  2. 500 + 250 = 750.
  3. The combined resistance is 750 Ω or 0.75 kΩ.

750 Ω or 0.75 kΩ.

Look for these in your work

  • Use consistent units.
  • Do not average resistances.
3 · Transfer

An alleged 5 Ω series total contains 3 Ω, 4 Ω and one ordinary positive resistor. Find the implied unknown and assess the data.

Give me a hint

Subtract known contributions and compare the sign with the model.

Compare my reasoning
  1. Known resistance totals 7 Ω.
  2. The implied unknown is 5 − 7 = −2 Ω.
  3. That contradicts the stated positive-resistor series model.

Implied −2 Ω; the supplied data or series assumption is inconsistent.

Look for these in your work

  • Show the subtraction.
  • Reject a negative ordinary-resistor interpretation.

Common mistakes

  • Series resistance adds, not averages.
  • Convert kΩ and Ω first.
  • A negative ordinary-resistor result needs an inconsistency check.
Recall without your notes

What can you explain now?

What is the series resistance rule?

Compare with the explanation

Add the component resistances.

Every stated positive series resistor contributes to the total.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Add series resistances in consistent units and check missing-value signs.

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Physical Science

Course scope and assessment

This page maps the 2025–2026 Cambridge IGCSE Physical Science 0652 syllabus: 38 Chemistry sections and 46 Physics sections. Every section links to original 0620/0625 teaching or one of five new 0652 lessons, with guided, independent and transfer practice and explained answers. Cambridge limits entry to Administrative Zone 3 and identifies November 2026 as the final exam series. The section crosswalk is complete as a first coverage map; a numbered-objective and Core/Supplement audit, independent subject-teacher review and cumulative assessment are still needed. Written practice does not award a Cambridge grade. Online activities cannot demonstrate supervised laboratory competence or complete Paper 5/6.

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