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IGCSE · Cambridge (CIE) · 0625

Calculate two resistor parallel equivalents

Use reciprocal or product-over-sum resistance and check that the result is below either positive branch resistance.

Section 4.3.2 · Series and parallel circuits

Series and parallel circuits pathway · 4.3.2

Extended only (Supplement); Physics 0625, 2026–2028. Use the stated source, operating conditions and meter ranges. Written practice is self-checked.

Before you begin

  • Use common branch p.d. and current sums.
  • Rearrange reciprocal expressions.

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

Quick readiness check

What p.d. do parallel branches share?

What you will learn

  • Calculate the equivalent of two positive parallel resistors.
  • Find an unknown branch resistance and check its feasibility.
Original pair of finite positive 3-ohm and 6-ohm parallel resistors has equivalent 2 ohms, below both. Product over sum is for exactly two resistors; open branches and wire shorts are different limiting cases.Common junction pair: A and BAB3 Ω6 ΩEquivalent: 2 Ω(3 × 6) / (3 + 6) = 2 Ω
Original pair of finite positive 3-ohm and 6-ohm parallel resistors has equivalent 2 ohms, below both. Product over sum is for exactly two resistors; open branches and wire shorts are different limiting cases.

Derive and apply the rule

For two constant resistors in parallel, the common p.d. V gives currents V/R1 and V/R2. Their sum is V/Req, so 1/Req = 1/R1 + 1/R2. For exactly two resistors this rearranges to Req = R1R2/(R1 + R2). Use consistent resistance units.

For 3 Ω and 6 Ω, Req = 18/9 = 2 Ω. It is below both branch values, as expected for finite positive resistors. Two equal R branches give R/2; they do not give 2R. For more than two branches use the full reciprocal sum rather than extending the two-resistor product expression arbitrarily.

Check an unknown branch

From a supplied equivalent, rearrange 1/Runknown = 1/Req − 1/Rknown. If Req = 2 Ω and Rknown = 6 Ω, the difference is 1/3, giving Runknown = 3 Ω. The inferred conductance must be positive for an ordinary finite positive resistor.

An equivalent equal to the known branch implies the added ideal branch is open in the limiting model; an equivalent larger than the known branch contradicts an additional finite positive parallel resistor. A direct wire short is a different limit, not a value to insert carelessly into the reciprocal rule as 1/0.

Pause and explain

What is the parallel equivalent of 3 Ω and 6 Ω?

Put the idea to work

Worked example

Calculate the equivalent of 3 Ω and 6 Ω connected in parallel.

Show the worked solution
  1. Use Req = R1R2/(R1 + R2).
  2. Req = 3 × 6/(3 + 6) = 18/9.
  3. Req = 2 Ω, below both values.

Answer 2 Ω.

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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

Calculate the equivalent of 3 Ω and 6 Ω in parallel.

Give me a hint

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

Compare my reasoning
  1. R1R2 = 18 Ω².
  2. R1 + R2 = 9 Ω.
  3. Req = 18/9 = 2 Ω.

2 Ω.

Look for these in your work

  • Use parallel connections.
  • Check below both values.
2 · Independent

A parallel pair has equivalent 2 Ω and one branch 6 Ω. Find the other branch.

Give me a hint

Subtract known conductance from total conductance.

Compare my reasoning
  1. 1/Rother = 1/2 − 1/6.
  2. The difference is 1/3.
  3. Rother = 3 Ω.

3 Ω.

Look for these in your work

  • Rearrange reciprocals.
  • Check 2 Ω is below both branches.
3 · Transfer

A pair is claimed to have equivalent 8 Ω with one branch 6 Ω and another finite positive resistor. Assess the claim.

Give me a hint

Compare the equivalent with the known branch before trusting the numbers.

Compare my reasoning
  1. An added finite positive parallel branch lowers the equivalent below 6 Ω.
  2. Eight ohms is larger than 6 Ω.
  3. The implied conductance 1/8 − 1/6 is negative, contradicting the model.

The claim is inconsistent with an additional finite positive parallel resistor.

Look for these in your work

  • Use the qualitative bound.
  • Explain the negative-conductance check.

Common mistakes

  • Product over sum is for exactly two resistors.
  • The result must be below either finite positive branch value.
  • Check the sign when inferring an unknown.
Recall without your notes

What can you explain now?

What is the equivalent of two equal R resistors in parallel?

Compare with the explanation

R/2.

Two equal branch currents double the total at the same p.d., halving the equivalent.

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Make it stick

Calculate parallel equivalents with consistent units and verify the below-either condition.

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 Physics

Course scope and assessment

491 sequenced Physics lessons provide 1473 guided, independent and application problems. 58 of 58 syllabus sections have full linked teaching sequences; 0 sections remain pending. Three earlier starter overviews remain available but are not counted as full sequences. The map covers all six subject areas for 2026–2028 and keeps review and assessment limits visible. This is available content, not full programme completion, an exam grade or certified mastery. Written practice is self-checked. Supervised practical experience, cumulative assessment and human subject-teacher review remain pending. Paper method exercises do not demonstrate practical competence or complete either practical-assessment route.

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