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
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028. Use the stated source, operating conditions and meter ranges. Written practice is self-checked. Human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
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.
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?
Worked example
Calculate the equivalent of 3 Ω and 6 Ω connected in parallel.
Show the worked solution
- Use Req = R1R2/(R1 + R2).
- Req = 3 × 6/(3 + 6) = 18/9.
- Req = 2 Ω, below both values.
Answer 2 Ω.
Try a topic test
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Test this topic →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.
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
- R1R2 = 18 Ω².
- R1 + R2 = 9 Ω.
- Req = 18/9 = 2 Ω.
2 Ω.
Look for these in your work
- Use parallel connections.
- Check below both values.
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/Rother = 1/2 − 1/6.
- The difference is 1/3.
- Rother = 3 Ω.
3 Ω.
Look for these in your work
- Rearrange reciprocals.
- Check 2 Ω is below both branches.
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
- An added finite positive parallel branch lowers the equivalent below 6 Ω.
- Eight ohms is larger than 6 Ω.
- 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.
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.