Derive and use combined resistance in parallel
Use equal branch p.d. and current addition to derive and apply the parallel-resistance rule.
AS Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 10. These original circuit diagrams and calculations are study aids; full timed assessment, supervised practical competence and human subject-teacher review remain outstanding.
Before you begin
- Use current, potential difference, resistance, energy and charge with consistent circuit directions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is common to branches connected in parallel?
What you will learn
- Derive the parallel-resistance equation using Kirchhoff's laws
- Use 1/Rₚ = 1/R₁ + 1/R₂ + …
- Relate the equivalent resistance to each positive branch resistance
Parallel branches share the same potential difference
Resistors connected across the same two junctions have the same p.d. V. The total current is the sum of branch currents: I = I₁ + I₂ + … .
Using I = V/R for each branch gives V/Rₚ = V/R₁ + V/R₂ + … . Divide by common non-zero V to derive 1/Rₚ = 1/R₁ + 1/R₂ + … .
Calculate the equivalent and check its size
For two parallel resistors, Rₚ = R₁R₂/(R₁ + R₂). For example, 6 Ω in parallel with 3 Ω gives 18/9 = 2 Ω.
With positive finite resistors the equivalent parallel resistance is smaller than the smallest branch resistance, because adding a branch provides another route for charge and increases total conductance.
Pause and explain
For two positive resistors in parallel, how does Rₚ compare with the smaller branch resistance?
Worked example
A 6.0 Ω and a 3.0 Ω resistor are in parallel. Find equivalent resistance.
Show the worked solution
- 1/Rₚ = 1/6.0 + 1/3.0 = 1/2.0.
- Therefore Rₚ = 2.0 Ω.
- Check it is below the smaller branch resistance, 3.0 Ω.
Answer Rₚ = 2.0 Ω.
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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 resistance of 4.0 Ω and 12 Ω in parallel.
Give me a hint
Use the two-resistor product-over-sum form.
Compare my reasoning
- Rₚ = (4.0 × 12)/(4.0 + 12).
- Rₚ = 48/16.
- Check the result is below 4.0 Ω.
Rₚ = 3.0 Ω.
Look for these in your work
- Parallel rule selected
- Value calculated
- Smallest-branch check passed
Three resistors of 6.0 Ω each connect in parallel. Find equivalent resistance.
Give me a hint
For equal parallel branches, divide one branch resistance by the number of branches.
Compare my reasoning
- 1/Rₚ = 1/6 + 1/6 + 1/6 = 3/6.
- Rₚ = 2.0 Ω.
- Check the total is below 6.0 Ω.
Rₚ = 2.0 Ω.
Look for these in your work
- All branch conductances added
- Reciprocal interpreted
- Size check stated
A 5.0 Ω resistor is connected in parallel with an unknown positive resistor. The equivalent is 2.0 Ω. Could the unknown resistance be 8.0 Ω? Explain or calculate it.
Give me a hint
Use 1/Rₚ = 1/R₁ + 1/R₂ and compare conductances.
Compare my reasoning
- 1/2.0 = 1/5.0 + 1/R₂.
- 1/R₂ = 0.50 − 0.20 = 0.30 Ω⁻¹.
- R₂ = 1/0.30 ≈ 3.3 Ω, not 8.0 Ω.
No. The unknown is approximately 3.3 Ω; 8.0 Ω would give an equivalent 3.1 Ω, not 2.0 Ω.
Look for these in your work
- Reciprocal equation used
- Unknown conductance found
- Proposed value rejected with check
Common mistakes
- Adding parallel resistances directly
- Forgetting current splits between parallel paths
- Accepting an equivalent larger than every branch for positive resistors
What can you explain now?
What is common to branches connected in parallel?
Compare with the explanation
The potential difference
They connect across the same pair of junctions.
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Make it stick
Why does adding a parallel branch lower an equivalent resistance?
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.