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4.3.5Electrons, Waves and Photons

Resistors in Series and Parallel Circuits — OCR A Level Physics A H556 Revision Notes & Topical Questions

5 min read 5 parts Lesson 31 of 62 Unit quiz

📚 OCR A Level Physics A H556 Hub / Topical Revision
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Part 1 of 5Resistors in Series

4.3.5 Resistors in Series and Parallel Circuits

How resistors are connected decides whether the combined resistance rises or falls.

Rule
  • The combined resistance is the sum of the individual resistances: R = R₁ + R₂ + R₃ and so on.
Reason
  • The same current passes through all of them, and the p.d.s across them add up to the supply value.
Effect
  • Adding another resistor in series always increases the total resistance and so reduces the current drawn.
Largest value
  • The total is always larger than the biggest single resistance in the chain.

Part 2 of 5Resistors in Parallel

Resistors in Parallel
Rule
  • The reciprocal of the combined resistance equals the sum of the reciprocals: 1/R = 1/R₁ + 1/R₂ + 1/R₃.
Reason
  • Each branch has the same p.d. and the currents add, so more branches mean more routes for charge.
Effect
  • Adding another branch always decreases the total resistance below the smallest individual value.
Two resistors
  • For two resistors only, R = R₁R₂ / (R₁ + R₂); identical resistors R in parallel give R/2.
PropertySeriesParallel
Combined resistanceSum of resistancesLess than smallest resistor
CurrentSame in eachSplits between branches
P.d.Splits between resistorsSame across each branch

Part 3 of 5Mixed Networks

1
Look for the simplest group of resistors that are clearly in series or clearly in parallel.
2
Replace that group with a single equivalent resistor and redraw the circuit.
3
Repeat until the whole network is reduced to a single resistance connected across the source.
4
Use the total resistance with the e.m.f. to find the current, then work back through the stages to find p.d.s and branch currents.
5
Sanity-check the answer: a parallel combination must be smaller than its smallest branch.

Part 4 of 5Worked Example

Question

  • A 6.0 Ω and a 12 Ω resistor are connected in parallel, and this pair is connected in series with a 5.0 Ω resistor.
  • Find the total resistance.

  1. Parallel pair1/R = 1/6.0 + 1/12 = 2/12 + 1/12 = 3/12, so R = 4.0 Ω.
  2. Add the series resistor4.0 + 5.0 = 9.0 Ω.

Answer Total resistance = 9.0 Ω.

Part 5 of 5Exam Checklist & Tips

4.3.5 – Key Points for the Exam

  • Series Components connected end to end in one path, so they share the same current.
  • Parallel Components connected across the same two points, so they share the same p.d.
  • Combined resistance The single resistance equivalent to a group of resistors.
  • Reciprocal
    • One divided by a quantity;
      • the parallel rule adds reciprocals of resistance.

Common Mistakes

  • Forgetting to invert the sum of reciprocals at the end of a parallel calculation.
  • Adding resistances directly for a parallel pair.
  • Giving a parallel total that is larger than one of the branch resistors.
  • Treating a mixed network as all series or all parallel instead of reducing it stage by stage.

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