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Effective resistance in series and parallel

Key idea: Resistances in series add up. Resistors in parallel have a combined resistance smaller than the smallest one, because each extra branch gives the current another path.

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Enlarged diagram

A supply cannot tell how the resistors connected to it are arranged. All it “sees” is how much current it has to drive for the p.d. across it. So any group of resistors can be replaced, on paper, by one resistor that draws the same current. Finding that single resistance is usually the first step in working out the current from a supply. Adding resistors in series makes it larger, but adding them in parallel makes it smaller, and the series and parallel rules explain why.

What effective resistance means

The effective resistance R of a group of resistors is the resistance of a single resistor that would draw the same current from the supply for the same p.d. across it.

Series and parallel circuit rulesTwo circuit diagrams. The series circuit has two resistors in one loop labelled with the same current I and potential differences V1 and V2 that add to the supply. The parallel circuit has two resistor branches, both across supply V, with currents I1 and I2 that add to the source current I.Series circuitone pathR₁, p.d. V₁R₂, p.d. V₂II is the same; V₁ + V₂ = VParallel circuitseparate branches between the same two pointsR₁R₂I₁I₂V₁ = V₂ = V; I = I₁ + I₂
Series and parallel circuit rulesTwo circuit diagrams. The series circuit has two resistors in one loop labelled with the same current I and potential differences V1 and V2 that add to the supply. The parallel circuit has two resistor branches, both across supply V, with currents I1 and I2 that add to the source current I.Series circuitone pathR₁, p.d. V₁R₂, p.d. V₂II is the same; V₁ + V₂ = VParallel circuitseparate branches between the same two pointsR₁R₂I₁I₂V₁ = V₂ = V; I = I₁ + I₂
The current and p.d. rules for series and parallel connections lead directly to the two effective-resistance formulae.

Resistors in series

In series the same current I passes through each resistor, and their p.d.s add up to the p.d. V across the whole group. Dividing by the current:

R = V/I = (V₁ + V₂ + …)/I = V₁/I + V₂/I + …

so

R = R₁ + R₂ + …

Each resistor adds to the opposition that the same current has to get through, so the effective resistance is larger than any of the individual resistances.

Worked example 1

Three resistors in series

Problem

Resistors of 3.0 Ω, 5.0 Ω and 12 Ω are connected end to end, with no branches between them. Find their effective resistance.

Show full solution

  1. Identify the connection

    Method

    Decide that the three resistors are in series.

    Reason

    There is only one path through them, so the same current passes through each.

    Working

    Answer

    One path, no junctions → series.

  2. Add the resistances

    Method

    Add all three resistances.

    Reason

    For resistors in series, R = R₁ + R₂ + R₃.

    Working

    Answer

    R = 3.0 + 5.0 + 12 = 20 Ω

Resistors in parallel

In parallel each branch has the same p.d. V, and the branch currents add up to the current I in the main circuit. Dividing by the p.d.:

1/R = I/V = (I₁ + I₂ + …)/V = I₁/V + I₂/V + …

so

1/R = 1/R₁ + 1/R₂ + …

Each extra branch gives the charge another path. The current in the branches already there does not change, because their p.d. has not changed, so the total current increases. A larger current for the same p.d. means a smaller resistance. That is why the effective resistance of resistors in parallel is smaller than the smallest branch resistance.

Two resistors in parallel

For exactly two resistors in parallel, the formula rearranges to

R = R₁R₂/(R₁ + R₂)

This “product over sum” works for two resistors only.

Guided practice 1

Two resistors in parallel

About 5 min

Problem

Resistors of 6.0 Ω and 3.0 Ω are connected in parallel. Calculate 1/R in inverse ohms, then find their effective resistance.

Find 1/R first, then invert

Show the steps Optional working
Unit: Ω⁻¹
Unit: Ω

Hints

Hint 1

set up the formula

Which quantity adds when resistors are connected in parallel?
Hint 2

finish

That sum gives 1/R, not R. Invert it.
Show solution step by step

  1. Add the reciprocals

    Method

    Use the parallel formula.

    Reason

    Both resistors are connected between the same two junctions.

    Working

    Answer

    1/R = 1/6.0 + 1/3.0 = 1/6.0 + 2/6.0 = 3/6.0 = 1/2.0

  2. Invert

    Method

    Take the reciprocal.

    Reason

    The calculation so far gives 1/R.

    Working

    Answer

    R = 2.0 Ω

Spot the mistake 1

Forgetting to invert

About 5 min

Learner response

Resistors of 4.0 Ω, 6.0 Ω and 12 Ω are connected in parallel. A learner writes 1/4.0 + 1/6.0 + 1/12 = 0.50, so the effective resistance is 0.50 Ω. Find the error and correct it.

Decide what the 0.50 represents

What does the sum 0.50 represent?
Show the steps Optional working
Unit: Ω⁻¹
Unit: Ω
Show solution step by step

  1. Identify what was calculated

    Method

    Recognise that the sum of the reciprocals is 1/R.

    Reason

    Each term is 1/R for one resistor, so the sum has the unit Ω⁻¹, not Ω.

    Working

    Answer

    1/R = 3/12 + 2/12 + 1/12 = 6/12 = 0.50 Ω⁻¹

  2. Invert

    Method

    Take the reciprocal to find R.

    Reason

    The question asks for the resistance, not its reciprocal.

    Working

    Answer

    R = 1/0.50 = 2.0 Ω

Check your understanding 1 Adding a second branch

A 6.0 Ω resistor is connected across a 12 V supply. A second 6.0 Ω resistor is then connected in parallel with it. What happens to:

  1. the current in the first resistor;
  2. the current from the supply;
  3. the effective resistance?
Unit: A
Show the steps Optional working
Unit: A
Unit: A
Show the steps Optional working
Unit: A
Unit: A
Unit: Ω
Show answer
  1. It stays at 12/6.0 = 2.0 A, because the p.d. across the first resistor is still 12 V.
  2. It doubles from 2.0 A to 4.0 A, because the second branch also carries 2.0 A.
  3. It halves from 6.0 Ω to 12/4.0 = 3.0 Ω.

Using effective resistance to find the current

Once a group of resistors has been replaced by its effective resistance, use I = V/R with the p.d. across the whole group to find the current into it.

Try it yourself 1

Finding the supply current

Minimal support

Problem

Resistors of 10 Ω and 15 Ω are connected in parallel across a 12 V supply. Find (i) their effective resistance and (ii) the current from the supply.

Combine the resistors before using I = V/R

Unit: Ω
Show the steps Optional working
Unit: Ω
Unit: A

Hints

Hint 1

combine

Find the effective resistance of the two parallel resistors first.
Hint 2

whole circuit

The whole 12 V is across that effective resistance.
Show solution step by step

  1. Find the effective resistance

    Method

    Combine the two parallel resistors.

    Reason

    The supply sees the two branches as a single load.

    Working

    Answer

    1/R = 1/10 + 1/15 = 3/30 + 2/30 = 1/6.0 R = 6.0 Ω

  2. Find the supply current

    Method

    Apply I = V/R to the whole circuit.

    Reason

    The 12 V supply is across the 6.0 Ω effective resistance.

    Working

    Answer

    I = 12/6.0 = 2.0 A

Check your understanding 2 Estimate first

Resistors of 8.0 Ω and 12 Ω are connected in parallel. Before calculating, say whether the effective resistance is more or less than 8.0 Ω. Then calculate it.

Prediction relative to 8.0 Ω
Show the steps Optional working
Unit: Ω⁻¹
Unit: Ω
Show answer

Less than 8.0 Ω, because a parallel combination is always smaller than its smallest branch resistance.

R = (8.0 × 12)/(8.0 + 12) = 96/20 = 4.8 Ω

Common mistakes

  • Stopping at 1/R and giving it as the resistance. The sum of reciprocals is in Ω⁻¹; invert it.
  • Adding resistances that are in parallel, or using reciprocals for resistances in series.
  • Using product over sum for three or more resistors in parallel.
  • Using one resistor’s resistance with the supply p.d. to find the supply current, instead of the effective resistance of the whole circuit.
  • Not checking the answer: a series total must be larger than each resistance, and a parallel total smaller than the smallest.

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