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Effective resistance and whole-circuit calculations

Key idea: Series resistances add; parallel resistances combine by adding their reciprocals. Reduce each group to find the supply current, then use series and parallel rules to work back through the circuit.

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A circuit with several resistors can be replaced, as far as the supply is concerned, by one resistor that draws the same current. Finding that effective resistance first, then working back out with the series and parallel rules, gives every current and p.d. in the circuit. The calculations use R = V/I from E.m.f., potential difference and resistance.

What effective resistance means

The effective resistance of a group of resistors is the resistance of the single resistor that could replace the whole group. With the same p.d. across it, the single resistor would draw the same current as the group. Assume constant resistances, negligible wire resistance and a supply that maintains its stated voltage.

First find the group’s two end points and trace the paths between them. The formula you use depends on those connections, so decide what is in series and what is in parallel before you calculate.

Resistors in series: add the resistances

R = R₁ + R₂ + R₃ + …

Why. In series the current I is the same in every resistor and the p.d.s add: V = IR₁ + IR₂ + IR₃. Dividing by I gives V/I = R₁ + R₂ + R₃. Each extra resistor makes the single path harder to drive charge through, so the effective resistance is larger than the largest resistor.

Worked example 1

Three resistors in series

Problem

Resistors of 4.0 Ω, 6.0 Ω and 10 Ω are in series across a 12 V supply. Find the effective resistance, the current and the p.d. across each resistor.

Show full solution

  1. Effective resistance

    Method

    Add the resistances.

    Reason

    In series the charge passes through every resistor in turn.

    Working

    Answer

    R = 4.0 + 6.0 + 10 = 20 Ω

  2. Current

    Method

    Divide the supply p.d. by the effective resistance.

    Reason

    The whole circuit has 12 V across 20 Ω, and the current is the same in all three resistors.

    Working

    Answer

    I = V/R = 12/20 = 0.60 A

  3. p.d. across each resistor

    Method

    Multiply the current by each resistance.

    Reason

    Each p.d. is V = IR for that resistor.

    Working

    Answer

    0.60 × 4.0 = 2.4 V, 0.60 × 6.0 = 3.6 V, 0.60 × 10 = 6.0 V

  4. Check

    Working

    Answer

    2.4 + 3.6 + 6.0 = 12 V. ✓ The largest resistor has the largest p.d.

Resistors in parallel: add the reciprocals

1/R = 1/R₁ + 1/R₂ + 1/R₃ + …

Why. In parallel the p.d. V is the same across every branch and the branch currents add: V/R = V/R₁ + V/R₂. Dividing by V gives the reciprocal rule.

For 6.0 Ω and 3.0 Ω: 1/R = 1/6.0 + 1/3.0 = 1/6.0 + 2/6.0 = 3/6.0, so R = 6.0/3 = 2.0 Ω. The last step, turning 1/R back into R, is easy to forget.

Each branch is another path for the charge. More charge flows each second for the same p.d., so V/I for the whole group is smaller. The effective resistance of a parallel group is smaller than the smallest branch resistance. If your answer is larger, check the connections and whether you took the final reciprocal.

Worked example 2

Three resistors in parallel

Problem

Resistors of 12 Ω, 6.0 Ω and 4.0 Ω are connected in parallel. Find the effective resistance.

Show full solution

  1. Add the reciprocals

    Method

    Use a common denominator of 12.

    Reason

    For parallel branches, the reciprocals of the resistances add.

    Working

    Answer

    1/R = 1/12 + 2/12 + 3/12 = 6/12

  2. Take the reciprocal

    Method

    Invert the sum.

    Reason

    The sum is 1/R, not R.

    Working

    Answer

    R = 12/6 = 2.0 Ω

  3. Check with currents

    Working

    Answer

    Across 12 V the branches carry 12/12 = 1.0 A, 12/6.0 = 2.0 A and 12/4.0 = 3.0 A, 6.0 A in total. 12/6.0 = 2.0 Ω. ✓ It is below the smallest branch, 4.0 Ω.

Two quick checks

  • Identical resistors. n identical resistors R in parallel have an effective resistance of R/n. Two 10 Ω resistors in parallel: 1/R = 1/10 + 1/10 = 2/10, so R = 5.0 Ω.
  • Adding a branch lowers the resistance. 6.0 Ω and 3.0 Ω in parallel give 2.0 Ω. Add another 6.0 Ω branch: 1/R = 1/6.0 + 2/6.0 + 1/6.0 = 4/6.0, so R = 1.5 Ω. Adding a resistor in parallel always lowers the effective resistance, even a large one.

Check your understanding 1

A 20 Ω resistor is in parallel with a 5.0 Ω resistor. Predict the range of their effective resistance before calculating it.

Predicted effective resistance
Show the steps Optional working
Unit: Ω⁻¹
Unit: Ω
Show answer

Greater than zero and below 5.0 Ω, the smaller branch. 1/R = 1/20 + 4/20 = 5/20, so R = 4.0 Ω.

Whole circuits: reduce the network, then work back out

Worked example 3

Reduce the network, then work back out

Problem

Trace a mixed resistor networkResistor R0 (4 ohms) joins A to B. Resistor R1 (6 ohms) and resistor R2 (6 ohms) each join B to C. An ideal 14 volt source has its positive terminal at A and negative terminal at C. All connecting wires are ideal.14 V+ABCR₀ = 4 ΩR₁ = 6 ΩR₂ = 6 Ω
Trace a mixed resistor networkResistor R0 (4 ohms) joins A to B. Resistor R1 (6 ohms) and resistor R2 (6 ohms) each join B to C. An ideal 14 volt source has its positive terminal at A and negative terminal at C. All connecting wires are ideal.14 V+ABCR₀ = 4 ΩR₁ = 6 ΩR₂ = 6 Ω
R₁ and R₂ share the same endpoints B and C. Replace that pair first, keeping B and C fixed; R₀ then carries the current entering the equivalent pair.

Find the current from the 14 V source and the current in each resistor.

Show full solution

  1. Trace the connections

    Method

    Start at B and follow both paths to C.

    Reason

    Resistors that share both end points are in parallel.

    Working

    Answer

    R₁ and R₂ (6 Ω each) are in parallel. R₀ joins A to B; it is not a third branch between B and C, even though it lies on the same line as R₁.

  2. Reduce the parallel pair

    Method

    Add the reciprocals.

    Reason

    The pair acts as one resistor.

    Working

    Answer

    1/R = 1/6 + 1/6 = 2/6, so the pair is 3 Ω

  3. Add the series resistor

    Method

    Add R₀ to the pair.

    Reason

    R₀ and the pair carry the same current one after the other.

    Working

    Answer

    4 + 3 = 7 Ω

  4. Find the source current

    Method

    Divide the source p.d. by the total resistance.

    Reason

    All of it passes through R₀.

    Working

    Answer

    I = V/R = 14/7 = 2.0 A

  5. Work back out

    Method

    Find the p.d. across R₀, then across the pair.

    Reason

    The p.d.s around the loop add up to 14 V.

    Working

    Answer

    R₀: 2.0 × 4 = 8.0 V; B to C: 14 - 8.0 = 6.0 V; each 6 Ω branch: 6.0/6 = 1.0 A

  6. Check

    Working

    Answer

    1.0 + 1.0 = 2.0 A at the junction. ✓

Check your understanding 2

In the shown network, a learner adds 4 + 6 + 6 = 16 Ω because all resistors appear in one drawing. Identify the first error and the correct grouping.

First error
Correct grouping
Show answer

The addition is correct, but the connection model is wrong. Charge has two paths from B to C, so neither 6 Ω resistor carries the whole source current. Reduce that parallel pair before adding R₀.

The method for a whole circuit reducible to series and parallel groups:

  1. Identify the series and parallel groups by tracing connections.
  2. Replace each group with its effective resistance, starting from the innermost group.
  3. Find the source current with I = V/R for the whole circuit.
  4. Work back out with the series and parallel rules to find each p.d. and current.
  5. Every time you use R = V/I, make sure V, I and R describe the same part of the circuit.
Trace a mixed resistor networkResistor R0 (4 ohms) joins A to B. Resistor R1 (6 ohms) and resistor R2 (3 ohms) each join B to C. An ideal 18 volt source has its positive terminal at A and negative terminal at C. All connecting wires are ideal.18 V+ABCR₀ = 4 ΩR₁ = 6 ΩR₂ = 3 Ω
Trace a mixed resistor networkResistor R0 (4 ohms) joins A to B. Resistor R1 (6 ohms) and resistor R2 (3 ohms) each join B to C. An ideal 18 volt source has its positive terminal at A and negative terminal at C. All connecting wires are ideal.18 V+ABCR₀ = 4 ΩR₁ = 6 ΩR₂ = 3 Ω
R₁ and R₂ share the same endpoints B and C. Replace that pair first, keeping B and C fixed; R₀ then carries the current entering the equivalent pair.

Check your understanding 3

For the shown 18 V circuit, R₀ = 4 Ω is in series with R₁ = 6 Ω and R₂ = 3 Ω in parallel. Find the parallel resistance, total resistance, p.d. across the branches and current in each resistor.

Show the steps Optional working
Unit: Ω⁻¹
Unit: Ω
Show the steps Optional working
Unit: Ω
Unit: Ω
Show the steps Optional working
Unit: Ω
Unit: A
Show the steps Optional working
Unit: A
Unit: V
Unit: V
Show the steps Optional working
Unit: V
Unit: A
Show the steps Optional working
Unit: V
Unit: A
Show answer
  1. The parallel pair: 1/R = 1/6 + 1/3 = 1/2, so it is 2 Ω.
  2. Total resistance: 4 + 2 = 6 Ω.
  3. Source current: 18/6 = 3.0 A. This current passes through R₀.
  4. p.d. across R₀: 3.0 × 4 = 12 V, so the parallel pair has 18 - 12 = 6.0 V across it.
  5. Branch currents: 6.0/6 = 1.0 A in R₁ and 6.0/3 = 2.0 A in R₂. They add up to the 3.0 A source current. ✓

Finding an unknown resistor

Worked example 4

An unknown series resistor

Problem

A 6.0 V supply drives a 4.0 Ω resistor in series with an unknown resistor R. The current is 0.50 A. Find R.

Show full solution

  1. Whole circuit

    Method

    Divide the supply p.d. by the current.

    Reason

    This gives the effective resistance of the whole loop.

    Working

    Answer

    Rₜₒₜₐₗ = V/I = 6.0/0.50 = 12 Ω

  2. Series rule

    Method

    Subtract the known resistor.

    Reason

    Series resistances add.

    Working

    Answer

    12 = 4.0 + R, so R = 8.0 Ω

  3. Check another way

    Working

    Answer

    The 4.0 Ω resistor has 0.50 × 4.0 = 2.0 V across it, leaving 6.0 - 2.0 = 4.0 V across R. R = 4.0/0.50 = 8.0 Ω. ✓

Check your understanding 4

An unknown resistor in parallel with a 12 Ω resistor has an effective resistance of 4.0 Ω. Find the unknown resistance.

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

1/R = 1/4.0 - 1/12 = 3/12 - 1/12 = 2/12, so R = 6.0 Ω. Check: 6.0 Ω and 12 Ω in parallel give 1/6 + 1/12 = 3/12, which is 4.0 Ω. ✓

For a whole-circuit calculation, use the supply p.d. with the total resistance and the supply current. For one component, use the p.d. across that component and the current through it.

Check your understanding 5 Recalculate when an internal branch opens

In the 18 V network, R₀ = 4 Ω is in series with parallel R₁ = 6 Ω and R₂ = 3 Ω. R₂’s branch opens. Find the source current, R₁’s current and its p.d. Which condition would be needed for R₁’s current to stay unchanged?

Show the steps Optional working
Unit: Ω
Unit: A
Unit: A
Show the steps Optional working
Unit: A
Unit: V
Condition for unchanged R₁ current
Show answer

The remaining path has R₀ and R₁ in series, so total resistance = 4 + 6 = 10 Ω. Source current = 18/10 = 1.8 A, and that same 1.8 A passes through R₁. Its p.d. is 1.8 × 6 = 10.8 V; R₀ has 1.8 × 4 = 7.2 V, checking the 18 V total. The earlier unchanged-branch rule assumes a maintained p.d. across those branches. Here the parallel group was behind R₀, so its p.d. was not fixed directly by the source. Opening R₂ changes the source current and R₀’s voltage drop, leaving more p.d. across R₁. R₂ carries zero current; the remaining branch is not unchanged.

Common mistakes

  • Adding parallel resistances directly. Add the reciprocals, then take the reciprocal of the sum.
  • Stopping at 1/R. If 1/R = 0.5, then R = 2 Ω, not 0.5 Ω.
  • Using the supply p.d. with one series resistor. Use the p.d. across that resistor.
  • Adding every resistor because they are all in one drawing. Resistors that share both end points are in parallel; trace the connections first.
  • Accepting a parallel answer larger than a branch. A parallel group is always smaller than its smallest branch.

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