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Mixed resistor circuits

Key idea: Solve a mixed circuit one group at a time: combine each parallel group first, then add the series resistances. Every branch of a parallel group has the group's p.d., not the full supply p.d.

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Real circuits are rarely all series or all parallel. A car’s headlamps sit in parallel with each other but in series with a fuse and a switch. To find the current anywhere in such a circuit, replace one simple group at a time with its effective resistance until a single resistor is left. That gives the current from the supply. Then work back through the groups, using the series and parallel rules, to find the current and p.d. for each component.

Recognising the series and parallel parts

How the circuit is drawn does not tell you which rule to use. Follow the wires and mark every junction.

  • Components are in series if the same current must pass through one and then the other, with no junction between them.
  • Components are in parallel if each one is a complete branch between the same two junctions, so they have the same p.d.

A group can be replaced by a single resistor only when it is wholly series or wholly parallel. In the calculations below, the supply is ideal and the connecting wires have no resistance.

Reduce, then work back

  1. Find a group that is wholly in series or wholly in parallel.
  2. Replace it with its effective resistance and redraw the circuit.
  3. Repeat until one resistor is left. Use I = E/R to find the current from the supply.
  4. Work back through the groups to find the p.d. across, and the current in, each one.
  5. Check that the currents add up at each junction and the p.d.s add up to E round each loop.
Reduce a mixed resistor networkThree circuit diagrams show the same 12 V source. In the first, a 3 ohm series resistor precedes parallel branches of 6 and 3 ohms. In the second, those parallel branches are replaced by one 2 ohm resistor in series with the unchanged 3 ohm resistor. In the third, the entire load is replaced by one 5 ohm resistor. All three draw the same total 2.4 A from the ideal source. Component voltages and branch currents require returning to the original circuit.1 Identify the group12 V3 Ω6 Ω3 ΩThese branches shareboth junctions.6 Ω ∥ 3 Ω = 2 Ω2 Replace the group12 V3 Ω2 ΩThe equivalent groupis now in series.3 Ω + 2 Ω = 5 Ω3 Find total resistance12 V5 ΩUse the total resistancefor the supply current.I = 12 V ÷ 5 Ω = 2.4 AEach replacement preserves the p.d. and total current at that group’s terminals.Return to the original circuit to find individual resistor p.d.s and branch currents.
Reduce a mixed resistor networkThree circuit diagrams show the same 12 V source. In the first, a 3 ohm series resistor precedes parallel branches of 6 and 3 ohms. In the second, those parallel branches are replaced by one 2 ohm resistor in series with the unchanged 3 ohm resistor. In the third, the entire load is replaced by one 5 ohm resistor. All three draw the same total 2.4 A from the ideal source. Component voltages and branch currents require returning to the original circuit.1 Identify the group12 V3 Ω6 Ω3 ΩThese branches shareboth junctions.6 Ω ∥ 3 Ω = 2 Ω2 Replace the group12 V3 Ω2 ΩThe equivalent groupis now in series.3 Ω + 2 Ω = 5 Ω3 Find total resistance12 V5 ΩUse the total resistancefor the supply current.I = 12 V ÷ 5 Ω = 2.4 AEach replacement preserves the p.d. and total current at that group’s terminals.Return to the original circuit to find individual resistor p.d.s and branch currents.
Constructed example: replace only an identified group at each step. The 6 Ω and 3 Ω parallel branches become 2 Ω; that group is in series with 3 Ω, giving 5 Ω overall.

Worked example 1

Currents in a mixed circuit

Problem

A 6.0 Ω resistor and a 3.0 Ω resistor are connected in parallel. This pair is in series with another 3.0 Ω resistor across a supply of e.m.f. 12 V. Find the effective resistance of the circuit, the current from the supply, the p.d. across each part and the current in each branch.

Show full solution

  1. Replace the parallel pair

    Method

    Combine the two resistors that share both junctions.

    Reason

    They are wholly in parallel, so they can be replaced by one resistor.

    Working

    Answer

    Rₚ = (6.0 × 3.0)/(6.0 + 3.0) = 2.0 Ω

  2. Find the current from the supply

    Method

    Add the series resistor, then use I = E/R.

    Reason

    All the supply current passes through the separate resistor and then through the parallel group.

    Working

    Answer

    R = 3.0 + 2.0 = 5.0 Ω, so I = 12/5.0 = 2.4 A

  3. Find the p.d. across each part

    Method

    Use V = IR for the series resistor, then subtract from E.

    Reason

    The two parts are in series, so their p.d.s add up to 12 V. The parallel pair gets only what is left.

    Working

    Answer

    V = 2.4 × 3.0 = 7.2 V across the series resistor; 12-7.2 = 4.8 V across the parallel pair

  4. Find the branch currents

    Method

    Divide the pair’s p.d. by each branch resistance.

    Reason

    Both branches have 4.8 V across them.

    Working

    Answer

    4.8/6.0 = 0.80 A in the 6.0 Ω branch; 4.8/3.0 = 1.6 A in the 3.0 Ω branch

  5. Results

    Working

    Answer

    Effective resistance 5.0 Ω; supply current 2.4 A. The series resistor has 7.2 V across it. Each parallel branch has 4.8 V; the 6.0 Ω and 3.0 Ω branches carry 0.80 A and 1.6 A.

Check your understanding 1 Find the branch p.d.

A parallel pair of resistors is in series with a 5.0 Ω resistor across an ideal 15 V supply. The current from the supply is 1.2 A. Find the p.d. across the pair and the current in its 18 Ω branch.

Show the steps Optional working
Unit: V
Unit: V
Unit: A
Show answer

The series resistor has a p.d. of V = IR = 1.2 × 5.0 = 6.0 V. The parallel pair therefore has 15-6.0 = 9.0 V across it. Each branch has this p.d., so the current in the 18 Ω branch is I = 9.0/18 = 0.50 A. Using the full supply p.d. for that branch would ignore the energy transferred in the series resistor.

Check your understanding 2 Choose the first reduction

A 6.0 Ω resistor and a 3.0 Ω resistor are in series within one branch. That branch is in parallel with 2.0 Ω. Before combining the branches, what resistance replaces the series pair?

Unit: Ω
Show answer

The series pair becomes 6.0 + 3.0 = 9.0 Ω. It remains one branch in parallel with 2.0 Ω; the 2.0 Ω resistor is not part of this first addition.

Guided practice 2

Series inside a parallel branch

About 6 min

Problem

A 6.0 Ω resistor and a 3.0 Ω resistor are connected in series in one branch. This branch is in parallel with a 2.0 Ω resistor. Find the effective resistance of the combination.

Reduce the series pair first

Show the steps Optional working
Unit: Ω
Unit: Ω

Hints

Hint 1

series first

Trace the current path in the branch. Is there a junction between the two resistors?
Hint 2

then parallel

That branch is in parallel with 2.0 Ω. Use the parallel formula and remember to invert.
Show solution step by step

  1. Reduce the series branch

    Method

    Add 6.0 Ω and 3.0 Ω.

    Reason

    There is no junction between them, so they are in series.

    Working

    Answer

    Rₛ = 6.0 + 3.0 = 9.0 Ω

  2. Combine the two branches

    Method

    Combine 9.0 Ω in parallel with 2.0 Ω.

    Reason

    These are now two complete branches between the same two junctions.

    Working

    Answer

    1/R = 1/9.0 + 1/2.0 = 2/18 + 9/18 = 11/18 R = 18/11 = 1.6 Ω

Exam-style question 1

A whole-circuit calculation

5 marks

Examination question

A 4.0 Ω resistor and a 12 Ω resistor are connected in parallel. This pair is in series with a 5.0 Ω resistor across a supply of e.m.f. 16 V. Calculate (a) the effective resistance of the circuit, (b) the current from the supply and (c) the current in the 12 Ω resistor. [5 marks]

Show each reduction and each use of V = IR

Work on paper or here. Compare the complete calculation with the model and its mark points; you do not need to transcribe your paper working.

Show solution step by step

  1. (a) Effective resistance

    2 marks

    Method

    Combine the parallel pair, then add the series resistor.

    Reason

    The 4.0 Ω and 12 Ω resistors share both junctions; the pair is then in series with 5.0 Ω.

    Working

    Answer

    1/Rₚ = 1/4.0 + 1/12 = 3/12 + 1/12 = 1/3.0 Rₚ = 3.0 Ω R = 3.0 + 5.0 = 8.0 Ω

  2. (b) Current from the supply

    1 mark

    Method

    Use I = E/R for the whole circuit.

    Reason

    The e.m.f. drives current through the whole 8.0 Ω.

    Working

    Answer

    I = 16/8.0 = 2.0 A

  3. (c) Current in the 12 Ω resistor

    2 marks

    Method

    Find the p.d. across the parallel pair, then divide by 12 Ω.

    Reason

    The whole 2.0 A flows through the pair’s 3.0 Ω effective resistance, and both branches have the pair’s p.d.

    Working

    Answer

    Vₚ = 2.0 × 3.0 = 6.0 V I₁₂ = 6.0/12 = 0.50 A

Check your understanding 3 Checking the answer

In the examination question above, find the p.d. across the 5.0 Ω resistor and the current in the 4.0 Ω resistor. Then enter the loop p.d. total and the sum of the branch currents to check the answers.

Unit: V
Unit: A
Show the steps Optional working
Unit: V
Unit: A
Unit: V
Show the steps Optional working
Unit: A
Unit: A
Show answer

V = 2.0 × 5.0 = 10 V, and 10 + 6.0 = 16 V, which equals the e.m.f.

I = 6.0/4.0 = 1.5 A, and 1.5 + 0.50 = 2.0 A, which equals the supply current.

Check your understanding 4 One branch is disconnected

In the 16 V circuit above, the 12 Ω branch is disconnected. The 4.0 Ω branch and the separate 5.0 Ω resistor remain connected to the ideal supply. Find the new total resistance, supply current and p.d. across the 4.0 Ω resistor.

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

The disconnected branch carries no current. The remaining resistors are in series:

R = 4.0 + 5.0 = 9.0 Ω, I = 16/9.0 = 1.78 A.

Using the unrounded current, the p.d. across the 4.0 Ω resistor is

V₄ = 16/9.0 × 4.0 = 7.11 V.

It rises from the original 6.0 V even though the supply current falls. The remaining branch now carries all the current rather than sharing it with the 12 Ω branch. Check the new current path before reusing the earlier parallel resistance.

Common mistakes

  • Adding all the resistances before checking whether any are in parallel.
  • Giving each parallel branch the full supply p.d. when another resistor in series with the group also has a p.d. across it.
  • Using the supply current for each parallel branch instead of splitting it.
  • Forgetting to invert 1/R after adding reciprocals.
  • Expecting the effective resistance of a whole mixed circuit to be less than its smallest resistor. That rule applies only to a group that is wholly in parallel.

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Syllabus and review details

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