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First year Lesson 7 of 14 About 18 min

Combination circuits

What you’ll learn

  • How to spot which parts of a circuit are in series and which are in parallel.
  • The reduce-and-redraw method that turns any combination circuit into one resistance.
  • How to work back out to the current and voltage of every part.
  • Why the wire feeding parallel loads makes every real circuit a combination circuit.

A combination circuit has some parts in series and some in parallel. You solve it by shrinking it: replace each simple group with one equivalent resistor, redraw, and repeat until one resistor is left — then work back out. It is just the two previous lessons used one after the other.

1 4 Ω 12 Ω 6 Ω 24 V 2 4 Ω 4 Ω 12 ∥ 6 = 72 ÷ 18 = 4 Ω 3 8 Ω 4 + 4 = 8 Ω · I = 24 ÷ 8 = 3 A
Reduce the parallel pair to one resistor, then add the series pieces. Three redraws, one answer.

The method: in, then back out

  1. Reduce. Find a group that is plainly series (one path through them) or plainly parallel (same two end points). Replace it with its equivalent. Redraw.
  2. Repeat until one resistance is left. That is RT.
  3. Supply current: IT = V ÷ RT.
  4. Work back out through your redraws: in a series piece the current is known, so V = I × R; in a parallel piece the voltage is known, so I = V ÷ R.
  5. Check: branch currents add to what feeds them; drops around any loop add to the supply.

1. The circuit in the drawing

  1. 12 Ω ∥ 6 Ω = (12 × 6) ÷ (12 + 6) = 72 ÷ 18 = 4 Ω.
  2. RT = 4 + 4 = 8 Ω. IT = 24 ÷ 8 = 3 A.
  3. Series 4 Ω: V = 3 × 4 = 12 V. Parallel pair: V = 3 × 4 = 12 V. Check 12 + 12 = 24 V ✓
  4. Branches: 12 ÷ 12 = 1 A and 12 ÷ 6 = 2 A. Check 1 + 2 = 3 A ✓

2. The wire is a resistor too

  1. Two 24 Ω heaters in parallel at the end of a run. The hot and the neutral each have 0.25 Ω of resistance — practice figures; the real ones come from the conductor size and length.
  2. Heaters: 24 ∥ 24 = 12 Ω. Conductors, in series with them, out and back: 0.25 + 0.25 = 0.5 Ω.
  3. RT = 12 + 0.5 = 12.5 Ω, so I = 120 ÷ 12.5 = 9.6 A.
  4. Voltage at the heaters: 9.6 × 12 = 115.2 V. Lost in the wire: 9.6 × 0.5 = 4.8 V. Each heater draws 115.2 ÷ 24 = 4.8 A.
  5. That lost 4.8 V is voltage drop — a second-year topic, but this is all it is.

3. A series pair inside a parallel branch

  1. 60 V across two branches: branch A is 10 Ω + 20 Ω in series, branch B is 30 Ω.
  2. Branch A = 30 Ω. Two 30 Ω branches in parallel: RT = 15 Ω, IT = 60 ÷ 15 = 4 A.
  3. Each branch has the full 60 V: branch A carries 60 ÷ 30 = 2 A, branch B 2 A.
  4. Inside branch A: 2 × 10 = 20 V and 2 × 20 = 40 V. Check 20 + 40 = 60 V ✓
Try one: 2 Ω in series with (10 Ω ∥ 10 Ω ∥ 5 Ω), on 12 V

10 ∥ 10 = 5 Ω; 5 ∥ 5 = 2.5 Ω. RT = 2 + 2.5 = 4.5 Ω, I = 12 ÷ 4.5 ≈ 2.67 A. The parallel group has 2.67 × 2.5 ≈ 6.67 V across it; the 2 Ω drops ≈ 5.33 V (together 12 V).

Sur le chantier

Tips from the job

  • Redraw it. Every journeyman who is good at this redraws the circuit as a simple ladder before doing any arithmetic.
  • Check as you go: currents into a junction equal currents out, and drops around a loop add up to the supply.
  • Seeing less voltage at a load than at the panel? The difference is being dropped across something in series — the conductors or a connection.
  • The series & parallel calculator checks your reductions.

Common mistakes

  • Treating two resistors as in series when a branch joins between them — then they are not carrying the same current.
  • Using the supply voltage across a parallel section that has a series resistor in front of it.
  • Stopping at the total resistance and forgetting the question asked for a branch current.
  • Rounding too early; keep three or four figures until the end.

Check yourself

7 questions. Tap an answer — it is marked straight away and the reason opens under it. Get 80% or better and the lesson is ticked off on the year page.

Not started yet.

  1. What is the first step in solving a combination circuit?

    Answer and why

    B. Reduce from the inside out: replace each simple series group or parallel group with its single equivalent resistance, redraw, and repeat until one resistor is left.

  2. A 12 Ω and a 6 Ω are in parallel, and that pair is in series with a 4 Ω. Total resistance?

    Answer and why

    B. 12 ∥ 6 = (12 × 6) ÷ 18 = 4 Ω. In series with 4 Ω: 4 + 4 = 8 Ω.

  3. That circuit is connected to 24 V. What current flows through the 4 Ω series resistor?

    Answer and why

    C. I = 24 ÷ 8 = 3 A from the supply, and all of it passes through the series resistor.

  4. What voltage is across the 12 Ω ∥ 6 Ω pair?

    Answer and why

    B. The pair acts like 4 Ω carrying 3 A: V = 3 × 4 = 12 V. The series 4 Ω takes the other 12 V (3 × 4), and 12 + 12 = 24 V ✓.

  5. How much current flows in the 6 Ω branch?

    Answer and why

    B. The 6 Ω branch has 12 V across it: 12 ÷ 6 = 2 A. The 12 Ω branch carries 12 ÷ 12 = 1 A. Together 3 A — the supply current ✓.

  6. (10 Ω + 20 Ω) in parallel with 30 Ω. Total resistance?

    Answer and why

    C. The series branch is 10 + 20 = 30 Ω. Now two equal 30 Ω branches in parallel: 30 ÷ 2 = 15 Ω.

  7. Why is a real branch circuit a combination circuit?

    Answer and why

    B. The loads are parallel branches, but the hot and neutral conductors feeding them are resistances in series with the lot. That series resistance is where voltage drop comes from.

Questions fréquentes

How do I tell if two resistors are in series?

They are in series if all the current through one has to go through the other — nothing joins or leaves at the point between them. If both ends of two resistors connect to the same two points, they are in parallel. If neither is true, reduce something else first and look again.

Do I need Kirchhoff’s laws for this?

You are already using them. “Currents into a point equal currents out” and “voltage drops around a loop add up to the supply” are Kirchhoff’s two laws in plain words. For the circuits in the first year, series and parallel reduction plus those two checks is all you need.

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