F. Electricity and Magnetism · Electricity and Magnetism

Analysing Mixed Series-Parallel Circuits

F2.2 F3.4 A1.12About 75 minutes

Learning goals

I can recognize which resistors in a mixed circuit are in series and which are in parallel.
I can reduce a mixed circuit step by step to a single equivalent resistance.
I can work backwards to find the current, potential difference and power for every resistor.
I can check my solution using Kirchhoff's current and voltage laws.

Words to know

mixed circuit
A circuit with some parts in series and some in parallel; also called a series-parallel or combination circuit.

equivalent resistance
The single resistance that could replace a group of resistors without changing the current drawn from the source.

branch
One of the separate paths between two junctions in a parallel section.

power dissipated
The rate at which a resistor turns electrical energy into thermal energy: P = VI = I²R = V²/R.

Real circuits are mixed

In your home, the outlets and lights on one circuit are in parallel with each other, but the whole group is in series with the breaker and the wires back to the panel. Car headlights, a stereo's speaker system and a string of LED lights are all mixed circuits. There is no single formula for a mixed circuit. Instead you use a strategy: simplify, solve, then rebuild.

The strategy

Find the innermost group that is purely in series or purely in parallel. Two resistors are in series if the same current must pass through both, with no junction between them. They are in parallel if both ends of each connect to the same two junctions.
Replace that group with its equivalent resistance. Redraw if it helps.
Repeat until only one resistance, R_T, remains.
Find the total current: I_T = V_T ÷ R_T.
Work backwards through your redrawn circuits. Current through series parts equals the current into that part; potential difference across parallel parts is shared.
Check with KCL at every junction and KVL around every loop. Optionally check that the powers add to P_T = V_T I_T.

A table with columns R, I, V and P keeps everything organized.

Example 1's circuit. R₂ and R₃ connect to the same two junctions (dots), so they are in parallel. That group is in series with R₁, because all the current through R₁ must pass through the group.

Worked example

Example 1: One resistor in series with a parallel pair

In the circuit above, a 24 V battery is connected to R₁ = 4.0 Ω in series with a parallel group of R₂ = 12 Ω and R₃ = 6.0 Ω. Find R_T, and the current through and potential difference across each resistor.

Parallel group: 1/R_p = 1/12 + 1/6.0 = 1/12 + 2/12 = 3/12, so R_p = 4.0 Ω.
Series with R₁: R_T = 4.0 + 4.0 = 8.0 Ω.
Total current: I_T = 24 ÷ 8.0 = 3.0 A. All of it passes through R₁.
V₁ = I_T R₁ = (3.0)(4.0) = 12 V. By KVL, the parallel group gets 24 − 12 = 12 V (or V_p = I_T R_p = (3.0)(4.0) = 12 V).
Branch currents: I₂ = 12 ÷ 12 = 1.0 A; I₃ = 12 ÷ 6.0 = 2.0 A.
KCL check: 1.0 + 2.0 = 3.0 A ✓.

R_T = 8.0 Ω; I₁ = 3.0 A, V₁ = 12 V; I₂ = 1.0 A, I₃ = 2.0 A, V₂ = V₃ = 12 V.

Example 2's circuit. R₂ and R₃ are in series with each other (one branch). That branch is in parallel with R₄. R₁ and R₅ carry the total current, so they are in series with the whole group.

Worked example

Example 2: A five-resistor circuit with powers

Use the circuit above (18 V source). Find R_T and the current, potential difference and power for every resistor. Check that the powers add up to the power supplied.

Step 1: R₂ and R₃ are in series in one branch: 4.0 + 2.0 = 6.0 Ω.
Step 2: That 6.0 Ω branch is in parallel with R₄ = 3.0 Ω: R_p = (6.0 × 3.0) ÷ (6.0 + 3.0) = 18 ÷ 9.0 = 2.0 Ω.
Step 3: R₁, R_p and R₅ are in series: R_T = 3.0 + 2.0 + 1.0 = 6.0 Ω.
I_T = 18 ÷ 6.0 = 3.0 A, which flows through R₁ and R₅. V₁ = (3.0)(3.0) = 9.0 V; V₅ = (3.0)(1.0) = 3.0 V.
The parallel group: V_p = I_T R_p = (3.0)(2.0) = 6.0 V. (KVL check: 9.0 + 6.0 + 3.0 = 18 V ✓.)
R₄ branch: I₄ = 6.0 ÷ 3.0 = 2.0 A. R₂–R₃ branch: I = 6.0 ÷ 6.0 = 1.0 A, so V₂ = (1.0)(4.0) = 4.0 V and V₃ = (1.0)(2.0) = 2.0 V. (KCL: 2.0 + 1.0 = 3.0 A ✓.)
Powers (P = VI): P₁ = (9.0)(3.0) = 27 W; P₂ = (4.0)(1.0) = 4.0 W; P₃ = (2.0)(1.0) = 2.0 W; P₄ = (6.0)(2.0) = 12 W; P₅ = (3.0)(3.0) = 9.0 W. Sum = 54 W. Supplied: P_T = (18)(3.0) = 54 W ✓ (conservation of energy).

R_T = 6.0 Ω, I_T = 3.0 A. See the table below for every resistor.

Resistor	R (Ω)	I (A)	V (V)	P (W)
R₁	3.0	3.0	9.0	27
R₂	4.0	1.0	4.0	4.0
R₃	2.0	1.0	2.0	2.0
R₄	3.0	2.0	6.0	12
R₅	1.0	3.0	3.0	9.0
Total	6.0	3.0	18	54

The completed analysis table for Example 2.

The reduction steps for Example 2. Simplify from the inside out, then reverse the steps to find each current and potential difference.

Three forms of the power equation

Combining P = VI with V = IR gives P = I²R and P = V²/R. Use whichever needs the values you already have. P = I²R is handy for series parts (same I); P = V²/R is handy for parallel parts (same V).

Just for fun

Why was the resistor in series so stressed out?

Because every single bit of current had to go through it. The parallel resistors just split the work.

Hands-on activity

Three-resistor puzzle

You need: paper, a pencil, a calculator, and (optional) the free PhET "Circuit Construction Kit: DC" simulation.

Imagine you have three identical 6.0 Ω resistors and must use all three every time.
Sketch every different way to connect them between two terminals. (Hint: there are four distinct arrangements.)
Calculate R_T for each. Order them from smallest to largest.
Optional: build each one in the simulation with a 12 V battery and check that I_T = 12 ÷ R_T.

What to look for: all in parallel, 2.0 Ω; two in parallel with the third in series, 3.0 + 6.0 = 9.0 Ω; two in series with the third in parallel, (12 × 6.0) ÷ 18 = 4.0 Ω; all in series, 18 Ω. With 12 V, the currents in the same order are 6.0 A, 1.3 A, 3.0 A and 0.67 A.

Your turn

A 6.0 Ω resistor is in series with a parallel pair of two 10 Ω resistors. Find R_T.

A branch with 2.0 Ω and 4.0 Ω in series is connected in parallel with a 3.0 Ω resistor. Find the equivalent resistance.

A 12 V battery is connected to R₁ = 2.0 Ω in series with a parallel pair of 6.0 Ω and 3.0 Ω. Find I_T, V₁, the potential difference across the pair, and the current in each branch.

In the last question, which resistor dissipates the most power? Show that the powers add up to the power supplied.

A 9.0 V battery is connected to a 1.0 Ω resistor in series with two 4.0 Ω resistors in parallel. Find I_T and the current in each 4.0 Ω resistor.

In the last question, one of the 4.0 Ω resistors is removed (its branch is opened). Find the new I_T and the new potential difference across the remaining 4.0 Ω resistor. Did it go up or down?

More practice: warm-up

Find the equivalent resistance of 3.0 Ω and 6.0 Ω in parallel.

A 2.0 Ω resistor is in series with the pair from the last question. Find R_T.

Two 10 Ω resistors in parallel are in series with a 5.0 Ω resistor. Find R_T.

When you begin analysing a mixed circuit, what should you look for first?

True or false: resistors in the parallel part of a mixed circuit always carry the same current.

More practice: on your own

A 20 V source is connected to R₁ = 5.0 Ω, then a parallel pair of two 20 Ω resistors, then R₄ = 5.0 Ω, all in series. Find I_T, V₁, V₄, the voltage across the pair, and the current in each 20 Ω resistor.

A 30 V source is connected to two parallel branches. Branch A has 10 Ω and 20 Ω in series. Branch B has 15 Ω and 15 Ω in series. Find R_T, I_T and the potential difference across each resistor.

Three identical lamps: L₁ is in series with a parallel pair L₂ and L₃. Which lamp is brightest? Compare its power with each of the others, assuming constant resistance.

A total current of 2.0 A flows through R₁ = 4.0 Ω and then into a parallel pair: a 6.0 Ω resistor carrying 1.5 A and an unknown resistor R. Find R and the source voltage.

What source voltage is needed to drive 4.0 A through a 2.0 Ω resistor in series with a parallel pair of 3.0 Ω and 6.0 Ω?

Challenge

In the three-lamp circuit above (L₁ in series with L₂ ∥ L₃), lamp L₃ burns out. Using a source voltage V and lamp resistance R, explain what happens to the brightness of L₁ and L₂.

A 24 V source is connected to R₁ = 2.0 Ω in series with a parallel group of three resistors: 6.0 Ω, 12 Ω and 12 Ω. Find I_T and the current in each parallel resistor.

Show that for the circuit in "On your own" question 1 the powers of all four resistors add up to the power supplied.

Check your understanding

Quiz

Two resistors are in parallel when…

A 4.0 Ω resistor is in series with a parallel pair of 8.0 Ω and 8.0 Ω. What is R_T?

The circuit in the last question is connected to 16 V. What is the current in each 8.0 Ω resistor?

After finding R_T and I_T for a mixed circuit, what is the best next step?

In a mixed circuit, the powers of all the resistors add to 48 W. The source is 12 V. What is the total current?

Which expression gives the power of a resistor when you know only its current and resistance?

For parents and teachers

Big idea: Mixed circuits (F2.2) are solved with a strategy, not a single formula: reduce step by step, find the total current, then work backwards using Ohm's law and Kirchhoff's laws (F3.4).

How to help: Ask your teen to redraw the circuit after each simplification step, and to fill in an R–I–V–P table. Checking KCL at junctions and KVL around loops catches almost every mistake.

Common difficulties:

Calling two resistors "in series" just because they are next to each other on paper. Ask: must the same current go through both, with no junction in between?
Giving every resistor the full source voltage, or splitting the total current equally among unequal branches.
Losing track during the backwards stage. A redrawn sketch at each stage helps a lot.
Rounding intermediate values too much; keep at least three digits until the end.