Series-Parallel Combinations

DC Circuits · 9 min read

Your car's radio, climate control, and lights all run off the same 12 V battery. A single fuse protects them all — that fuse is in series with everything else.

The loads themselves are wired in parallel, each with its own private path to ground.

The real world is almost never purely series or purely parallel. It's a mix — and that mix has a name: a series-parallel combination.

What “Series-Parallel” Means

A series-parallel circuit has some components in series (same current) and some in parallel (same voltage) — mixed together in the same loop.

You don't need new laws. You need a method for reducing the mix back down to something Ohm's Law can solve in one step.

R₁ is in series with the source; R₂ and R₃ share the remaining voltage in parallel.

Reduction in Pictures

Any series-parallel circuit collapses to a single equivalent resistance in two steps.

Original circuit

↓ Step 1: Collapse R₂∥R₃ into one equivalent

R_{par}

↓ Step 2: Series sum gives

R_{total} = R_1 + R_{par}

The Four-Step Method

Collapse the parallel pair. Replace R₂∥R₃ with a single equivalent: $R_{par} = R_2 R_3 / (R_2 + R_3)$ .
Sum the series resistors. $R_{total} = R_1 + R_{par}$ .
Find source current. $I_{source} = V / R_{total}$ .
Expand outward. Voltage drops: $V_{R1} = I_{source} \times R_1$ , $V_{par} = V - V_{R1}$ . Branch currents: $I_2 = V_{par}/R_2$ , $I_3 = V_{par}/R_3$ .

The Formulas

All four steps in equation form:

R_{par} = \frac{R_2 R_3}{R_2 + R_3}

R_{total} = R_1 + R_{par}

I_{source} = \frac{V}{R_{total}}

V_{R1} = I_{source} \cdot R_1 \qquad V_{par} = V - V_{R1}

I_2 = \frac{V_{par}}{R_2} \qquad I_3 = \frac{V_{par}}{R_3}

Worked Example

R₁ = 4 Ω in series with (R₂ = 6 Ω ∥ R₃ = 12 Ω), source = 24 V.

Collapse: $R_{par} = (6 \times 12)/(6+12) = 72/18 = 4\text{ Ω}$
Total: $R_{total} = 4 + 4 = 8\text{ Ω}$
Source current: $I_{source} = 24/8 = 3\text{ A}$
Expand: $V_{R1} = 3 \times 4 = 12\text{ V}$ , $V_{par} = 24 - 12 = 12\text{ V}$ , $I_2 = 12/6 = 2\text{ A}$ , $I_3 = 12/12 = 1\text{ A}$
Verify: $I_2 + I_3 = 3\text{ A}$ ✓ $V_{R1}+V_{par} = 24\text{ V}$ ✓

Series-parallel problems need no new laws — just collapse inward for

R_{total}

, then expand outward for branch quantities.

Think Before Calculating

Predict before running the numbers — then verify with the four steps.

Scenario 1: R1=10 Ω, R2=20 Ω, R3=20 Ω at 12 V. Find total resistance and source current.

Scenario 2: R1=4Ω, R2=6Ω, R3=12Ω at 24V. Double R1 to 8 Ω — does source current increase or decrease?

Scenario 3: With R1 doubled to 8 Ω, what happens to the parallel voltage and branch currents?

Common Mistakes

Reduction Visualiser

Drag the sliders — the top bar shows how R₁ and the parallel pair split the source voltage; the bottom bar shows how the parallel pair splits the current.

Voltage24.0 V

1 V24 V

R₁4 Ω

R₂6 Ω

R₃12 Ω

R_{par} = R_2 \| R_3

4 Ω

R_{total}

8 Ω

I_{source}

3.000 A

Voltage drops

Current split in parallel pair

✅ Identify purely series or purely parallel clusters first
✅ Collapse inward: parallel pair → $R_{par}$ , then add series R₁
✅ Find $I_{source}$ , then expand outward for voltages and branch currents
✅ $V_{par}$ is always less than $V_{source}$ (R₁ drops some voltage)
✅ Always verify with KCL and KVL at the end