2.2 Series, Parallel, and Combination Circuits

After Ohm's law, the next General Knowledge skill is recognizing how loads connect. Arkansas Journeyman items may show a simple schematic or describe "three 60 Ω resistors in parallel" and ask for total resistance or total current from a stated supply voltage. Getting series versus parallel rules backward is one of the fastest ways to miss an otherwise easy calculation point.

Series Circuits — One Path

In a series circuit there is only one current path. Every ampere that leaves the source returns through every component.

Rules:

Voltage divider: The voltage across one series resistor is proportional to its share of total resistance:

E_x = E_total × (R_x / RT)

Worked Example — Series Resistances

Three resistors — 10 Ω, 20 Ω, and 30 Ω — are connected in series across 120 V.

RT = 10 + 20 + 30 = 60 Ω

I = 120 / 60 = 2 A (same through each)

Voltage across the 30 Ω resistor:

E30 = 2 × 30 = 60 V

Or by divider: E30 = 120 × (30/60) = 60 V. Check: 20 + 40 + 60 = 120 V across the three parts.

Parallel Circuits — Shared Voltage

In a parallel circuit each branch sees the full source voltage. Currents split among branches and recombine at the return.

Rules:

Important check: Parallel RT is always smaller than the smallest individual branch resistance. If your answer is larger than the smallest R, you added instead of paralleling.

Two-Resistor Parallel Shortcut

For exactly two resistors:

RT = (R1 × R2) / (R1 + R2)

Equal resistors: two 40 Ω in parallel → RT = 20 Ω. Three equal resistors → RT = R / 3.

Worked Example — Parallel Branches

Two loads on a 120 V circuit: 40 Ω and 60 Ω in parallel.

RT = (40 × 60) / (40 + 60) = 2,400 / 100 = 24 Ω

I_total = 120 / 24 = 5 A

Branch currents:

I40 = 120 / 40 = 3 A

I60 = 120 / 60 = 2 A

3 + 2 = 5 A — confirms Kirchhoff's current law.

Three Unequal Parallel Resistors

Find RT for 10 Ω, 15 Ω, and 30 Ω in parallel:

1/RT = 1/10 + 1/15 + 1/30 = 0.100 + 0.067 + 0.033 = 0.200

RT = 1 / 0.200 = 5 Ω

Note 5 Ω < 10 Ω (smallest branch) — the sanity check passes.

On 120 V, I_total = 120 / 5 = 24 A. Individual branches: 12 A, 8 A, and 4 A respectively.

Combination Circuits

Real exam diagrams often mix series and parallel. Strategy:

1. Identify pure parallel groups and reduce each group to a single equivalent resistance.
2. Redraw: those equivalents now sit in series (or another parallel group).
3. Continue until one RT remains, then apply Ohm's law for total current.
4. Work backward: use series current or parallel voltage to find individual branch values asked in the stem.

Worked Example — Combination

A 10 Ω resistor is in series with a parallel combination of 20 Ω and 30 Ω. Supply is 120 V.

Step 1 — Parallel pair:

R_par = (20 × 30) / (20 + 30) = 600 / 50 = 12 Ω

Step 2 — Series with 10 Ω:

RT = 10 + 12 = 22 Ω

Step 3 — Total current:

I_total = 120 / 22 ≈ 5.45 A

Voltage across the parallel pair:

E_par = 5.45 × 12 ≈ 65.5 V

(The 10 Ω series resistor drops about 54.5 V; 54.5 + 65.5 = 120 V.)

Current through the 20 Ω branch:

I20 = 65.5 / 20 ≈ 3.28 A

Current through the 30 Ω branch:

I30 = 65.5 / 30 ≈ 2.18 A

3.28 + 2.18 ≈ 5.46 A ≈ I_total through the series 10 Ω — consistent.

Open and Short Concepts (Theory Wording)

• Open in a series string stops all current (I = 0). An open in one parallel branch kills only that branch; other branches still see source voltage.
• Short across a resistor forces nearly zero voltage across that path and can dump large current limited only by remaining circuit resistance and source capacity — overcurrent devices exist for this reason.

Exam Timing Tip

On the Prov Arkansas JW exam you have about three minutes per question on average. Series/parallel items should take under a minute if you:

1. Label whether the unknown is RT, I, or E_x.
2. Write the one-line formula before punching the calculator.
3. Run the "RT < smallest R?" check on every parallel answer.

Power in Series vs Parallel

Because power is additive for separate loads, total watts equal the sum of each resistor's dissipation. In series at 2 A through 10 Ω, 20 Ω, and 30 Ω: P = I²R gives 40 W + 80 W + 120 W = 240 W, which matches P = E × I = 120 × 2. In parallel at 120 V across 40 Ω and 60 Ω: P40 = E²/R = 360 W and P60 = 240 W for 600 W total — matching 120 V × 5 A. If your power totals disagree with E × I_total, revisit the resistance reduction.

Lighting strings marketed as "series holiday lights" behave like series filaments: one open extinguishes the string. Building receptacles and most utilization equipment are parallel on the branch so each device receives full circuit voltage — that mental model helps you reject series answers on dwelling wiring theory items.

These habits protect the nine General Knowledge points and free clock time for Article 250 and Article 430 lookups later in the session.