Electrical

Series and Parallel Circuits

Last updated: March 2026 · Beginner
Before you start

You should be comfortable with:

Real-world applications
Electrical

Voltage drop, wire sizing, load balancing

Circuits on the job are either series, parallel, or a combination of both. Understanding how resistance adds up in each configuration is essential for troubleshooting, calculating loads, and passing your journeyman exam.

Series Circuits

In a series circuit, components are connected end-to-end in a single path. Current has only one way to flow.

Series Rules

  1. Current is the same through every component: Itotal=I1=I2=I3I_{\text{total}} = I_1 = I_2 = I_3
  2. Voltage divides across each component: Vtotal=V1+V2+V3V_{\text{total}} = V_1 + V_2 + V_3
  3. Resistance adds up directly:

Rtotal=R1+R2+R3+R_{\text{total}} = R_1 + R_2 + R_3 + \ldots

Example 1: Three Resistive Loads in Series

Three heating elements of 4 Ω\Omega, 6 Ω\Omega, and 10 Ω\Omega are wired in series on a 240V circuit.

Total resistance:

Rtotal=4+6+10=20 ΩR_{\text{total}} = 4 + 6 + 10 = 20 \text{ }\Omega

Total current:

I=VR=24020=12 AI = \frac{V}{R} = \frac{240}{20} = 12 \text{ A}

Voltage across each element:

V1=12×4=48 VV2=12×6=72 VV3=12×10=120 VV_1 = 12 \times 4 = 48 \text{ V} \quad V_2 = 12 \times 6 = 72 \text{ V} \quad V_3 = 12 \times 10 = 120 \text{ V}

Check: 48+72+120=24048 + 72 + 120 = 240 V. Correct — the individual voltages add up to the source voltage.

Parallel Circuits

In a parallel circuit, components are connected across the same two points. Each component has its own path for current.

Parallel Rules

  1. Voltage is the same across every component: Vtotal=V1=V2=V3V_{\text{total}} = V_1 = V_2 = V_3
  2. Current divides among the branches: Itotal=I1+I2+I3I_{\text{total}} = I_1 + I_2 + I_3
  3. Resistance follows the reciprocal formula:

1Rtotal=1R1+1R2+1R3+\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots

The total resistance in a parallel circuit is always less than the smallest individual resistance.

The Product-Over-Sum Shortcut (Two Resistors Only)

When you have exactly two resistors in parallel, skip the reciprocal formula and use:

Rtotal=R1×R2R1+R2R_{\text{total}} = \frac{R_1 \times R_2}{R_1 + R_2}

This is faster on the exam and on the job.

Example 2: Two Parallel Branch Circuits

Two 120V branch circuits each feed a different baseboard heater. Heater A has 16 Ω\Omega resistance, Heater B has 24 Ω\Omega.

Total resistance (product-over-sum):

Rtotal=16×2416+24=38440=9.6 ΩR_{\text{total}} = \frac{16 \times 24}{16 + 24} = \frac{384}{40} = 9.6 \text{ }\Omega

Total current from the panel:

Itotal=1209.6=12.5 AI_{\text{total}} = \frac{120}{9.6} = 12.5 \text{ A}

Current through each heater:

IA=12016=7.5 AIB=12024=5 AI_A = \frac{120}{16} = 7.5 \text{ A} \qquad I_B = \frac{120}{24} = 5 \text{ A}

Check: 7.5+5=12.57.5 + 5 = 12.5 A. Correct — the branch currents add up to the total.

Example 3: Three Parallel Loads

A 120V circuit feeds three receptacle loads: 20 Ω\Omega, 30 Ω\Omega, and 60 Ω\Omega (all in parallel).

Total resistance using the reciprocal formula:

1Rtotal=120+130+160=360+260+160=660=110\frac{1}{R_{\text{total}}} = \frac{1}{20} + \frac{1}{30} + \frac{1}{60} = \frac{3}{60} + \frac{2}{60} + \frac{1}{60} = \frac{6}{60} = \frac{1}{10}

Rtotal=10 ΩR_{\text{total}} = 10 \text{ }\Omega

Total current:

I=12010=12 AI = \frac{120}{10} = 12 \text{ A}

Answer: The combined resistance is 10 ohms and the total current is 12 amps.

Series vs. Parallel Comparison

PropertySeriesParallel
CurrentSame through all componentsDivides among branches
VoltageDivides across componentsSame across all branches
Total resistanceRT=R1+R2+R_T = R_1 + R_2 + \ldots1RT=1R1+1R2+\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots
Total resistance vs. individualGreater than the largest RRLess than the smallest RR
If one component opensEntire circuit stopsOther branches continue
Common job-site exampleOld-style Christmas lights, some control circuitsReceptacles on a branch circuit, parallel heater elements

Quick Reference: Equal Resistors Shortcut

When all resistors in parallel are the same value:

Rtotal=RnR_{\text{total}} = \frac{R}{n}

where nn is the number of equal resistors. Three 30 Ω\Omega loads in parallel:

Rtotal=303=10 ΩR_{\text{total}} = \frac{30}{3} = 10 \text{ }\Omega

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Problem 1: Two resistors of 10 Ω\Omega and 15 Ω\Omega are in series on a 120V circuit. What is the total current?

Rtotal=10+15=25 ΩR_{\text{total}} = 10 + 15 = 25 \text{ }\Omega

I=12025=4.8 AI = \frac{120}{25} = 4.8 \text{ A}

Answer: The current is 4.8 amps.

Problem 2: Two 12 Ω\Omega resistors are wired in parallel. What is the total resistance?

Using the equal-resistor shortcut:

Rtotal=122=6 ΩR_{\text{total}} = \frac{12}{2} = 6 \text{ }\Omega

Or product-over-sum: 12×1212+12=14424=6 Ω\frac{12 \times 12}{12 + 12} = \frac{144}{24} = 6 \text{ }\Omega

Answer: The total resistance is 6 ohms — exactly half of one resistor.

Problem 3: A 240V circuit feeds two parallel heaters: 16 Ω\Omega and 48 Ω\Omega. What is the total current drawn from the panel?

Rtotal=16×4816+48=76864=12 ΩR_{\text{total}} = \frac{16 \times 48}{16 + 48} = \frac{768}{64} = 12 \text{ }\Omega

I=24012=20 AI = \frac{240}{12} = 20 \text{ A}

Answer: Total current is 20 amps. You can verify: I1=24016=15I_1 = \frac{240}{16} = 15 A, I2=24048=5I_2 = \frac{240}{48} = 5 A, and 15+5=2015 + 5 = 20 A.

Problem 4: Four identical 40 Ω\Omega lamps are wired in parallel on a 120V circuit. What current does the circuit draw?

Rtotal=404=10 ΩR_{\text{total}} = \frac{40}{4} = 10 \text{ }\Omega

I=12010=12 AI = \frac{120}{10} = 12 \text{ A}

Answer: The circuit draws 12 amps.

Problem 5: In a series circuit with a 10 Ω\Omega and 20 Ω\Omega resistor on 120V, how much voltage drops across the 20 Ω\Omega resistor?

Rtotal=10+20=30 ΩR_{\text{total}} = 10 + 20 = 30 \text{ }\Omega

I=12030=4 AI = \frac{120}{30} = 4 \text{ A}

V20=I×R=4×20=80 VV_{20} = I \times R = 4 \times 20 = 80 \text{ V}

Answer: The 20 Ω\Omega resistor drops 80 volts. The 10 Ω\Omega drops the remaining 40V.

Common Mistakes to Avoid

  1. Adding resistors in parallel like series. In parallel, resistance goes down, not up. Two 10 Ω\Omega resistors in parallel equal 5 Ω\Omega, not 20 Ω\Omega.
  2. Using product-over-sum with three or more resistors. The shortcut only works for exactly two resistors. For three or more, use the reciprocal formula.
  3. Forgetting to flip the reciprocal. The formula gives you 1Rtotal\frac{1}{R_{\text{total}}} — you must take the reciprocal of your result to get RtotalR_{\text{total}}.
  4. Assuming current is equal in all parallel branches. Current divides inversely with resistance — the branch with less resistance carries more current.

Key Takeaways

  • In series: resistance adds, current is constant, voltage divides
  • In parallel: resistance decreases, voltage is constant, current divides
  • The product-over-sum shortcut (R1×R2R1+R2\frac{R_1 \times R_2}{R_1 + R_2}) saves time for two-resistor parallel combinations
  • For nn equal parallel resistors: Rtotal=R/nR_{\text{total}} = R / n
  • Most branch circuits in a building are parallel — each outlet and fixture has the same voltage but draws its own current
  • If one load opens in parallel, the others keep working; in series, everything stops

Math for Electricians

Last updated: March 28, 2026