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How to Calculate Series and Parallel Resistors

August 29, 2026 6 min read
Quick answer

Resistors in series add up: R = R1 + R2 + … (the value increases). Resistors in parallel combine as 1/R = 1/R1 + 1/R2 + … (the value is smaller than the smallest resistor). For two resistors the handy shortcut is R = (R1 × R2) / (R1 + R2). For example, 100 Ω and 220 Ω give 320 Ω in series and ~68.75 Ω in parallel.

By wiring resistors in series or parallel you can obtain a value you do not have on hand and set how current and power are shared. The two arrangements change the total resistance in opposite directions: series increases it, parallel decreases it. Below are the formulas, handy shortcuts and worked examples.

Series resistors: the values add up

In a series connection the resistors are placed end to end and the same current flows through all of them. The total (equivalent) resistance is simply the sum:

Rtotal = R1 + R2 + … + Rn

The result is always larger than the biggest resistor. In a series circuit the source voltage is shared between the resistors in proportion to their values (series vs parallel circuits).

Parallel resistors: reciprocals decrease the value

In a parallel connection the resistor terminals are common and the same voltage is applied to all of them; the current splits between branches. The equivalent resistance is the reciprocal of the sum of reciprocals:

1 / Rtotal = 1/R1 + 1/R2 + … + 1/Rn

The result is always smaller than the smallest resistor. Two common shortcuts:

CaseShortcut
Two resistors (R1, R2)R = (R1 × R2) / (R1 + R2)
n EQUAL resistors (R)Rtotal = R / n

Formula summary

ConnectionFormulaTotal resistanceShared quantity
SeriesR = R1 + R2 + …IncreasesSame current
Parallel1/R = 1/R1 + 1/R2 + …DecreasesSame voltage

Worked examples

  • Series: 100 Ω + 220 Ω = 320 Ω.
  • Parallel (equal): 100 Ω ‖ 100 Ω = 100 / 2 = 50 Ω.
  • Parallel (unequal): 1 kΩ ‖ 2 kΩ = (1000 × 2000) / (1000 + 2000) = 2,000,000 / 3000 ≈ 667 Ω.

How to solve a mixed (series + parallel) circuit

In mixed circuits simplify from the inside out: first reduce parallel groups to a single resistor, then add the series values. Example: 100 Ω in series with a (300 Ω ‖ 600 Ω) parallel group.

  1. Parallel group: (300 × 600) / (300 + 600) = 180,000 / 900 = 200 Ω.
  2. Series total: 100 Ω + 200 Ω = 300 Ω.

Common mistakes

  • Forgetting the reciprocals in parallel: the parallel result is not the average of the two values; work with reciprocals.
  • Using the shortcut for more than two resistors: R1×R2/(R1+R2) is for TWO resistors only; use the full reciprocal formula for three or more.
  • Ignoring power sharing: in a series circuit the largest resistor dissipates the most power; check each resistor's power rating.

Use the calculator

To enter several resistors in series or parallel and get the equivalent value instantly, use our Parallel/Series Resistor Calculator.

Frequently Asked Questions

How do you calculate series resistors?+
Series resistors add up: R = R1 + R2 + … + Rn. For example 100 Ω + 220 Ω = 320 Ω. The result is larger than the biggest resistor.
How do you calculate parallel resistors?+
Use 1/R = 1/R1 + 1/R2 + …. For two resistors the shortcut is R = (R1 × R2) / (R1 + R2). The result is smaller than the smallest resistor.
What is the shortcut for two parallel resistors?+
R = (R1 × R2) / (R1 + R2). For example 1 kΩ ‖ 2 kΩ = 2,000,000 / 3000 ≈ 667 Ω. This shortcut works only for two resistors.
What happens when equal resistors are in parallel?+
For n equal resistors R in parallel the total is R/n. For example two 100 Ω in parallel = 50 Ω, four = 25 Ω.
How do you solve a mixed series-parallel circuit?+
Simplify from the inside out: first reduce parallel groups to a single resistor, then add the series values.