How to Add Resistors in Series and in Parallel

Resistors in series sit one after another on a single path. Resistors in parallel sit side by side between the same two points, giving the current several paths. The total resistance is worked out differently for each.

For learning and checking coursework only. Electrical work can kill. Follow the electrical code where you live and have a licensed electrician do or inspect real wiring.

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The steps

Series: add them: R = R₁ + R₂ + R₃ + …
Parallel: add the reciprocals, 1/R = 1/R₁ + 1/R₂ + 1/R₃ + …, then flip the answer over.
For just two in parallel, the shortcut is R = (R₁ × R₂) ÷ (R₁ + R₂).
For a mixed circuit, work out the innermost group first, replace it with its total, and keep going.
Check: a series total is more than the largest resistor; a parallel total is less than the smallest.

A worked example

Find the total resistance of 100 Ω, 220 Ω and 470 Ω (a) in series and (b) in parallel.

(a) Series: 100 + 220 + 470 = 790 Ω.
(b) Parallel: 1/R = 1/100 + 1/220 + 1/470 = 0.01 + 0.004545 + 0.002128 = 0.016673.
Flip it over: R = 1 ÷ 0.016673 ≈ 59.98 Ω. (Exactly 25,850/431 Ω.)
Check: 790 Ω is more than 470 Ω, and 59.98 Ω is less than 100 Ω. ✓

790 Ω in series and about 59.98 Ω in parallel.

Why it works

In series, the current has to push through every resistor in turn, so the resistances add up. In parallel, each extra path lets more current through: what adds is the conductance, 1/R, how easily each path conducts. Adding the conductances and flipping back gives the total resistance.

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Work it out: 100, 220 and 470 Ω in parallelOpens The Circuit Board with this example

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Common mistakes

Forgetting to flip. 0.016673 is 1/R, not R.
Adding parallel resistors. Each parallel path makes it easier for current to flow, so the total goes down, not up.
Using the two-resistor shortcut on three at once. Combine two first, then combine that result with the third.
Mixing kΩ and Ω. Put every resistor in the same unit before adding.

Questions people ask

What is 100 Ω in parallel with 220 Ω?

(100 × 220) ÷ (100 + 220) = 22,000 ÷ 320 = 68.75 Ω.

What about three equal 10 kΩ resistors in parallel?

For equal resistors in parallel, divide one of them by how many there are: 10 kΩ ÷ 3 ≈ 3.33 kΩ.

How do I work out 100 and 220 in parallel, in series with 470?

The parallel pair is 68.75 Ω. Add the 470 in series: 68.75 + 470 = 538.75 Ω.

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