How this calculator works
Series wiring puts resistors end to end, so the same current crosses all of them and their resistances simply add. Parallel wiring gives current multiple paths at once, so the combined resistance is found from the reciprocals — and it always ends up lower than any single resistor in the group.
- R1, R2 — the two required resistors
- R3 — optional third resistor — leave at 0 to use only R1 and R2
- R_series — total resistance wired end to end
- R_parallel — total resistance wired side by side
More detail
The gap between series and parallel, in numbers
Take 100 Ω, 200 Ω, and 300 Ω. Wired in series they add straight up to 600 Ω. Wired in parallel the total drops to 600/11 ≈ 54.55 Ω — under a fifth of the series figure, and lower than even the smallest single resistor (100 Ω) in the set. Every extra parallel path gives current another way through, so adding a resistor in parallel can only pull the total down, never up.
Build tip. Don't have the exact resistor value a design calls for? Two resistors from a standard E12/E24 kit, combined in series or parallel, will often land closer to the target than any single stocked part.
Frequently asked questions
What's the combined resistance of 100 Ω and 200 Ω in parallel?
1/(1/100 + 1/200) = 1/0.015 ≈ 66.67 Ω. Leave R3 at 0 above to reproduce this with just two resistors.
What's the combined resistance of 100 Ω, 200 Ω, and 300 Ω in series?
Series just adds them: 100 + 200 + 300 = 600 Ω.
Why is the parallel total always smaller than the smallest resistor?
For 100 Ω, 200 Ω, and 300 Ω the parallel total is 600/11 ≈ 54.55 Ω, below even the 100 Ω resistor alone — because parallel wiring gives current three separate paths instead of one, and more paths always means less overall opposition.
Do capacitors follow the same rule?
No — capacitors are the reverse. Capacitors in series combine like resistors in parallel (1/C_total = 1/C1 + 1/C2 + …), and capacitors in parallel combine like resistors in series (C_total = C1 + C2 + …).