Parallel & Series Resistor Calculator

The parallel resistance formula

For any number of resistors in parallel, the reciprocals add:

1 / Rt = 1 / R1 + 1 / R2 + … + 1 / Rn

For exactly two, that rearranges into the form worth memorising — product over sum:

Rt = R1 × R2 / (R1 + R2)

And for n identical resistors it collapses to Rt = R / n: two 1 kΩ resistors in parallel are 500 Ω, three are 333 Ω, four are 250 Ω. In series there is no reciprocal at all — Rt = R1 + R2 + … + Rn, and two 1 kΩ resistors are 2 kΩ.

Two things follow from the formula that are worth having as instincts. The total in parallel is alwayssmaller than the smallest resistor in the set, so if an answer comes out larger than one of its legs, the arithmetic went wrong somewhere. And the result is dominated by the smallest leg: 1 kΩ with 100 kΩ across it is 990 Ω, a 1% change, which is inside the tolerance of a ±5% part.

Both directions of the same question

Forwards is the formula everyone learns: reciprocals add in parallel, values add in series. The panel at the top does that for any number of resistors, and accepts the shorthand you would write on a schematic —4k7, 1M, 220R.

Backwards is the one you actually need at a bench. You have worked out that you want 3.3 kΩ, you have a drawer of E12 parts, and 3.3 k is not one of them. The second panel searches the series for the pair that lands closest.

Why reciprocals

Conductance — the reciprocal of resistance — is what actually adds. Each parallel path offers the current another route, and the routes sum. Writing it as 1/Rt = 1/R1 + 1/R2 is saying the total conductance is the sum of the conductances, which is the physical statement; the reciprocal at the end just converts back to the unit we are used to.

Seen that way, two consequences stop being surprises. The total is always below the smallest leg, because adding conductance can only increase conductance. And a leg much larger than the others barely matters: 1 MΩ across 1 kΩ changes the total by 0.1%, which is less than the tolerance of either part.

Worked examples

Two in parallel

1 kΩ and 2.2 kΩ: 1000 × 2200 / (1000 + 2200) = 687.5 Ω. Below the smaller of the two, as it must be.

Reaching a value you cannot buy

Suppose you need 3.3 kΩ from E12 parts. The nearest single values are 2.7 k (18% low) and 3.9 k (18% high). Pair 3.9 k with 22 k and you get3900 × 22000 / 25900 = 3313 Ω — 0.4% high. The pairing error has gone from 18% to under half a percent without leaving the drawer.

Splitting dissipation

You need 50 Ω carrying 0.5 W and have only quarter-watt parts. Two 100 Ω parts in parallel give 50 Ω and, being equal, take half the power each — 0.25 W, which is exactly at the limit, so use four 200 Ω parts instead and run each at 0.125 W.

What pairing does not buy you

It is tempting to read a pair that lands 0.1% from the target as a 0.1% resistor. It is not. Two ±5% parts are still ±5% parts; the combination can sit anywhere in roughly that band, and the precision of the arithmetic has no bearing on it. Pairing solves availability — getting a value that is not manufactured — not accuracy.

Two further costs are worth knowing. Two parts occupy more board than one and give twice the opportunity for a bad joint. And their temperature coefficients do not cancel; if anything a mismatched pair drifts in a way a single part would not. Where accuracy is the goal, a single ±1% part from E96 beats a clever pair from E12 every time.

Series, briefly

Resistances in series add directly, and the tool reports that total alongside the parallel one because the list you typed answers both questions. Series is the way to reach a value above anything in the drawer, and it shares dissipation the same way parallel does — but in series the current is common and the largest resistor takes the most power, which is the opposite of the parallel case.