How this instrument works
Parallel means both parts bridge one pair of nodes, so both sit under an identical voltage. Each draws its own current, V⁄R₁ and V⁄R₂, and those two streams merge at that junction — which is why an honest statement of this rule concerns conductance rather than resistance. Conductances add, plainly: G = G₁ + G₂, in siemens, a reciprocal ohm that carried the joke name mho until 1971, when the 14th General Conference on Weights and Measures settled on siemens. Invert that sum and product over sum drops out.
Node bookkeeping belongs to Gustav Kirchhoff, who laid out his two circuit laws in an 1845 paper written as a student at Königsberg, barely twenty-one years old. His current law — whatever charge arrives at a junction must leave it — is what licenses that addition; Ohm's relation only supplies each branch its share. One practical corollary is familiar to anyone who has stitched extra copper onto a ground plane: a second route can never raise resistance, only lower it, because charge has somewhere else to go.
Product over sum is exact, though only for exactly two branches and only under assumptions worth naming aloud. Wiring is treated as zero ohms, which stops being fair around milliohm current shunts, where a few centimetres of trace outweigh those parts themselves. Both resistors are assumed ohmic and sitting at one temperature, so self-heating that nudges either value quietly nudges your total. Past a few megahertz, lead inductance and stray capacitance turn each resistor into a little network of its own, and no single number describes that pair any more.
- Enter your first value in Resistor 1. Its unit menu takes kΩ and MΩ as well, so a 4.7 kΩ pull-up needs no hand conversion.
- Enter your second in Resistor 2 — order never matters here, because parallel combination is commutative.
- Read Total resistance, returned to six figures. Sanity check it: your answer must land below whichever input is smaller.
- For three or more branches, fold them a pair at a time — put each running total back into Resistor 1 and your next part into Resistor 2.
Worked example — two 100 Ω parts making a 50 Ω load
Radio gear expects 50 ohms, and 50 is not a figure you can buy off an E12 shelf: 47 and 56 sit either side of that gap. Bench builders reach for a pair of 100 Ω metal film parts instead. Put 100 into Resistor 1 and 100 into Resistor 2, and Total resistance returns (100 × 100) ⁄ (100 + 100) = 10000 ⁄ 200 = 50 ohms, exactly. Equal branches always halve, one case anybody can verify without an instrument at all.
Pairing buys more than a value. Each part now carries half of that current, so together they shed twice as much heat as one could survive — two quarter-watt bodies handle half a watt. Their tolerances partly cancel as well, since two independent 1% errors average toward about 0.7%. Two small packages side by side also present less lead inductance than one large body, which explains why a dummy load still behaving at 200 MHz is almost never built from a single part.
Questions
Why is the total always smaller than either resistor?
Because you have added a route, not an obstacle. Charge that previously had to cross R₁ now finds R₂ open alongside it, so a pair conducts more than either branch alone and its equivalent resistance falls below whichever value is smaller. That inequality doubles as your quickest error check: enter 100 Ω beside 400 Ω, and any answer above 100 is arithmetic gone astray. Correct here is 80 Ω.
Does product over sum work for three resistors?
No, and mistaking it for a general rule is this page's most common slip. R₁R₂R₃ ⁄ (R₁+R₂+R₃) is simply not equivalent to three branches; product over sum is what a reciprocal sum collapses to when exactly two terms are present. With three or more, either add reciprocals directly or fold them pairwise — combine R₁ with R₂, then combine that result with R₃. Pairwise folding is legitimate because parallel combination is associative, and it is exactly what this instrument is built around.
How do two parallel resistors split the current?
In inverse proportion to their resistances, which is why a divider expression carries its opposite resistor upstairs: I₁ = I_total · R₂ ⁄ (R₁ + R₂). Writing R₁ in that numerator is a routine mistake. Set 100 Ω beside 400 Ω and your smaller branch takes four fifths of all current — and, since one voltage spans both, four fifths of that heat too. Never assume unequal branches share a load evenly; rate whichever part is smaller for what it will genuinely see.
What happens if one of the resistances is zero?
Total resistance collapses to zero. A dead short across that pair wins regardless of what sits beside it, and product over sum reports this correctly, since R₁ = 0 empties its numerator. An opposite edge case is quietly useful too: an open branch behaves as infinite resistance, and infinity in parallel changes nothing whatsoever, which is how a disconnected part can hide inside an apparently working circuit until somebody goes looking.
What units should I enter, and what does my answer come back in?
Whatever matches your parts — each field accepts ohms, kilohms, and megohms, and every entry converts to ohms before arithmetic runs. An ohm is SI's derived unit of resistance, one volt per ampere; since a 2019 revision it has been realised from quantum Hall resistance, where von Klitzing's constant h⁄e² sits at exactly 25812.80745 Ω. For mental checks, remember that two branches sharing a prefix keep it: 10 kΩ beside 10 kΩ gives 5 kΩ.
Where does this combination actually show up in practice?
Anywhere two things hang off one rail. Bus pull-ups are a classic trap — a serial bus carrying 10 kΩ of pull-up at its controller plus another on each of two plug-in modules is really running about 3.3 kΩ, and its drivers may struggle to pull a clean low. CAN wiring exploits this deliberately: 120 Ω terminators at both ends read 60 Ω on a meter, which is a standard bench check that both are fitted. Current-sense shunts get stacked in parallel to reach milliohm values nobody stocks.