Two Resistors in Parallel Calculator
Product over sum for a pair of resistors, R1 times R2 divided by R1 plus R2: 1000 Ω with 2200 Ω gives 687.5 Ω of equivalent resistance.
Rp (Ω)
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Two resistors in parallel: product-over-sum
When you have exactly two resistors in parallel, the general formula collapses to something handy: R = (R₁·R₂) / (R₁ + R₂). That comes straight out of 1/R = 1/R₁ + 1/R₂. Whatever you get, it will be smaller than the smaller of the two resistors. Both see the same voltage, and the current splits inversely with resistance (I ∝ 1/R). A couple of examples: 100 Ω || 100 Ω works out to (100·100)/200 = 50 Ω, and 1 kΩ || 2.2 kΩ lands around 687.5 Ω.
Applications
It shows up when you need to split dissipated power across several parts, build in some redundancy on power-supply paths, or hit a non-standard value that the E12/E24 series doesn't offer. Parallel sense resistors for current measurement in switching regulators are another common case.
FAQ
Why is the parallel total smaller than the smallest resistor? Every extra path you give the current pulls the overall resistance down.
How does the current divide? It goes inversely with each resistor, so the smaller R carries more current at the same voltage.
How do I choose the wattage? Each resistor dissipates P = V²/R. The smaller ones burn more power, so rate them accordingly.
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The results provided by this tool are for general informational and educational purposes only and do not constitute professional, financial, medical, legal, tax or accounting advice. Always confirm important decisions with a qualified professional and official sources.