When resistors sit side by side across the same two points, they're in parallel, and their combined resistance follows the reciprocal rule: 1/Rt = 1/R₁ + 1/R₂ + …. The key intuition is that every extra path gives the current another way through, so the total resistance always drops below the smallest branch. For just two resistors there's a neat shortcut — product over sum — and for equal resistors it's simply R divided by how many there are.
Reviewed: June 20, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: the parallel-resistance reciprocal rule and product-over-sum identity.
The parallel resistance equations
Conductance (1/R) is what adds in parallel, so sum the reciprocals of each resistor and invert the total to get Rt. The product-over-sum form is just the two-resistor case of that rule, handy for mental maths. For three or more, either use the full reciprocal sum or apply product-over-sum pairwise. Whatever the mix, the answer is always smaller than the smallest individual resistor.
Worked example — mixed resistors
Scenario: A 100 Ω and a 300 Ω resistor are connected in parallel. What is the total resistance?
The pair combine to 75 Ω — below the 100 Ω branch, as expected. Swap the 300 Ω for another 100 Ω and the total drops to 50 Ω (equal resistors, R/n); use a large 1000 Ω instead and it barely moves to 90.9 Ω, because a high-value resistor adds little conductance. Add a third 100 Ω branch to the original pair and the reciprocal sum gives 1/100 + 1/300 + 1/100 ≈ 0.02333, for Rt ≈ 42.9 Ω.
Frequently Asked Questions
1/Rt = 1/R₁ + 1/R₂ + …, then invert. Two 100 Ω → 50 Ω.
For two resistors: Rt = R₁R₂/(R₁+R₂). e.g. (100×300)/400 = 75 Ω.
Each parallel path adds a route for current, so Rt drops below the smallest branch.
R/n. Two 100 Ω = 50; three = 33.3; four = 25 Ω.
Same voltage; current splits — the smallest resistor carries the most. (Series is the opposite.)