Add up to 6 resistors in parallel and get the total resistance instantly. Includes the two-resistor shortcut, the reverse solver and rules of thumb.
| Input | Result |
|---|---|
| 100 Ω ∥ 200 Ω | 66.67 Ω |
| Three 10 Ω in parallel | 3.33 Ω |
| 1 kΩ ∥ 100 Ω ∥ 10 Ω | 9.01 Ω |
For exactly two resistors, the reciprocal formula collapses to product over sum: R = R₁ × R₂ ÷ (R₁ + R₂). 100 Ω and 200 Ω: 100×200 ÷ 300 = 66.67 Ω. For three or more, the reciprocal sum this calculator uses is the only practical way.
You have R₁ in the drawer and want a target Rt. Solve the two-resistor formula for the partner: R₂ = R₁ × Rt ÷ (R₁ − Rt). Example: want 75 Ω and have a 100 Ω: R₂ = 100×75 ÷ 25 = 300 Ω. Note R₁ must be larger than the target — parallel resistance is always below every branch.
| Combination | Result |
|---|---|
| n equal resistors of R | R ÷ n (three 10 Ω → 3.33 Ω) |
| One resistor much bigger than the other | Total ≈ the smaller one (1 kΩ ∥ 10 Ω → 9.90 Ω) |
| Any parallel network | Always lower than the smallest branch |
The reciprocals add: 1/R_total = 1/R₁ + 1/R₂ + … This comes from Kirchhoff's laws: every branch sees the same voltage, so the branch currents add, and conductances (1/R) therefore add too.
Each added resistor is one more path for current. More total current at the same voltage means lower effective resistance — it can never exceed the easiest path alone.
Use the reverse formula with the resistor you already have: R₂ = R₁ × R_target ÷ (R₁ − R_target). To get 75 Ω from a 100 Ω you need 300 Ω in parallel. The resistor you start with must be bigger than the target.
Series is the simple case: values just add (R = R₁ + R₂). Parallel and series combine — solve ladder networks by collapsing one pair at a time.
Within their tolerance: a standard 5% (E24) resistor can sit anywhere in ±5%, and the parallel combination inherits roughly that spread. For precision dividers use 1% (E96) parts.