Method and source check
Last checked:
Automated source and implementation review, including worked examples and documented numerical limits. This review does not provide professional advice or specialist approval.
Method
For two to ten ideal positive resistors, sum 1/R for each parallel branch and invert the sum to get equivalent resistance.
R_eq = 1 / Σ(1/R_i)
Variables and units
- R_i
- each resistance in Ω
- R_eq
- equivalent resistance in Ω
Worked example
100Ω in parallel with 100Ω gives 50Ω.
Limitations
The calculation retains floating-point precision, and the UI displays up to six significant digits; the tested 0.00005 Ω resistance remains visible.
- Ideal positive-resistor arithmetic only; component tolerance, heating, and physical installation are outside this calculation.
- The form schema rejects derived overflow and unsupported numerical combinations before showing results; the calculation helper expects schema-validated input.
- 100Ω || 100Ω → 50Ω
- 10Ω || 1000Ω → about 9.90099Ω
- 0.0001Ω || 0.0001Ω → 0.00005Ω
- Two 1e-320 Ω resistors are rejected on a resistor input because their reciprocal sum overflows.