Parallel Resistor Calculator
Enter two or more resistor values to get the parallel equivalent — always smaller than the smallest branch, with the conductance sum shown.
Last updated: 2026-09-28
How the calculation works
- Each parallel branch shares the same voltage; the currents add, so conductances (1/R) add and the equivalent is their reciprocal.
- The equivalent resistance is always LOWER than the smallest resistor — adding parallel paths always increases total current capability.
- For exactly two resistors the product-over-sum shortcut gives the same answer without fractions.
- Equal resistors divide: n equal R in parallel give R/n — two 1 kΩ give 500 Ω, four give 250 Ω.
Formula
1/R_eq = 1/R₁ + 1/R₂ + ... + 1/Rₙ two resistors shortcut: R_eq = R₁ × R₂ / (R₁ + R₂)
| Symbol | Meaning | Unit |
|---|---|---|
R_eq | Equivalent parallel resistance | Ω |
G | Conductance sum (1/R) | S |
Worked example
Interpreting the result
Parallel resistors divide current (and power) — two equal resistors each dissipate half the total, which is how you stretch a power rating (with derating for sharing mismatch). They also make non-standard values: parallel a 10 kΩ with 100 kΩ to trim down 9.1%. For current sharing to be accurate, resistors should be same tolerance and same technology.
Assumptions
- Ideal resistors — no stray capacitance or inductance (matters only at high frequency).
- Same voltage across every branch (true parallel connection).
Limitations
- Power sharing assumes equal-value, equal-tolerance resistors; unequal values dissipate proportionally to their conductance.
- Not for series-parallel networks — break those into stages and solve each.
Frequently asked questions
What is the formula for two resistors in parallel?
R_eq = R₁ × R₂ ÷ (R₁ + R₂) — product over sum. Two 10 Ω resistors give 5 Ω; a 10 Ω and 30 Ω give 7.5 Ω.
Why is parallel resistance always lower?
Every parallel branch is an additional current path at the same voltage — total current can only increase, so the equivalent resistance (V/I) can only decrease.