Article — Parallel Resistor Calculator
Parallel resistor calculator and formula
For resistors connected in parallel, the reciprocal of the total resistance equals the sum of the reciprocals of each branch: 1/R_total = sum(1/R_i). Two 10 kΩ resistors in parallel give 5 kΩ; three 1 kΩ resistors give 333 Ω. The total is always less than the smallest individual resistor in the network.
Parallel resistors are everywhere: power supply current sharing, voltage dividers, audio attenuators, terminator networks on data buses. The formula scales from two resistors to dozens, and the same equation governs parallel conductors in a wire bundle and parallel inductors in an RF tank circuit.
What does parallel resistance mean
Two resistors are in parallel when both ends are tied to the same pair of nodes. Current entering the top node splits between the two paths, then recombines at the bottom node. Each resistor sees the same voltage across its terminals, but the currents through them differ unless the resistors are identical.
Compare to series: in series, the same current flows through both resistors and the voltage divides between them. In parallel, the same voltage appears across both and the current divides. These are dual topologies, and you can convert between them via the reciprocal transformation.
Two equal resistors in parallel cut the resistance in half. Three give a third. Ten give a tenth. For n equal resistors of value R in parallel, R_total = R / n. Useful when you need a non-standard value: ten 1 kΩ gives 100 Ω with the power handling of ten 1/4 W parts.
The parallel resistor formula
The general equation sums reciprocals. The two-resistor case has a popular shortcut. Both produce the same numerical result.
1/R_t = 1/R_1 + 1/R_2 +... + 1/R_nR_t = R_1 R_2 / (R_1 + R_2) (two only)R_t = R / n (n equal resistors)Worked examples:
- Two 1 kΩ — R_t = (1 × 1) / (1 + 1) = 0.5 kΩ = 500 Ω
- 1 kΩ and 4.7 kΩ — R_t = (1 × 4.7) / (1 + 4.7) = 4.7 / 5.7 = 824 Ω
- 10 Ω, 20 Ω, 30 Ω — 1/R_t = 0.1 + 0.05 + 0.0333 = 0.1833. R_t = 1 / 0.1833 = 5.45 Ω
- Six 100 Ω — R_t = 100 / 6 = 16.67 Ω. Total power capacity is 6× a single resistor.
Two-resistor parallel shortcut
The product-over-sum form is convenient for hand calculation. It comes directly from algebraic manipulation of 1/R_t = 1/R_1 + 1/R_2.
R_t = (R_1 × R_2) / (R_1 + R_2). Multiply the two resistor values, divide by their sum. Done. For 470 Ω and 220 Ω in parallel: (470 × 220) / (470 + 220) = 103,400 / 690 = 150 Ω.
For three or more resistors, you can iterate the two-resistor formula: combine R1 and R2, then combine that result with R3, and so on. It is slower than the reciprocal-sum form but easier on a four-function calculator.
Parallel resistors vs series resistors
The two topologies do opposite things to total resistance.
Series resistors share current and divide voltage. The current through each is identical; the voltage across each is proportional to its resistance. Two 10 kΩ across 10 V drop 5 V each, carrying 0.5 mA total.
Parallel resistors share voltage and divide current. The voltage across each is identical; the current through each is inversely proportional to its resistance. Two 10 kΩ across 10 V carry 1 mA each, totaling 2 mA, and behave as a single 5 kΩ load.
Practical applications of parallel resistors
Engineers use parallel combinations for several reasons.
Reaching non-standard values. Standard E24 resistors come in steps like 1.0, 1.1, 1.2, 1.3, 1.5, 1.6, 1.8. Need 850 Ω? Put a 1 kΩ in parallel with a 5.6 kΩ (=847 Ω) or two 1.5 kΩ with a 2.2 kΩ in series. Many vintage designs use this technique to hit a precise filter cutoff or gain.
Power sharing. A single 100 Ω 1/4 W resistor handles 250 mW. Four 100 Ω resistors in series-parallel (two pairs in series, then in parallel) form a single 100 Ω with 1 W rating. The current divides equally and each part dissipates a quarter of the total.
Redundancy. If one of three parallel 1 kΩ resistors opens, the remaining two still conduct. The total rises from 333 Ω to 500 Ω, a degradation but not a failure. Critical circuits like medical defibrillator load networks and high-voltage bleeder strings exploit this.
The 50 Ω characteristic impedance used in RF coax came from a 1929 Bell Labs trade-off: 77 Ω minimizes attenuation in air, 30 Ω maximizes power handling, and the geometric mean is around 50. Two 100 Ω in parallel terminate a 50 Ω line.
Power dissipation in parallel resistors
Each branch dissipates P = V²/R using the common voltage. Total power is the sum of branch powers, which also equals V²/R_total or I_total × V.
A 10 V supply across a 100 Ω and a 1 kΩ in parallel: 100 Ω dissipates 10²/100 = 1 W, 1 kΩ dissipates 10²/1000 = 0.1 W. Total 1.1 W. Note the 100 Ω carries 10× more current and dissipates 10× more power. If you must mix wattage ratings in parallel, oversize the lowest-value (and therefore hottest) resistor.
Power dissipation in a parallel branch is V²/R. Halving R doubles the power. Always check the small-value resistor first when sizing wattage in a parallel network. A 0.25 W rating on a 100 Ω resistor across 5 V is right at the limit.
Common parallel resistor mistakes
The formula is straightforward. Most errors are setup errors.
- Adding values directly — that is the series formula. Parallel uses reciprocals. Two 10 kΩ in parallel is 5 kΩ, not 20 kΩ.
- Mixing units — convert everything to ohms before summing reciprocals, or convert everything to kΩ. 1 kΩ with 470 Ω is 1000 with 470, not 1 with 470.
- Expecting a result larger than the smallest member — impossible. The smallest individual resistor sets an upper bound for the parallel total.
- Ignoring tolerance stack — five 1 percent resistors in parallel do not give 0.2 percent total. Use root-sum-square for independent tolerances: 1 / √5 = 0.45 percent.
- Forgetting the lowest-value branch heats up most — the small resistor in a mixed-value parallel network is the one most likely to fail. Size its wattage first.