Resistors in Series and Parallel Calculator

Series resistances add, parallel resistances combine as 1 ÷ (1/R₁ + 1/R₂ + …), and a divider output is Vin × R₂ ÷ (R₁ + R₂).

Ω
Ω
Third resistor
Ω
V
Total in series
3kΩ
Total in parallel666.667 Ω
Divider output across R₂ (R₁ over R₂)8 V
Current in the series circuit4 mA
Power in the series circuit48 mW

Show the math

Series = 1,000 + 2,000 = 3,000 Ω
Parallel = 1 ÷ (1/1,000 + 1/2,000) = 666.667 Ω
Divider = 12 V × 2,000 ÷ (1,000 + 2,000) = 8 V
Series current = 12 ÷ 3,000 = 0.004 A; power = 12² ÷ 3,000 = 0.048 W

Rounded the same way as the result above.

How it works

Resistors in series carry the same current, so their voltage drops add and their resistances add. Resistors in parallel share the same voltage, so their conductances (1/R) add and the total is the reciprocal of that sum, always less than the smallest resistor.

A voltage divider is two resistors in series: the output taken across the second resistor is the source voltage times R₂ divided by the total. It assumes nothing else loads the output.

Series: R = R₁ + R₂ + R₃ Parallel: 1 ÷ R = 1 ÷ R₁ + 1 ÷ R₂ + 1 ÷ R₃ Divider: Vout = Vin × R₂ ÷ (R₁ + R₂)

Worked example

Resistors of 1 kΩ and 2 kΩ on a 12 V source:

  1. Series = 1,000 + 2,000 = 3,000 Ω
  2. Parallel = 1 ÷ (1/1,000 + 1/2,000) = 666.667 Ω
  3. Divider = 12 V × 2,000 ÷ (1,000 + 2,000) = 8 V
  4. Series current = 12 ÷ 3,000 = 0.004 A; power = 12² ÷ 3,000 = 0.048 W
InputValue
Resistor 11000 Ω
Resistor 22000 Ω
Third resistorNot used
Resistor 3 (if used)3000 Ω
Source voltage12 V
ResultValue
Total in series3 kΩ
Total in parallel666.667 Ω
Divider output across R₂ (R₁ over R₂)8 V
Current in the series circuit4 mA
Power in the series circuit48 mW

Assumptions and limits

  • Ideal resistors; tolerances, temperature coefficients, and power ratings are not modeled.
  • The divider output is unloaded. A load on Vout lowers it, since the load is in parallel with R₂.
  • The series current and power are for all resistors connected in one loop across the source.

Common questions

How do I find the total resistance in parallel?

Add the reciprocals of the resistances and take the reciprocal of the sum. For two resistors this is R₁R₂ ÷ (R₁ + R₂).

Why is parallel resistance always smaller?

Each added path gives the current another way through, so the total opposition falls below the smallest single resistor.

How does a voltage divider work?

Two series resistors split the source voltage in proportion to their resistances; the output across R₂ is Vin × R₂ ÷ (R₁ + R₂).

Does it account for loading?

No. Put the load’s resistance in parallel with R₂ and use the parallel result as the lower resistor.

Sources

  • Series and parallel resistance and the voltage-divider rule, from standard circuit theory (Kirchhoff’s laws with Ohm’s law).

Updated 2026-09-30