# Series and parallel resistor calculator

> Series and parallel resistor calculator: total resistance of any number of resistors, plus capacitors or inductors in series or parallel.

Interactive version: https://www.calcopenly.com/science/series-parallel-calculator
Subject: Science calculators

Resistors add in series, R = R₁ + R₂ + …, while in parallel their reciprocals add, 1/R = 1/R₁ + 1/R₂ + … . Inductors follow the same two rules when their magnetic fields do not interact. Capacitors do the opposite: they add in parallel and combine by reciprocals in series. Enter any number of values in one unit; 4.7k is read as 4,700.

The default, 100 Ω, 220 Ω and 470 Ω in parallel, gives 59.98 Ω, below the smallest part as every parallel network must be; the same three in series give 790 Ω. Combining parts is how a value outside the E-series is built, such as 500 Ω from two 1 kΩ resistors in parallel.

The parts are treated as ideal: no lead or contact resistance, no capacitor leakage and no mutual inductance between coils.

## Inputs

- **Components** (options: Resistors, Capacitors, Inductors)
- **Connected in** (options: Series, Parallel)
- **Values**: Separate with commas or spaces; 4.7k works for kilo
- **Unit** (options: mΩ, Ω, kΩ, MΩ)
- **Unit** (options: pF, nF, µF, mF, F)
- **Unit** (options: nH, µH, mH, H)

## Results

- Equivalent value — main result
- In base units
- Number of components

## Formula

$$
\text{Series: } R = \sum R_i,\ L = \sum L_i,\ \tfrac1C = \sum \tfrac1{C_i}\qquad \text{Parallel: } \tfrac1R = \sum \tfrac1{R_i},\ \tfrac1L = \sum \tfrac1{L_i},\ C = \sum C_i
$$

## Worked examples

### 100, 220, 470 Ω in series

- Components: Resistors
- Connected in: Series
- Values: 100, 220, 470
- Unit: Ω
- **Equivalent value: 790 Ω**
- Checked against: Python 3.8 decimal: 100 + 220 + 470

### 100, 220, 470 Ω in parallel

- Components: Resistors
- Connected in: Parallel
- Values: 100, 220, 470
- Unit: Ω
- **Equivalent value: 59.9768 Ω**
- Checked against: Python 3.8 fractions: 1/(1/100 + 1/220 + 1/470) = 25850/431 = 59.976798…

### Two equal 1 kΩ in parallel

- Components: Resistors
- Connected in: Parallel
- Values: 1k 1k
- Unit: Ω
- **Equivalent value: 500 Ω**
- Checked against: R/n for n equal resistors (OpenStax UP2 §10.2)

### Three 100 nF in parallel

- Components: Capacitors
- Connected in: Parallel
- Values: 100 100 100
- Unit: nF
- **Equivalent value: 300 nF**
- Checked against: Sum: 300 nF

### 10 µF and 22 µF in series

- Components: Capacitors
- Connected in: Series
- Values: 10, 22
- Unit: µF
- **Equivalent value: 6.875 µF**
- **In base units: 6.875 × 10⁻⁶ F**
- Checked against: Python 3.8 fractions: 1/(1/10 + 1/22) = 6.875 µF

### 10 mH and 4.7 mH in series

- Components: Inductors
- Connected in: Series
- Values: 10, 4.7
- Unit: mH
- **Equivalent value: 14.7 mH**
- **In base units: 1.47 × 10⁻² H**
- Checked against: Sum: 14.7 mH (no mutual coupling)

## Questions

### How do you calculate resistors in parallel?

Add the reciprocals and invert: 1/R = 1/R₁ + 1/R₂ + … . For two resistors this reduces to product over sum, R = R₁R₂/(R₁ + R₂), so 100 Ω and 220 Ω in parallel give 22,000/320 = 68.75 Ω. The total is always smaller than the smallest resistor, because each extra path carries more current.

### What is the resistance of equal resistors in parallel?

Divide one resistor's value by the number of resistors: n equal resistors R in parallel give R/n. Two 1 kΩ resistors give 500 Ω and four 100 Ω resistors give 25 Ω. They share the current equally, so the power ratings add too: four ¼ W resistors in parallel can dissipate 1 W between them.

### How do you add capacitors in series and in parallel?

In parallel, capacitances add directly: three 100 nF capacitors give 300 nF. In series, the reciprocals add, 1/C = 1/C₁ + 1/C₂, so 10 µF and 22 µF give 6.875 µF, less than the smaller part. Series capacitors carry the same charge, so the smaller capacitor takes the larger share of the voltage.

### Do inductors add like resistors?

Yes, when their magnetic fields do not couple: L = L₁ + L₂ in series and 1/L = 1/L₁ + 1/L₂ in parallel, so 10 mH and 4.7 mH in series give 14.7 mH. Coils that share flux add or subtract a mutual-inductance term of 2M in series, depending on winding direction, which this calculator does not model.

### How accurate is the series and parallel resistor calculator?

Accuracy depends on your inputs and the method's assumptions. Decimal arithmetic uses 50 significant digits, but estimates, numerical methods and source data can be less precise; the displayed rounding does not remove those limits. It is checked against 7 worked examples whose answers come from independent sources; for example, “100, 220, 470 Ω in series” is checked against Python 3.8 decimal: 100 + 220 + 470.

### Where does the method come from?

OpenStax University Physics Volume 2, §10.2 Resistors in series and parallel; §8.2 Capacitors in series and in parallel; HyperPhysics — Inductors in series and parallel (no mutual inductance).

## Sources

- [OpenStax University Physics Volume 2, §10.2 Resistors in series and parallel; §8.2 Capacitors in series and in parallel](https://openstax.org/books/university-physics-volume-2/pages/10-2-resistors-in-series-and-parallel)
- [HyperPhysics — Inductors in series and parallel (no mutual inductance)](http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indcom.html)