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.

Actualizado 28 de septiembre de 2026Ejemplos verificados: 7

Equivalent value

Cifras significativas: 6; Al más cercano; empates alejándose de cero

Number of components: 3

3 resistors in parallel act like one of 59.9768 Ω — always smaller than the smallest single part.

3 resistors in parallel: 59.9768 Ω

Cómo se calcula S

Resistors in parallel
$\frac{1}{R_{\text{eq}}} = \frac{1}{100} + \frac{1}{220} + \frac{1}{470} \;\Rightarrow\; R_{\text{eq}} = 59.9768\ \mathrm{\Omega}$
Reciprocals add.

Acerca de Series and parallel resistor calculator

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.

Ejemplos resueltos

100, 220, 470 Ω in series

Components: Resistors
Connected in: Series
Values: 100, 220, 470
Unidad: Ω

Equivalent value: 790 Ω

Fuente de comprobación: Python 3.8 decimal: 100 + 220 + 470

100, 220, 470 Ω in parallel

Components: Resistors
Connected in: Parallel
Values: 100, 220, 470
Unidad: Ω

Equivalent value: 59.9768 Ω

Fuente de comprobación: 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
Unidad: Ω

Equivalent value: 500 Ω

Fuente de comprobación: R/n for n equal resistors (OpenStax UP2 §10.2)

Three 100 nF in parallel

Components: Capacitors
Connected in: Parallel
Values: 100 100 100
Unidad: nF

Equivalent value: 300 nF

Fuente de comprobación: Sum: 300 nF

Preguntas

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.

¿Qué precisión tiene «Series and parallel resistor calculator»?

La precisión depende de tus datos y de los supuestos del método. El cálculo decimal usa 50 cifras significativas, pero las estimaciones, los métodos numéricos y los datos de origen pueden ser menos precisos; el redondeo mostrado no elimina esos límites. Ejemplos resueltos comprobados con fuentes independientes: 7. Por ejemplo, «100, 220, 470 Ω in series» se comprueba con Python 3.8 decimal: 100 + 220 + 470.

¿De dónde procede el método?

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).

Acerca de esta calculadora

\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

Fuentes

Verificado con las referencias

Esta calculadora incluye 7 ejemplos resueltos con respuestas de fuentes independientes. Forman parte del conjunto de pruebas y también puedes ejecutarlos aquí.