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.

Atualizado 28 de setembro de 2026Exemplos verificados: 7

Equivalent value

Algarismos significativos: 6; Ao mais próximo; empates afastando-se de zero

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

Como é calculado 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.

Sobre 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.

Exemplos resolvidos

100, 220, 470 Ω in series

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

Equivalent value: 790 Ω

Fonte de verificação: Python 3.8 decimal: 100 + 220 + 470

100, 220, 470 Ω in parallel

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

Equivalent value: 59.9768 Ω

Fonte de verificação: 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
Unidade: Ω

Equivalent value: 500 Ω

Fonte de verificação: R/n for n equal resistors (OpenStax UP2 §10.2)

Three 100 nF in parallel

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

Equivalent value: 300 nF

Fonte de verificação: Sum: 300 nF

Perguntas

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.

Qual é a precisão de “Series and parallel resistor calculator”?

A precisão depende dos dados inseridos e das hipóteses do método. O cálculo decimal usa 50 algarismos significativos, mas estimativas, métodos numéricos e dados de origem podem ter menor precisão; o arredondamento exibido não elimina essas limitações. Exemplos resolvidos verificados com fontes independentes: 7. Por exemplo, “100, 220, 470 Ω in series” é verificado com Python 3.8 decimal: 100 + 220 + 470.

De onde vem o 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).

Sobre 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

Fontes

Verificado com as referências

Esta calculadora inclui 7 exemplos resolvidos com respostas de fontes independentes. Eles fazem parte do conjunto de testes e você também pode executá-los aqui.