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.

Aggiornato Esempi verificati: 7

Separate with commas or spaces; 4.7k works for kilo
Prova
Equivalent value
Ω
Equivalent value: 59.9768 Ω
Cifre significative: 6; Al più vicino; a parità lontano da 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

AB100 Ω220 Ω470 Ω
Come si calcola S
  1. Resistors in parallel

    1Req=1100+1220+1470  ⇒  Req=59.9768 Ω\frac{1}{R_{\text{eq}}} = \frac{1}{100} + \frac{1}{220} + \frac{1}{470} \;\Rightarrow\; R_{\text{eq}} = 59.9768\ \mathrm{\Omega}

    Reciprocals add.

Informazioni su 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.

Esempi svolti

100, 220, 470 Ω in series

Components
Resistors
Connected in
Series
Values
100, 220, 470
Unità
Ω
Equivalent value
790 Ω

Fonte di verifica: Python 3.8 decimal: 100 + 220 + 470

100, 220, 470 Ω in parallel

Components
Resistors
Connected in
Parallel
Values
100, 220, 470
Unità
Ω
Equivalent value
59.9768 Ω

Fonte di verifica: 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
Unità
Ω
Equivalent value
500 Ω

Fonte di verifica: R/n for n equal resistors (OpenStax UP2 §10.2)

Three 100 nF in parallel

Components
Capacitors
Connected in
Parallel
Values
100 100 100
Unità
nF
Equivalent value
300 nF

Fonte di verifica: Sum: 300 nF

Domande

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.

Quanto è preciso «Series and parallel resistor calculator»?

La precisione dipende dai dati inseriti e dalle ipotesi del metodo. Il calcolo decimale usa 50 cifre significative, ma stime, metodi numerici e dati di origine possono essere meno precisi; l’arrotondamento visualizzato non elimina questi limiti. Esempi svolti verificati con fonti indipendenti: 7. Per esempio, «100, 220, 470 Ω in series» viene verificato con Python 3.8 decimal: 100 + 220 + 470.

Da dove proviene il metodo?

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).

Informazioni su questa calcolatrice

Series: R=∑Ri, L=∑Li, 1C=∑1CiParallel: 1R=∑1Ri, 1L=∑1Li, C=∑Ci\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

Fonti

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

Verificato con le fonti

Questa calcolatrice include 7 esempi svolti con risposte da fonti indipendenti. Fanno parte della suite di test e puoi eseguirli anche qui.

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