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.

Aktualisiert 28. September 2026Geprüfte Beispiele: 7

Equivalent value

Signifikante Stellen: 6; Zum nächsten Wert, bei Gleichstand von null weg

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

So wird gerechnet 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.

Über 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.

Durchgerechnete Beispiele

100, 220, 470 Ω in series

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

Equivalent value: 790 Ω

Prüfquelle: Python 3.8 decimal: 100 + 220 + 470

100, 220, 470 Ω in parallel

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

Equivalent value: 59.9768 Ω

Prüfquelle: 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
Einheit: Ω

Equivalent value: 500 Ω

Prüfquelle: R/n for n equal resistors (OpenStax UP2 §10.2)

Three 100 nF in parallel

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

Equivalent value: 300 nF

Prüfquelle: Sum: 300 nF

Fragen

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.

Wie genau arbeitet „Series and parallel resistor calculator“?

Die Genauigkeit hängt von Ihren Eingaben und den Annahmen der Methode ab. Die Dezimalrechnung nutzt 50 signifikante Stellen, doch Schätzungen, numerische Verfahren und Quelldaten können ungenauer sein. Die angezeigte Rundung beseitigt diese Grenzen nicht. Anhand unabhängiger Quellen geprüfte Rechenbeispiele: 7. Beispielsweise wird „100, 220, 470 Ω in series“ anhand von Python 3.8 decimal: 100 + 220 + 470 geprüft.

Woher stammt die Methode?

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).

Über diesen Rechner

\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

Quellen

Anhand von Quellen geprüft

Dieser Rechner enthält 7 Rechenbeispiele mit Ergebnissen aus unabhängigen Quellen. Sie laufen in der Testsuite und können auch hier ausgeführt werden.