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.

更新于 2026年9月28日已验证的示例：7

Equivalent value

有效数字：6；取最近值，等距时远离零

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

计算方法 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.

关于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.

计算示例

100, 220, 470 Ω in series

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

Equivalent value: 790 Ω

核验来源：Python 3.8 decimal: 100 + 220 + 470

100, 220, 470 Ω in parallel

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

Equivalent value: 59.9768 Ω

核验来源：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
单位: Ω

Equivalent value: 500 Ω

核验来源：R/n for n equal resistors (OpenStax UP2 §10.2)

Three 100 nF in parallel

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

Equivalent value: 300 nF

核验来源：Sum: 300 nF

常见问题

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.

“Series and parallel resistor calculator”有多准确？

准确性取决于输入值和方法的假设。十进制运算使用50位有效数字，但估算、数值方法和源数据的精度可能较低；显示时的舍入并不能消除这些限制。已按独立来源核验的计算示例：7。例如，“100, 220, 470 Ω in series”根据Python 3.8 decimal: 100 + 220 + 470进行核验。

这种方法出自哪里？

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).

关于此计算器

\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

来源

已对照来源验证

此计算器包含 7 个已解示例，答案来自独立来源。这些示例会在测试套件中运行，你也可以在此运行验证。