Series and parallel calculator
Enter your resistors or capacitors, get the totals.
How do I add resistors in series?
Add them: R = R₁ + R₂ + … 100 Ω, 200 Ω and 300 Ω in series make 600 Ω.
How do I add resistors in parallel?
1/R = 1/R₁ + 1/R₂ + …, so R = 1 ÷ (1/100 + 1/200 + 1/300) = 54.55 Ω — always less than the smallest resistor. Two equal resistors in parallel give half the value.
Why is it the other way around for capacitors?
Capacitors in parallel add up, C = C₁ + C₂ + …, while in series 1/C = 1/C₁ + 1/C₂ + … — the opposite of resistors.
How do I enter the values?
Type them separated by commas or spaces and pick the unit: 100, 200, 300 in Ω, or 10, 22 in µF.
What about a mix of series and parallel?
Work in steps: combine each parallel group first, then add the results in series (or the other way round) — enter each group separately.
Is there a shortcut for two resistors in parallel?
Yes: multiply them and divide by their sum, R = R₁R₂ ÷ (R₁ + R₂). 220 Ω and 330 Ω give 72,600 ÷ 550 = 132.0 Ω. Identical resistors are easier still: n of them in parallel give one resistor's value divided by n. The result is always smaller than the smallest resistor.
What stays the same in series and in parallel?
In a series string the same current flows through every part, and the voltages add up to the supply. In parallel every part has the same voltage across it, and the currents add. For capacitors it is the charge that is shared in series and the voltage in parallel (OpenStax University Physics, sections 10.2 and 8.2). OpenStax's parallel example, 1.00, 2.00 and 2.00 Ω on 3.00 V, draws 3.00 + 1.50 + 1.50 = 6.00 A.
Why the rules are opposite for resistors and capacitors
Resistors in series add because the same current has to pass through each of them, so their voltage drops add; in parallel the current has several paths, so the conductances (1/R) add. Capacitors behave the other way around: in parallel they act like one larger capacitor, so the values add; in series the total is smaller than the smallest. The calculator takes every value in one list in the same unit, rejects zero or negative entries, and gives the series and parallel results together.
Worked example
Three resistors of 220, 330 and 470 Ω make 1020 Ω in series and 103.1 Ω in parallel, below the smallest of them. A mixed network, 100 Ω in series with a 220 Ω and 330 Ω pair in parallel, is solved in two steps: the pair first, 132.0 Ω, then 132.0 + 100 = 232.0 Ω. For capacitors, OpenStax's 1.000, 5.000 and 8.000 µF give 0.755 µF in series and 14.0 µF in parallel; with the first two in series and that pair in parallel with the 8.000 µF, the total is 8.833 µF.
Common mistakes
Swapping the capacitor rule for the resistor rule is the usual error. A quick check catches it: a parallel group of resistors, or a series string of capacitors, must come out below its smallest part. The unit applies to the whole list, so with kΩ selected a 470 Ω resistor is entered as 0.47. A resistor of zero ohms, a wire, is rejected; in parallel it would make the whole group 0 Ω. For networks, work from the inside out and enter one group at a time, as in the example.
Sources: OpenStax, University Physics Volume 2, section 10.2, “Resistors in Series and Parallel,” and section 8.2, “Capacitors in Series and in Parallel” (openstax.org, read October 27, 2026), including Examples 10.2, 10.3, 8.4 and 8.5; the two-resistor formula is an algebraic rearrangement. Related: Ohm's law calculator, resistor color code calculator, voltage divider calculator, capacitor calculator. How we check our calculators.