Enter the values of resistors, capacitors or inductors to get their total in series and in parallel. You can also see how close two standard E24 parts in series or parallel can get to a value that no single E24 part provides.
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Closest combinations (E24 series)
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E24
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Empty rows are ignored. Magnetic coupling between inductors (mutual inductance), component tolerance and wiring resistance are not included. The combination search uses E24 values from 1 Ω to 10 MΩ for resistors, 1 pF to 10 µF for capacitors and 1 nH to 10 mH for inductors.
See it as a graph
Other inputs stay at their current values while the chosen input varies from 1/4 to 4 times its current value. The red dot is the current value.
Series and parallel formulas
In series, the same current flows through every part and their voltages add. For resistors, V = I·R1 + I·R2 + …, so the total is the sum R = R1 + R2 + …. In parallel, every part sees the same voltage and their currents add. Since I = V/R1 + V/R2 + …, the total follows 1/R = 1/R1 + 1/R2 + …. Each parallel resistor adds another path for current, so the total is smaller than the smallest resistor. For example, 100 Ω, 200 Ω and 300 Ω give 600 Ω in series and about 54.5 Ω in parallel. Inductors follow the same formulas as resistors, because in series the same current flows and their voltages V = L·dI/dt add.
Why capacitors are the other way round
Capacitors add in parallel and combine as the reciprocal of the sum of reciprocals in series, the reverse of resistors. A parallel-plate capacitor has C = εS/d (S: plate area, d: plate spacing). Connecting capacitors in parallel is like enlarging the plate area, while putting identical capacitors in series is like widening the spacing. In terms of the equations, capacitors in series all hold the same charge Q and their voltages V = Q/C1 + Q/C2 + … add, which gives 1/C = 1/C1 + 1/C2 + …. For example, 10 µF and 22 µF give 32 µF in parallel and about 6.88 µF in series. Note also that in series, the smaller capacitor takes the larger share of the voltage.
Product over sum for two, R/n for equal values
Rearranging 1/R = 1/R1 + 1/R2 for two resistors in parallel gives R = R1R2/(R1 + R2), product over sum. For 6 Ω and 3 Ω, that is 18 ÷ 9 = 2 Ω. n equal resistors R in parallel give R/n: two 10 kΩ resistors make 5 kΩ, and four make 2.5 kΩ. Two very different values in parallel give roughly the smaller one. 100 Ω in parallel with 10 kΩ is about 99.0 Ω, so a large resistor can be used to trim a smaller one down slightly. Two capacitors in series follow the same form, C1C2/(C1 + C2), so two 100 nF capacitors in series make 50 nF.
E series, and combining two values to hit a target
Standard resistor and capacitor values come from the E series, sequences that are close to geometric. The E24 series divides each decade (factor of 10) into 24 values, with an average ratio of about 1.1 (the 24th root of 10) between neighbors. The spacing is not uniform, though: 1.3 and 1.5 are about 15% apart, so the nearest single value to 1.4 kΩ is 1.3 kΩ or 1.5 kΩ, an error of about 7%. With two parts, 1.2 kΩ + 200 Ω in series gives exactly 1.4 kΩ. This tool tries every pair of E24 values in its range, in series and in parallel, and lists them by error. Tested on targets from 100 Ω to 100 kΩ, roughly 90% of values come within 0.1%, and even the worst cases stay within about 0.33%.
Tolerance and power rating when combining parts
Even a pair that hits the target exactly will be off by the parts' tolerance. Two ±5% resistors in series or parallel still give a worst-case tolerance of ±5%, so combining parts does not improve accuracy. When accuracy matters, use resistors of ±1% or better, such as the E96 series. Watch the power, too. In series the same current flows, so the larger resistor dissipates more; in parallel the same voltage is applied, so the smaller resistor dissipates more. With 100 Ω and 1 kΩ in parallel across 5 V, the 100 Ω resistor dissipates 250 mW and the 1 kΩ resistor 25 mW. Two equal resistors in parallel split the power evenly, so two 1/4 W parts can handle 1/2 W, though you should still leave margin below the rating (derating). When capacitors are put in series for a higher voltage rating, a balancing resistor is commonly placed across each one so that differences in leakage current do not skew the voltage split.
The combined resistance of resistors in series is their sum: 100 + 200 + 300 = 600 Ω. The same current flows through all three, and the voltage divides in proportion to resistance. Applying 12 V, for example, gives a current of 12 V/600 Ω = 20 mA and voltages of 2 V, 4 V and 6 V across the resistors. With the same three in parallel, the combined resistance is about 55 Ω, smaller than the smallest resistor (100 Ω).
For two resistors in parallel, the combined resistance is R1R2/(R1 + R2) (product over sum): 6 × 3/(6 + 3) = 18/9 = 2 Ω. It is also the reciprocal of 1/6 + 1/3 = 1/2. A parallel combination is smaller than its smallest resistor (3 Ω). Both resistors see the same voltage and the current divides in inverse proportion to resistance, so the 3 Ω resistor carries twice the current of the 6 Ω one.
A parallel-plate capacitor has C = εS/d (ε: permittivity, S: plate area, d: spacing), proportional to area and inversely proportional to spacing. Doubling the area doubles it, and halving the spacing doubles it again, for 2 × 2 = 4 times in total. Filling the gap with a dielectric of relative permittivity εr multiplies the capacitance by εr. Multilayer ceramic capacitors stack many thin dielectric layers to reach large capacitances.
FAQ
How do I calculate resistors in parallel?
Add the reciprocals of the resistances and take the reciprocal of the total (1/R = 1/R1 + 1/R2 + …). For two resistors, use product over sum, R1R2/(R1 + R2): 6 Ω and 3 Ω in parallel give 2 Ω. For n equal resistors, the result is R/n, so two 10 kΩ resistors in parallel give 5 kΩ.
Why does adding a resistor in parallel lower the resistance?
In parallel, each resistor adds another path for current at the same voltage, so more total current flows. By Ohm's law, R = V/I, more current means a lower total resistance. Whatever value you add in parallel, the total is always smaller than the smallest resistor.
Why does capacitance decrease when capacitors are in series?
In series, each capacitor holds the same charge Q and their voltages add, so less charge is stored at a given voltage. The effect is the same as widening the plate spacing, giving 1/C = 1/C1 + 1/C2 + …. Two 100 nF capacitors in series make 50 nF. In parallel, by contrast, it is as if the plate area grew, and the capacitances simply add.
How can I make a resistor value I don't have?
Combine two values, for example from the E24 series, in series or parallel. 7.07 kΩ is not in E24, and the nearest value, 6.8 kΩ, is about 3.8% low, but 6.8 kΩ + 270 Ω in series gives exactly 7.07 kΩ. Enter a target in the second part of this tool to see the five combinations with the smallest error. The parts' tolerance still applies, so choose tighter-tolerance resistors when accuracy matters.
What should I watch for when combining inductors?
The resistor formulas hold only when the inductors are not magnetically coupled. Inductors placed close together pick up each other's flux, creating a mutual inductance M, and the series total becomes L1 + L2 ± 2M (the sign depends on whether the fluxes add or oppose). To avoid coupling, space them apart, set their axes at right angles or use shielded inductors. Also check the winding resistance and the rated current.