Capacitor Combination Calculator
Enter up to 3 capacitors and the connection mode to compute the equivalent capacitance. Series 1μF+1μF → 0.5μF; parallel 1μF+2μF+3μF → 6μF.
Input Data
Results
At a glance:Capacitor combination: the equivalent capacitance when multiple capacitors are connected in series or parallel. Series: capacitors connected end-to-end; the reciprocal of the equivalent capacitance is the sum of reciprocals 1/C_eq=Σ(1/Cᵢ), so C_eq is smaller than any single capacitor (voltage is divided, voltage rating increases); parallel: capacitors connected at the same two nodes; the equivalent capacitance is the sum C_eq=ΣCᵢ, so C_eq is larger than any single capacitor (charge storage increases). History: Volta invented the battery in 1800; the Leyden jar (1745) was the earliest capacitor; Faraday studied dielectrics in 1837. Classic example: two 1μF in series → 0.5μF; two 1μF in parallel → 2μF. Applications: (1) power-supply filtering — large parallel capacitors store energy; (2) resonant circuits — series/parallel tuning; (3) voltage dividers — series capacitors divide voltage; (4) energy storage — supercapacitors in parallel; (5) power-factor correction — parallel capacitors compensate.
Formula
Series equivalent: 1/C_eq = 1/C1 + 1/C2 + 1/C3 → C_eq = 1/(Σ1/Cᵢ)
Parallel equivalent: C_eq = C1 + C2 + C3
Two-capacitor series: C_eq = C1·C2 / (C1 + C2)
Series voltage division: V1 = V·C_eq/C1, V2 = V·C_eq/C2
Parallel charge division: Q1 = C1·V, Q2 = C2·V (same voltage)
$$\frac{1}{C_{\text{eq(series)}}} = \sum_{i=1}^{n} \frac{1}{C_i}, \quad C_{\text{eq(parallel)}} = \sum_{i=1}^{n} C_i$$How to Use
- Enter capacitances C1, C2, C3 (F); C3=0 means only 2 capacitors are used.
- Select the connection mode (series / parallel).
- The tool computes the equivalent capacitance C_eq (F, μF). Series 1μF+1μF → 0.5μF; parallel 1μF+2μF+3μF → 6μF.
Common Capacitor Combinations
| C1 (μF) | C2 (μF) | C3 (μF) | Series C_eq (μF) | Parallel C_eq (μF) |
|---|---|---|---|---|
| 1 | 1 | 0 | 0.5 | 2 |
| 10 | 10 | 0 | 5 | 20 |
| 1 | 2 | 3 | 0.545 | 6 |
| 100 | 100 | 100 | 33.3 | 300 |
| 0.1 | 0.01 | 0 | 0.00909 | 0.11 |
| 470 | 1000 | 0 | 319.7 | 1470 |
| 1 | 1 | 1 | 0.333 | 3 |
Series 1/C=Σ1/Cᵢ, parallel C=ΣCᵢ. Series C_eq is smaller than the smallest single capacitor; parallel C_eq is larger than the largest single capacitor. Two equal capacitors in series halve, in parallel double.
Case Studies
Power-Supply Filter Capacitor Design
5V switching-power-supply output filter: parallel 470μF electrolytic (low-freq) + 0.1μF ceramic (high-freq) → C_eq=470.1μF.
Electrolytic capacitors have high ESR (~0.1Ω); at 1MHz impedance Z=1/(2πfC)=1/(2π×1MHz×470μF)=0.34Ω, poor high-freq filtering.
Parallel 0.1μF ceramic ESR is low (~0.01Ω); high-freq impedance 1/(2π×1MHz×0.1μF)=1.6Ω, improving high-freq filtering. The combination covers the full band.
High-Voltage Voltage Divider
10 kV high-voltage measurement: 3 capacitors of 1 nF/4 kV in series → C_eq=0.333 nF, voltage rating 12 kV (each shares 3.33 kV).
Series voltage division V1=V×C_eq/C1=10000×0.333/1=3333 V; each capacitor bears 1/3 of the voltage.
Real designs must account for capacitance tolerance (±5%) causing uneven voltage; parallel bleeder resistors (10 MΩ) ensure uniform voltage sharing.
FAQ
What is the difference between series and parallel capacitors?
Series: 1/C=Σ1/Cᵢ, C_eq smaller than the smallest single capacitor, voltage rating increases (voltage is divided among capacitors); parallel: C=ΣCᵢ, C_eq larger than the largest single capacitor, voltage rating unchanged (same voltage). Series is used for high-voltage division and reducing capacitance; parallel for increasing capacitance and power-supply filtering. Two equal capacitors in series halve, in parallel double.
Why does the equivalent capacitance decrease in series?
In series the effective plate separation increases. 1/C=Σ(1/Cᵢ); adding reciprocals makes the total reciprocal larger, so C_eq decreases. Physically: in series the charge Q is the same on each (Q=CᵢVᵢ), total voltage V=ΣVᵢ=Q·Σ(1/Cᵢ), so C_eq=Q/V=1/(Σ1/Cᵢ). For example two 1μF in series → 0.5μF (effective plate distance doubled).
How to choose a filter capacitor combination?
Power-supply filters commonly use parallel combinations: (1) large capacitor (470~1000μF electrolytic) for low-frequency ripple; (2) medium capacitor (1~10μF ceramic) for mid-frequency; (3) small capacitor (0.1μF ceramic) for high-frequency. The parallel combination C_eq=ΣCᵢ provides a low-impedance path at every frequency. Switching supplies (100kHz~10MHz) need multi-stage filtering. Note that ESR and ESL affect high-frequency performance.
How do series capacitors divide voltage?
In series the charge Q is the same (Q=C_eq×V); each capacitor's voltage Vᵢ=Q/Cᵢ=C_eq×V/Cᵢ. The smaller the capacitance, the larger its share (inverse division). For example C1=1μF, C2=2μF in series on 12V → V1=12×(2/3)/1=8V, V2=12×(2/3)/2=4V. In practice add parallel bleeder resistors to compensate for tolerance.
How are supercapacitors combined?
Supercapacitors (1~3000F) have low rated voltage (2.7V) and need series connection to raise the voltage rating. Four 100F/2.7V in series → C_eq=25F/10.8V. But cell differences cause overvoltage, requiring balancing circuits (active or passive). Parallel increases capacity but voltage is unchanged. Electric-vehicle braking recuperation often uses supercapacitor banks to absorb instantaneous high power.
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References
Content reviewed by the Calculatorism editorial team. Results are for reference only; please refer to the relevant authorities for the official figures.