What this calculator does
Find the total value of a group of capacitors, and how much charge and energy they hold. A capacitor stores electric charge on two conductors with an insulator between them. Its capacitance says how much charge it holds for each volt. When you wire several capacitors together, the group acts like one capacitor. Its value is called the equivalent capacitance. This calculator finds it for capacitors connected in parallel (side by side) or in series (end to end).
Enter the voltage across the group, and it also works out the stored charge (Q = CV) and the stored energy (E = ½CV²). Here C is the capacitance and V is the voltage. It then lists the voltage, charge and energy for each capacitor. That breakdown matters most for series strings. There, the voltage doesn't split evenly, and the smallest capacitor can be pushed past its rating.
Where it's used
Why this math matters outside the classroom:
- Circuit repair. Technicians replace a missing 20 µF capacitor with two 10 µF parts wired in parallel, since parallel values simply add.
- Series pairs. Designers put two 100 µF capacitors in series to get 50 µF, which also splits the voltage across the pair.
- Camera flash. A camera flash capacitor of 100 µF charged to 300 V stores ½ × 0.0001 × 300² = 4.5 J.
- Stored charge. Physics students find that a 47 µF capacitor at 9 V holds Q = CV = 423 µC of charge.
- Power supplies. Power supply designers pick smoothing capacitors big enough to carry the load between the peaks of rectified AC.
How to use it
- Choose Parallel or Series to match the wiring.
- Type the capacitor values, separated by commas or spaces. Pick the unit they're in: pF, nF, µF, mF or F.
- If you like, enter the voltage across the whole group to get charge and energy.
- Read the total capacitance. Then check the table for each capacitor's share of the voltage, charge and energy.
What the inputs mean
- Capacitor values
- All values share the unit chosen below. If your values use different units, convert them first: 0.1 µF is 100 nF, and 1000 pF is 1 nF.
- Capacitance unit
- The prefixes are exact powers of ten: 1 pF = 10⁻¹² F, 1 nF = 10⁻⁹ F, 1 µF = 10⁻⁶ F, 1 mF = 10⁻³ F. F is the farad, the basic unit of capacitance.
- Voltage across the group (optional)
- The voltage across the whole group. In a series string, the capacitors share it. In a parallel bank, each one gets all of it.
Formula and method
Parallel: C = C₁ + C₂ + … + Cₙ Series: 1/C = 1/C₁ + 1/C₂ + … + 1/Cₙ Charge: Q = C × V Energy: E = ½ × C × V² Series part k: Vₖ = Q ÷ Cₖ Parallel part k: Qₖ = Cₖ × V
C is the total capacitance, and C₁, C₂ up to Cₙ are the separate capacitors. Q is the charge in coulombs, V is the voltage and E is the energy in joules. A small k marks one particular capacitor in the group, so Vₖ is its voltage and Qₖ is its charge.
Capacitors combine the opposite way to resistors. In parallel, each capacitor gets the same voltage and holds its own charge. So the charges add, and the capacitances add too. The total is bigger than any single part.
In series, charge pushed onto the first plate draws an equal charge through every capacitor in the chain. So each one holds the same charge Q. The voltages add, and each one is Q ÷ Cₖ. That means the reciprocals (1 divided by each capacitance) add. The total is smaller than the smallest part.
Energy grows with the square of the voltage. That's because each extra bit of charge takes more work to push on as the voltage builds. Charge and energy are worked out in farads, coulombs and joules. Then they're shown with a handy prefix, such as µC or mJ.
Worked example
Capacitors of 10 µF, 22 µF and 47 µF are connected in parallel across 12 V.
- Total capacitance: C = 10 + 22 + 47 = 79 µF, which is 0.000079 F.
- Charge: Q = 0.000079 × 12 = 0.000948 C, or 948 µC.
- Energy: E = 0.5 × 0.000079 × 12² = 0.005688 J, or 5.688 mJ.
- In series the same three capacitors would give only about 5.998 µF.
Result: The bank is 79 µF. Charged to 12 V, it holds 948 µC and 5.688 mJ.
Understanding the result
In a series string, the smallest capacitor carries the biggest share of the voltage. Make sure each part's voltage rating is above its share, ideally with room to spare. Real leakage can also make the split uneven over time. That's why high-voltage strings often add balancing resistors.
Parallel banks are used to raise capacitance and share ripple current (the small ups and downs in current). Each capacitor must be rated for the full voltage.
The energy figure shows what a capacitor can release in a short burst. Even a few joules at a few hundred volts can give a painful or dangerous shock. So discharge large capacitors safely before you handle them.
Assumptions and limitations
- The capacitors are ideal, with no leakage, series resistance or inductance.
- Each capacitor's value stays fixed. It doesn't change with voltage, temperature or frequency.
- Series capacitors start uncharged, so they all hold the same charge.
- Only pure series or pure parallel groups are handled. To solve a mixed network, work it out in stages.
- Real capacitors have tolerances that can be ±10% to ±20% or more. Some ceramic types lose a large part of their capacitance at high DC voltage.
- It doesn't work out charging time, reactance (opposition to AC current) at a frequency, or a capacitor's value from its plate size.
Frequently asked questions
Why do capacitors in series add like resistors in parallel?
Because capacitance is charge per volt. In series, all the capacitors hold the same charge while their voltages add. So the total charge per volt goes down. In math terms, the reciprocals add, exactly like conductances (1 ÷ resistance) in a parallel resistor bank.
What is the quick formula for two capacitors in series?
Multiply them, then divide by their sum: C = C₁C₂ ÷ (C₁ + C₂). For 2 µF and 3 µF, that's 6 ÷ 5 = 1.2 µF. Two equal capacitors in series give half of one.
Why connect capacitors in series at all?
To handle a higher voltage than one capacitor is rated for, or to make a smaller value than you have on hand. The trade-off is less total capacitance. You also need to check each part's share of the voltage.
Where does the ½ in ½CV² come from?
The voltage across a capacitor rises from zero as it charges. So on average, each bit of charge is moved through half of the final voltage. The total work is Q × V ÷ 2. Since Q = CV, that becomes ½CV².
How this calculator is checked
Method: Standard results for ideal capacitors: parallel sum, series reciprocal sum, Q = CV and E = ½CV². The calculation is covered by automated tests with known reference values and edge cases. Last updated 2026-10-18. Found a problem? Tell us.
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