Tooldeon

Capacitors in series and parallel calculator

Type the capacitor values separated by spaces, pick series or parallel, and enter the voltage across the whole group. You get the equivalent capacitance, the charge it holds, the energy stored and, for each capacitor, the voltage across it (series) or its own charge (parallel).

Separate values with spaces. Prefixes p, n, u or µ, m; F is optional, e.g. 10µ 4.7nF 100p.
Total capacitance5.998 µF
Capacitors
3
Total charge Q
71.97 µC
Stored energy
431.8 µJ
C1 = 10 µF
7.197 V
C2 = 22 µF
3.271 V
C3 = 47 µF
1.531 V

In series every capacitor holds the same charge Q; the list shows the voltage across each one.

Runs in your browser: nothing you enter is sent anywhere.

How to use it

  1. Choose series or parallel.
  2. Type the capacitances separated by spaces: 10µ 22µ 47µ, 4.7n 10n, 100p 220pF. A bare number is in farads.
  3. Enter the voltage across the whole group to see the charge, the energy and the share of each capacitor.

Formulas

  • Series: 1/C = 1/C₁ + 1/C₂ + …; each capacitor holds the same charge Q = C × V and takes Vᵢ = Q ÷ Cᵢ.
  • Parallel: C = C₁ + C₂ + …; each capacitor has the full voltage V and holds Qᵢ = Cᵢ × V.
  • Energy: E = ½ × C × V².

Results are rounded to four significant digits with the prefix that suits them.

Worked example

10 µF, 22 µF and 47 µF in series: 1/10 + 1/22 + 1/47 = 0.1 + 0.04545 + 0.02128 = 0.16673 per µF, so C = 5.998 µF. At 12 V the group holds Q = 5.998 µF × 12 V = 71.97 µC and stores 431.8 µJ. The 10 µF part takes 71.97 ÷ 10 = 7.197 V, the 22 µF part 3.271 V and the 47 µF part 1.531 V; together that is the full 12 V (11.999 after rounding).

The same three in parallel give 10 + 22 + 47 = 79 µF, 948 µC at 12 V and 5.688 mJ.

Tips

  • m is milli, not mega. 1m is 1 mF; type 1000µ if that is clearer.
  • Electrolytics in series need matched values or balancing resistors, or one of them takes too much voltage.
  • Tolerance matters. Ceramic and electrolytic capacitors are often ±10 % to ±20 %, so the real total varies by as much.

Frequently asked questions

How do capacitors add in series?

In series the reciprocals add, 1/C = 1/C₁ + 1/C₂ + …, so the total is always smaller than the smallest capacitor. Two equal capacitors in series give half of one.

And in parallel?

In parallel the capacitances simply add, C = C₁ + C₂ + … . It is the opposite of resistors, which add in series and combine by reciprocals in parallel.

Why put capacitors in series?

To share a voltage that is higher than one capacitor's rating. The smallest capacitor takes the largest share, Vᵢ = Q ÷ Cᵢ, so check every voltage in the list against the part's rating. Real parts also leak unevenly, which is why balancing resistors are used across series capacitors.

How is the energy found?

From the total capacitance and the voltage, E = ½ × C × V². The charge is Q = C × V.

More in Unit converters

All tools in this category: Unit converters ·Updated November 7, 2026

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