What it does
Many textbook problems give a network of capacitors joined to a battery and ask for the total capacitance, then the charge and voltage on each one. The method is always the same: find a group that is simply in series or simply in parallel, replace it with one capacitor, and repeat until only one is left. Then work backwards: the battery charges that one equivalent capacitor, and its charge and voltage are shared back out to the capacitors it replaced.
- In parallel, capacitors share the same voltage, and the capacitances add: C = C₁ + C₂.
- In series, capacitors carry the same charge, and the reciprocals add: 1/C = 1/C₁ + 1/C₂. The total is smaller than the smallest one.
This page does it with you, one step at a time. The circuit diagram colours the group being reduced, and each step is written out with the numbers. When the network is down to one capacitor, a table gives the charge, voltage and stored energy on every capacitor.
It follows the electrostatic potential and capacitance chapter of Class 12 Physics (combination of capacitors in series and in parallel, energy stored in a capacitor). Try a single capacitor with a dielectric first in the parallel-plate capacitor lab, compare with resistors in series and parallel, and for capacitors of real shapes use the capacitor inductor calculator.
How to use it
- Choose a network: two in series, two in parallel, three in series, one in series with a parallel pair, two parallel pairs in series, or a small ladder.
- Set each capacitor (1 to 12 µF) with its box, and the battery (1 to 24 V) with the slider.
- Press Back and Next step to go through the reduction one group at a time, or Show all steps.
- When the last step is shown, read the equivalent capacitance, the battery’s charge and energy, and the table for each capacitor.
Worked example
The page opens with C₁ = 4 µF in series with a parallel pair, C₂ = 2 µF and C₃ = 6 µF, on a 12 V battery.
- C₂ and C₃ are in parallel: 2 + 6 = 8 µF. Call it C₂₃.
- C₁ and C₂₃ are in series: 1/C = 1/4 + 1/8, giving 2.67 µF.
The battery supplies Q = 2.67 × 12 = 32 µC. C₁ carries all of it, so V₁ = 32 / 4 = 8 V. That leaves 4 V across the pair: C₂ holds 2 × 4 = 8 µC and C₃ holds 6 × 4 = 24 µC, which add back up to 32 µC. The energies are 128, 16 and 48 µJ, which add up to ½ × 2.67 × 12² = 192 µJ.
Tips
- For two in series, C = C₁C₂ / (C₁ + C₂) is quicker than reciprocals.
- In series, the smallest capacitor gets the largest share of the voltage, because V = Q/C and Q is the same for each.
- Always reduce the innermost group first, just like brackets in arithmetic.
FAQ
What about networks that are neither series nor parallel? Some, like a bridge, cannot be reduced this way. They need Kirchhoff’s rules (or a symmetry trick) and are not covered here.
Are the capacitors charged from zero? Yes: the values are for the network charged from a battery with all capacitors starting empty.
Is anything saved? No. Everything is worked out on this page and nothing leaves your device.
Where the numbers come from
- Capacitors in parallel: C = C₁ + C₂ + …; in series: 1/C = 1/C₁ + 1/C₂ + …; Q = CV and U = ½CV². Standard physics results.
- Checked against an independent Python script using exact fractions for every network, and by checking that the energies add up to ½CV² for the whole network.
Standard scientific values, such as constants, atomic weights and melting points, are facts found in every textbook; we name where ours were checked. If you are an author or publisher and want a value, a credit or a table changed or removed, tell us and we will act quickly.
Last updated 2026-11-09.