Capacitors in Series: the Smallest One Decides, and Then It Fails

Why anyone puts them in series at all

Not for capacitance, which the arrangement reduces, but for voltage rating: two identical parts each rated 400 V make a string that can stand 800 V, at half the capacitance. The catch is that the split is only even while the two are genuinely identical, and real parts differ in tolerance and in leakage. The one that leaks less ends up holding more voltage, which stresses it more, which usually makes it leak less still. That runaway is why series strings in production carry balancing resistors across each capacitor — components whose only job is to force the division to stay where the design put it.

The tolerance is wider than the calculation

Reporting 56.9 µF from three nominal values implies a precision that the parts do not have. Electrolytic capacitors commonly carry tolerances of twenty per cent or worse, and their value drifts with temperature and with age as the electrolyte dries, so the string is somewhere in a band rather than at a point. That is fine for a smoothing or decoupling job, where any value above a minimum works, and it is not fine for a timing or filter circuit, where the value sets a frequency. For those, a film or ceramic part with a tight tolerance is the component the design actually needs, and the arithmetic is worth carrying to a decimal only once such a part is in the circuit.

Frequently asked questions

And in parallel?

They simply add: 100 and 220 in parallel are 320 µF, and the voltage rating of the combination is that of the lowest-rated member rather than the sum. Parallel is the arrangement used when more capacitance is wanted, and it is also the reason a board carries several small decoupling capacitors instead of one large one — the small ones have lower internal inductance and respond faster, so the group behaves better at high frequency than any single part of the same total value.

Does the same formula work for any number of capacitors?

Yes — add the reciprocal of each and take the reciprocal of the sum, for two parts or for twenty. Two shortcuts are worth remembering because they come up constantly: n identical capacitors in series give one nth of the single value, so ten 100 µF parts make 10 µF, and two unequal ones can be done as the product over the sum. Beyond three parts the general form is quicker than any shortcut, and it is the one the calculation above uses.