How this instrument works
Wire two capacitors end to end and that middle node is electrically stranded: whatever charge leaves one plate has nowhere to go but its neighbour, so both components carry an identical charge Q. Voltage behaves in reverse — each part holds Q/C volts, and those volts add up to whatever drives them. Rearranging that one line of bookkeeping produces a reciprocal rule, and with it a result that catches people out every time: a series pair stores less charge per volt than its smaller member.
Faraday's name sits on this unit, but one farad is an absurd quantity of it. Bench parts run from picofarads of stray coupling between two adjacent traces, through nanofarad ceramics that quiet logic rails, up to tens of microfarads inside electrolytics smoothing rectifier ripple. Series stacking is how engineers get past a part's voltage rating: two 450 V electrolytics in a chain survive 900 V, paying for that headroom by halving capacitance and demanding balancing resistors across each can.
This rule assumes both parts genuinely share one charge, which fails wherever that middle node leaks — stray capacitance to chassis metal, a scope probe clipped on to watch that split, or DC leakage in aluminium electrolytics will divert charge and skew results. It also treats each part as a fixed C, which class-2 ceramics are not; an X7R part can shed a third of its marked value under DC bias. Nothing here describes series resistance or lead inductance, which govern how a stack behaves above a few megahertz.
- Type the first value into Capacitor 1 and pick its unit from the menu — pF, nF, uF, mF or F — rather than keying exponents by hand.
- Type the second into Capacitor 2. Order changes nothing; the expression is symmetric in both.
- Read Total capacitance, then sanity-check it: a correct answer always sits below the smaller of your two entries.
- For a chain of three or more, feed the running Total capacitance back into Capacitor 1 and put the next part in Capacitor 2.
Worked example — two 1 uF parts in a chain
A decoupling job asks for 0.5 uF and the parts drawer holds nothing but 1 uF ceramics. Put two of them nose to tail: C = (1e-6 × 1e-6) ⁄ (1e-6 + 1e-6) = 1e-12 ⁄ 2e-6 = 5e-7 F. Enter 1 uF into Capacitor 1, 1 uF into Capacitor 2, and Total capacitance reads 5e-7 F — 0.50 uF on the unit menu.
Equal parts always halve, which is the cleanest way to remember that series and parallel trade places between capacitors and resistors: two 1 kΩ resistors in a chain give 2 kΩ, two 1 uF ceramics give 0.5 uF. What you buy with that loss is headroom, since each ceramic now sees only half of whatever is applied across the pair.
Questions
Why is series capacitance smaller than either part?
Because charge is shared while voltage divides. Capacitance is charge per volt, so holding Q fixed while volts add can only push that ratio down. Geometrically it amounts to moving two plates further apart — series wiring effectively lengthens the gap any field must cross, and wider gaps hold less charge at any given voltage.
Do capacitors in series each see the same voltage?
Only if they are equal. Charge is common, so V = Q/C and whichever value is smaller takes the larger share. Put 1 uF and 3 uF across 12 V and 9 V lands on the 1 uF part, 3 V on the 3 uF one. Assuming an even split is how the smaller can gets destroyed, which is exactly why high-voltage stacks carry balancing resistors.
Does the product-over-sum shortcut work for three or more?
No. C₁C₂/(C₁+C₂) is exact for a pair and nothing else. For longer chains, sum reciprocals — 1/C = 1/C₁ + 1/C₂ + 1/C₃ — or apply the instrument repeatedly, folding each running answer back into Capacitor 1. Three matched parts land at one third of a single value, not two thirds.
How does this differ from capacitors in parallel?
Parallel is an additive case: plates sit side by side at a common voltage, charges add, and C = C₁ + C₂. Series is a reciprocal case. Notice how resistance mirrors it in reverse — resistors add in a chain and combine reciprocally side by side, precisely opposite. Swapping those two rules is a common slip in first-year circuit work.
Why do stacked electrolytics need balancing resistors?
Aluminium electrolytics leak, and no two leak at quite matching rates. Over seconds that middle node drifts toward whichever part leaks less, loading it with most of an applied supply until it passes its rating and vents. A matched resistor across each can — large enough to waste little power, small enough to swamp leakage current — pins that split near 50/50.
What happens to stored energy in a series pair?
It drops at a fixed supply voltage. Energy is ½CV², so halving C halves the joules held at the same volts. Stacking buys voltage rating, never storage. Supercapacitor banks make the bargain plain: six 2.7 V cells in a chain yield a 16.2 V unit holding one sixth of the farads of a single cell.