Series Capacitance Calculator

Series Capacitance Formula

A series chain of capacitors has one path, so the same charge is pushed onto every capacitor in it. Charge cannot accumulate on the isolated plates between two capacitors - whatever leaves one plate arrives on the next - so every capacitor in the chain stores the same charge Q, whatever its value.

The total voltage across the chain is the sum of the individual voltages, which is Kirchhoff's voltage law again:

The defining relationship for a capacitor is Q = C × V, so each voltage is the shared charge divided by that capacitor. Substituting gives:

The charge is common to every term, so it divides out - and what is left is the reciprocal formula for equivalent series capacitance:

This is the reverse of resistors, and the swap is the source of most errors with capacitors: series capacitors use reciprocals, parallel capacitors add. Physically it makes sense - putting capacitors in series is like making the dielectric thicker, and capacitance falls as separation grows.

Two capacitors: the product-over-sum shortcut

For exactly two capacitors the reciprocals clear to the product-over-sum form, which is faster by hand:

As with resistors, this is valid for two capacitors only. Using it on three gives a wrong answer silently.

Voltage distribution across capacitors in series

Because the charge is shared and V = Q/C, the smallest capacitor takes the largest share of the voltage. Each capacitor sees:

This is exactly backwards from a resistive divider, and it is the fact that damages parts. Two capacitors in series across a supply do not each see half the voltage unless they are equal. A 100 nF in series with a 10 nF puts about 91% of the applied voltage across the 10 nF part.

Worked example

The answer is smaller than 100 nF, the smaller of the two. If you got 320 nF you used the parallel rule by mistake — the single most common error with capacitors.

Why Put Capacitors in Series at All?

Series capacitance is smaller and costs more parts, so there is always a specific reason for it. There are three worth knowing.

Doubling the voltage rating

This is the main one. Two 400 V capacitors in series withstand 800 V, at half the capacitance. It is how high-voltage electrolytic banks are built when no single part has the rating - common in valve amplifiers, motor drives and mains-side supplies.

Reaching a value you cannot buy

Capacitors come in coarser steps than resistors - E6 or E12 for most types - so the gaps are wide. Two in series reach values between the standard ones, which matters in a filter or a timing circuit where the value sets a frequency.

DC blocking with a defined value

An AC coupling capacitor in series with a signal path blocks DC while passing AC. Its value and the following load resistance set a high-pass corner, so the series value is a filter design decision rather than an arbitrary large number.

Balancing Resistors for Series Capacitors

The voltage-sharing formula above assumes ideal capacitors. Real electrolytics have leakage current that varies part to part and rises with temperature and age. The capacitor with the lower leakage ends up holding more of the voltage, which increases its leakage further - a runaway that ends with one capacitor over its rating.

The fix is a balancing resistor - also called a bleeder or equalising resistor - across each capacitor. Their current dominates the leakage mismatch and forces the division to follow the resistors instead. A common rule of thumb is to size them for roughly ten times the worst-case leakage current:

They also discharge the bank when the equipment is switched off, which is a safety requirement in anything running at hazardous voltages. The trade-off is a constant drain and constant dissipation, so the resistors have to be rated for continuous duty at the voltage they sit across.

Series Capacitance in AC Coupling and Filters

A series capacitor into a resistive load is a first-order high-pass filter. The corner frequency, where the response is 3 dB down, is:

Two coupling capacitors in series make the effective C smaller, which pushes the corner up and cuts more bass from an audio path than either would alone. If two are in series for a voltage rating, the resulting equivalent capacitance is what goes into the filter calculation, not the individual value printed on the part.

Working the other way round is the more useful direction: decide the corner you want, calculate the capacitance it needs, and use the find-missing-value mode to work out the second capacitor when the first is fixed by what you have.

What the Ideal Formula Leaves Out

The reciprocal formula treats every capacitor as a pure capacitance. Real parts have three properties that matter in a series chain:

• ESR — equivalent series resistance — adds directly along the chain, so a series string has more loss than one capacitor. In a ripple-carrying application that is heat.
• Tolerance is wide. Electrolytics are often ±20% and can be −20/+80%, so the calculated series value is a nominal figure, not a measurement.
• Ceramic capacitors lose capacitance under DC bias. A Class 2 part such as X7R can lose half its marked value near its rated voltage, which no series formula accounts for.

For anything where the value has to be right - a timing circuit, a filter corner, a reference - use C0G/NP0 ceramics or film capacitors, and check the datasheet curve rather than the marking.

Common mistakes

• Adding capacitors in series. Series capacitors use reciprocals; it is parallel that adds. This is the most common capacitor error by a wide margin.
• Forgetting the final reciprocal, so the answer comes out inverted and enormous.
• Using product-over-sum for three or more capacitors. It is exact for two only.
• Assuming two capacitors in series each see half the voltage. They divide in inverse proportion to capacitance, so the smaller part takes more — and a mismatched pair can put most of the supply across one of them.
• Omitting balancing resistors across series electrolytics, and letting leakage mismatch push one capacitor past its rating.
• Trusting the marked value on a Class 2 ceramic under DC bias, where the real capacitance can be half of it.

Frequently asked questions

How do you calculate capacitors in series?

Add the reciprocals and invert: 1/Ceq = 1/C1 + 1/C2 + … + 1/Cn. For exactly two capacitors use Ceq = C1×C2/(C1+C2). The result is always smaller than the smallest capacitor in the chain.

Why does series capacitance decrease?

Every capacitor in the chain holds the same charge, and the voltages across them add. More voltage for the same charge means less capacitance, since C = Q/V. Physically it is like increasing the plate separation, which reduces capacitance.

How does voltage divide across capacitors in series?

In inverse proportion to capacitance: Vk = V × Ceq/Ck. The smallest capacitor takes the largest share of the voltage — the opposite of a resistive divider. Only equal capacitors split the voltage equally.

Do capacitors in series need balancing resistors?

Electrolytics in series do. Their leakage currents differ, so voltage drifts onto the lower-leakage part until it exceeds its rating. A resistor across each capacitor, passing about ten times the worst-case leakage, forces the division to follow the resistors and also discharges the bank at switch-off. Film and ceramic capacitors leak far less and usually do not need them.

What happens to the voltage rating of capacitors in series?

It adds, which is usually the reason for the series connection: two 400 V parts withstand 800 V. That only holds if the voltage actually divides as intended, which for electrolytics means fitting balancing resistors.

Are capacitors in series the same as resistors in parallel?

Mathematically yes — both use the sum of reciprocals, and both give a result smaller than the smallest element. It is a useful mnemonic: capacitors behave backwards compared with resistors in both arrangements.