Calculate equivalent capacitance, voltage distribution, and energy for capacitors in series. Visual voltage distribution and per-capacitor breakdown.
When capacitors are connected in series, the same charge accumulates on each capacitor, but the voltage divides between them in inverse proportion to their capacitance. The equivalent capacitance is always less than the smallest individual capacitor: 1/C_total = 1/C₁ + 1/C₂ + ... + 1/Cₙ.
Series connection is used when the voltage across the combination exceeds the rating of a single capacitor — for example, in voltage multiplier circuits, high-voltage power supplies, and AC coupling. Understanding the voltage distribution is critical for safety: a smaller capacitor in series develops a proportionally larger voltage across it, and if this exceeds its rating, it can fail catastrophically.
This calculator computes the equivalent capacitance, charge, voltage across each capacitor, and stored energy for any number of series capacitors. It includes visual voltage distribution charts and a detailed per-capacitor breakdown table. Check the example with realistic values before reporting. Use the steps shown to verify rounding and units. Cross-check this output using a known reference case.
Series capacitor calculations involve reciprocal sums and voltage division that are tedious for more than two capacitors. Verifying that no individual capacitor exceeds its voltage rating is critical for safety but easy to overlook in manual calculations.
This calculator instantly computes equivalent capacitance, per-capacitor voltage, charge, and energy for any number of capacitors. The visual voltage distribution chart makes it easy to verify safe operation at a glance.
Series: 1/C_total = 1/C₁ + 1/C₂ + ... + 1/Cₙ. Same charge on all: Q = C_total × V. Voltage on each: Vi = Q/Ci. Energy: E = ½CiVi². Total energy: E = ½C_total × V².
Result: 57.9 µF equivalent, voltages: 28.9V, 13.2V, 6.2V
1/C = 1/100 + 1/220 + 1/470 = 0.01 + 0.00455 + 0.00213 = 0.01667. C_total = 60.0 µF. Q = 60.0 × 50 = 3000 µC. V1 = 3000/100 = 30.0V, V2 = 3000/220 = 13.6V, V3 = 3000/470 = 6.4V.
The fundamental principle of series capacitors is charge conservation: the same charge Q accumulates on every capacitor because they share the same current path. The total voltage is the sum of individual voltages: V = V₁ + V₂ + ... + Vₙ = Q/C₁ + Q/C₂ + ... + Q/Cₙ = Q × (1/C₁ + 1/C₂ + ... + 1/Cₙ).
Since C_total = Q/V, we get the familiar formula: 1/C_total = 1/C₁ + 1/C₂ + ... + 1/Cₙ. For two capacitors: C_total = (C₁ × C₂)/(C₁ + C₂).
The voltage across each capacitor is inversely proportional to its capacitance: Vi = V_total × C_total/Ci. This means the smallest capacitor bears the largest voltage. If a 10 µF and 100 µF capacitor are in series at 100V, the 10 µF cap sees 90.9V while the 100 µF cap sees only 9.1V.
In practice, manufacturing tolerances (±20% for electrolytics) and different leakage currents cause the actual voltage distribution to differ from the calculated ideal. Balancing resistors across each capacitor ensure a predictable voltage distribution under all conditions.
**Voltage multiplier circuits** use series capacitors and diodes to multiply the input voltage (Cockcroft-Walton, Villard cascade). **AC coupling** in audio uses series capacitors to block DC and pass audio signals. **Energy storage** for high-voltage applications (defibrillators, pulsed power) uses series banks of lower-voltage capacitors to achieve the required total voltage.
In series, each capacitor must store the same charge. The total voltage is the sum of individual voltages. Since V = Q/C, smaller capacitors develop larger voltages. The equivalent capacitance C_total = Q/V_total is always less than any individual Ci because V_total > Vi for any individual cap.
Capacitors in series are analogous to resistors in parallel, and vice versa. In series, the same current (charge) flows through each, and voltages add — just like current divides in parallel resistors. The math follows the same reciprocal pattern.
The smallest capacitor gets the most voltage. V = Q/C, and since Q is the same for all series capacitors, smaller C means larger V. This is the opposite of resistor voltage dividers, where the largest resistance gets the most voltage.
Yes — the total voltage rating is the sum of individual ratings. Two 200V capacitors in series can handle 400V. However, the voltage distribution depends on capacitance values, and leakage currents can cause unequal voltage sharing. Balancing resistors (10-100 kΩ) across each cap are often used.
Each capacitor must handle the voltage that appears across it, which depends on its share of the total capacitance. Verify that Vi = Q/Ci does not exceed the rating of any individual capacitor.
In terms of equivalent capacitance, yes (50 µF). But the voltage rating doubles, and the physical size is larger. This trade-off is the main reason for using series capacitors — increased voltage handling.