Capacitor Voltage Divider Calculator and Formula

A capacitor voltage divider calculator finds the AC voltage across either capacitor in a series pair when you enter the capacitances and the applied voltage. It uses the capacitive reactance relationship, and returns the voltage across each capacitor plus a check that the two voltages add to the input.

Formula

For two ideal capacitors connected in series across an AC source, voltage divides inversely with capacitance. The voltage on one capacitor is the source voltage multiplied by the other capacitor’s value, divided by the sum of both capacitances:

V₁ = V × C₂ / (C₁ + C₂)
V₂ = V × C₁ / (C₁ + C₂)

• V is the applied AC voltage in volts RMS.
• C₁ and C₂ are the two capacitances. This calculator accepts microfarads (µF) for both.
• V₁ and V₂ are the respective capacitor voltages in volts RMS.

Using the other capacitor in each numerator may seem backward, but it follows from the fact that a smaller capacitance has a larger capacitive reactance and therefore takes a larger share of the series voltage. The two expressions add to V. The frequency does not appear because, for ideal capacitors, both reactances contain the same frequency factor, which cancels in the ratio.

Worked Example

With the calculator’s default values, C₁ = 10 µF, C₂ = 20 µF, and V = 12 V RMS:

1. Voltage across capacitor 1: V₁ = 12 × 20 / (10 + 20) = 240 / 30 = 8.00 V RMS.
2. Voltage across capacitor 2: V₂ = 12 × 10 / (10 + 20) = 120 / 30 = 4.00 V RMS.
3. Check: 8.00 + 4.00 = 12.00 V RMS, equal to the applied voltage.

The result also illustrates the inverse relationship: the 10 µF capacitor has twice the voltage of the 20 µF capacitor.

How a Capacitor Divider Works

Capacitors oppose changes in voltage through capacitive reactance, given by X_C = 1 / (2πfC), where X_C is in ohms, frequency f is in hertz, and capacitance C is in farads. In a series circuit, the same current flows through both components. The voltage across each is proportional to its reactance; therefore, it is inversely proportional to capacitance.

This is an AC divider, not a general-purpose replacement for a resistive divider. An ideal capacitor blocks steady DC after it has charged, so the formula does not predict a useful steady-state division of a DC supply. During a DC switching transient, the capacitors may divide voltage according to their capacitances initially, but leakage and other real-world effects can change their eventual voltages.

When to Use the Calculator

Use it for a first-pass calculation of the sinusoidal voltage distribution across two series capacitors. It is useful when examining coupling networks, AC signal paths, or a circuit in which capacitors are intentionally placed in series. Enter capacitances in the same unit and the source voltage as an RMS value; the output is also RMS.

• For equal capacitances, each capacitor receives half the source voltage.
• If one capacitance is smaller, that capacitor receives the larger voltage share.
• If one capacitance is much larger than the other, most of the total voltage appears across the smaller one.

The result is a voltage ratio, not a guarantee that a component is suitable for a circuit. Check the capacitor’s specified AC voltage, frequency behavior, tolerance, temperature range, and intended application. Do not treat the computed RMS value as a peak value: for a sine wave, peak voltage is RMS voltage multiplied by √2.

Real-World Limits and Common Mistakes

The calculation assumes ideal capacitors in series, a steady sinusoidal AC source, and no other circuit elements affecting the voltage. Real capacitors have tolerance, leakage, losses, and frequency-dependent behavior. A load connected across one capacitor can change the division, as can a parallel resistor or other path. In a more complex circuit, use impedance analysis rather than this two-capacitor ratio.

• Mixing units: Both capacitances must use the same unit. The calculator uses µF for both, so a value entered in nF must first be converted (1 µF = 1,000 nF).
• Entering peak voltage as RMS: Use the source’s RMS value if you want RMS outputs. For a sine wave, convert peak to RMS by dividing by √2.
• Assuming frequency changes the ideal ratio: Frequency cancels from the ideal two-capacitor ratio, but it still affects each capacitor’s reactance and circuit current. Real components also depart from ideal behavior.
• Ignoring tolerance: Actual capacitances can differ from their marked values. Because voltage division depends on the ratio, tolerance can shift the voltage on each part.

Capacitors can retain charge after a circuit is disconnected, and circuits connected to mains or other hazardous voltages can cause serious injury. Do not rely on this calculator to establish safe working conditions or component ratings. Follow appropriate discharge and measurement procedures and seek qualified help where needed.

Frequently Asked Questions

Does a capacitor voltage divider work with DC?
Not as a steady-state divider. Capacitors can share voltage while a DC source is changing or during charging, but an ideal capacitor blocks steady DC current after charging. Leakage and circuit details affect the long-term result.

Why does the smaller capacitor get more voltage?
Capacitive reactance is inversely proportional to capacitance. In a series pair the current is the same, so the capacitor with greater reactance—usually the smaller capacitance—has the greater voltage drop.

Does the AC frequency affect the voltage division?
It cancels from the ideal two-capacitor ratio because both reactances include the same frequency factor. Frequency still affects reactance and current, and real component behavior may make the actual division frequency-dependent.

Can I use this for three or more capacitors?
This calculator covers two capacitors. For an ideal series string, each capacitor’s voltage is proportional to its reactance, or inversely proportional to its capacitance; calculate each share from the full series impedance. Real-world leakage and tolerances may also matter.

Are the answers RMS or peak volts?
The calculator treats the entered source voltage as RMS and reports RMS outputs. For a sine wave, multiply an RMS result by √2 to obtain its peak value.

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in c1 | Capacitor 1 capacitance | uF | 10 | 0.000001
in c2 | Capacitor 2 capacitance | uF | 20 | 0.000001
in v | Applied AC voltage | V RMS | 12 | 0
out v1 | Voltage across capacitor 1 | V RMS | v * c2 / (c1 + c2) | 2
out v2 | Voltage across capacitor 2 | V RMS | v * c1 / (c1 + c2) | 2
out total | Sum of capacitor voltages | V RMS | v1 + v2 | 2