Enter the AC frequency and any two of the following (capacitance, RMS voltage across the capacitor, reactive power) into the calculator to determine the capacitor’s reactive power (in VAR) or the missing variable. Note: an ideal capacitor has ~0 W of real (average) power; this calculator focuses on reactive power.

Capacitor Reactive Power Calculator

Enter any 2 values (Capacitance, RMS Voltage, Reactive Power magnitude) to calculate the missing variable

Capacitor Reactive Power Formula

The following formula is used to calculate the capacitor’s reactive power magnitude in sinusoidal steady-state (using RMS voltage).

|Q_c| = 2\pi f C V^2
  • Where |Qc| is the reactive power magnitude (VAR)
  • f is the frequency (Hz)
  • C is the capacitance (F)
  • V is the RMS voltage across the capacitor (V)

Using the common sign convention, a capacitor supplies negative reactive power (Qc < 0), so Qc = −|Qc|. An ideal capacitor dissipates approximately 0 W of real (average) power; any nonzero real power is due to losses such as ESR and dielectric loss.

How to Calculate Capacitor Reactive Power?

The following two example problems outline how to calculate capacitor reactive power.

Example Problem #1:

  1. First, determine the AC frequency. In this example, the frequency is 60 Hz.
  2. Next, determine the capacitance and RMS voltage across the capacitor. For this example, C = 100 µF and V = 120 V.
  3. Finally, calculate the reactive power magnitude using the formula above.

|Qc| = 2π f C V2

Inserting the values from above and solving the equation with the given values gives:

|Qc| = 2π × 60 × (100 µF = 0.0001 F) × (120)2 ≈ 542.87 VAR (so Qc ≈ −542.87 VAR using the usual capacitive sign convention)

Example Problem #2: 

Using the same process as example problem #1, we first define the variables outlined by the formula. In this case, the values are:

frequency (Hz) = 50

voltage (RMS volts) = 230

reactive power magnitude |Qc| = 1 kVAR (= 1000 VAR)

Entering these values into the formula (solving for capacitance, C = |Q| / (2π f V2)) or using the calculator above gives:

C = 1000 ÷ (2π × 50 × 2302) ≈ 0.0000602 F ≈ 60.2 µF