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Capacitor Energy Calculator

Find the energy stored in a capacitor.

Result

33.840 mJ
5,640.00 µC

How it is calculated

A charged capacitor stores energy E = ½·C·V² and charge Q = C·V. Enter the capacitance C in microfarads (µF) and the voltage V in volts. The tool converts C to farads (× 10⁻⁶), then reports energy in millijoules and charge in microcoulombs. Because voltage is squared, doubling V quadruples the energy but only doubles the charge. Capacitance cannot be negative. The result is the ideal stored energy: it ignores leakage, the series resistance of the part, and the fact that you can never discharge a capacitor to exactly zero volts in a real circuit.

Worked example

A 470 µF capacitor charged to 12 V. C = 470 × 10⁻⁶ F = 0.00047 F. Energy = 0.5 × 0.00047 × 12² = 0.5 × 0.00047 × 144 = 0.03384 J, which is 33.840 mJ. Charge = 0.00047 × 12 = 0.00564 C, which is 5,640.00 µC. At 24 V the energy would be four times larger, about 135 mJ.

Try this example →

Common mistakes

  • Entering capacitance in farads or nanofarads. The field expects microfarads.
  • Forgetting the square: the energy depends on V², not on V.
  • Using the capacitor's voltage rating instead of the voltage it is actually charged to.
  • Reading millijoules as joules or confusing mJ with µC. They are different quantities.

About this calculator

This calculator applies the formula E = ½CV² to find the energy stored in a capacitor (in joules) from its capacitance (farads) and the voltage across it, and separately computes the stored charge using Q = CV (coulombs).

Electronics students use it for circuit theory homework, while hobbyists and engineers use it to size capacitors for power-supply filtering, camera flash circuits, and energy-storage applications where knowing exactly how much energy a capacitor can release matters for both performance and safety. Because energy scales with the square of voltage, the calculator makes clear why doubling a capacitor's rated voltage quadruples its stored energy rather than just doubling it.

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Comments (1)

  • Editorial team

    The output is in millijoules, so large values need converting: 470 µF charged to 400 V gives 37,600.000 mJ, which is 37.6 J. That is a serious amount of stored energy, enough to hurt, which is why high-voltage capacitors should be discharged before anyone handles them, with guidance from someone qualified. Which circuit are you sizing a capacitor for?

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