Capacitor Energy Calculator

Enter capacitance and voltage to get stored energy in joules and Wh, plus charge in coulombs and mAh — the numbers behind flash circuits and supercapacitor backup.

Capacitor Energy Calculator inputs

Last updated: 2026-09-28

How the calculation works

  • Energy follows E = ½·C·V²: charging a capacitor stores work equal to half the charge times the final voltage.
  • Charge follows Q = C·V directly — linear in voltage, unlike the energy's square relationship.
  • The mAh figure converts charge at 1 mAh = 3.6 C; it describes charge, not usable energy at a load voltage.

Formula

E = ½ × C × V²
Q = C × V
Formula variables
SymbolMeaningUnit
EStored energyJ
QStored chargeC
CCapacitanceF
VVoltage across the capacitorV

Worked example

A camera flash capacitor, 1000 µF at 300 V: E = 0.5 × 0.001 × 300² = 45 J — enough for the xenon tube's burst. The same capacitance at 12 V stores just 0.072 J: voltage dominates energy. A 10 F supercapacitor at 2.7 V: 36.5 J, about 0.01 Wh.

Interpreting the result

The V² law is the practical takeaway: doubling the voltage quadruples stored energy — and quadruples the arc-flash hazard on power-electronics bus capacitors, which stay charged long after power-off. Supercapacitors hold far less energy per kg than batteries (roughly 5-10 Wh/kg vs 100-250), but accept millions of cycles and deliver enormous power — that trade decides the application.

Assumptions

  • Ideal capacitor; real ESR dissipates a few percent on fast discharge.
  • Fully charged to V and discharged fully for the energy figure.

Limitations

  • Capacitor voltage rating must exceed the working voltage with margin (typically 20%+).
  • Series-connected capacitors share voltage unevenly without balancing resistors.

Frequently asked questions

How much energy is in a capacitor?

E = ½·C·V² joules. A 1000 µF cap at 12 V holds 0.072 J; at 300 V the same cap holds 45 J — 625× more, because voltage enters squared.

Why half of C·V²?

As the capacitor charges, its voltage rises from 0 to V, so the average voltage during charging is V/2. Energy = Q × V_avg = (C·V) × (V/2) = ½·C·V². The other half is dissipated in the charging resistance — a thermodynamic fact, not a design choice.

Related tools

Technical references