Capacitance and inductance
Capacitor energy
- where
- energy stored (J)
- capacitance (F)
- voltage (V)
Explain this equation
Works out the energy stored in a charged capacitor. Relevant to discharge hazards and to supply hold-up capacitors.
E is the stored energy in joules. C is the capacitance in farads and V is the voltage across it.
As a capacitor charges, its voltage rises from zero, so on average each coulomb is stored at half the final voltage. That gives the ½, and Q = CV gives the V².
Sense check: 100 µF at 12 V stores about 0.0072 J. Because V is squared, the same capacitor at 400 V stores about 8 J: double the voltage, four times the energy.
Watch out: this is not a safety assessment. Always prove a capacitor discharged before touching it.
Stored capacitor energy increases with the square of voltage. Double the voltage and the stored energy becomes four times as large at unchanged capacitance.
Use the actual terminal voltage and capacitance in farads. Stored energy must be managed through an appropriate safe procedure.
Enter your values to see the result.
Common mistake
Forgetting the half or failing to square voltage.
Limits
Ideal capacitor; not a discharge safety assessment.
Sources
The tools calculate the displayed equation. They do not automatically choose cable sizes, approve installations, supply current regulatory tables, certify equipment or replace inspection and testing.