Capacitor Energy Calculator
Calculate the energy stored in a capacitor given its capacitance and voltage. The energy stored follows the formula E = ½CV² and is measured in joules. See also our Capacitance Calculator and RC Circuit Calculator.
How to Calculate Capacitor Energy
The energy stored in a capacitor is proportional to its capacitance and the square of the voltage across it. This quadratic relationship with voltage means that doubling the voltage quadruples the stored energy. The energy is stored in the electric field between the capacitor plates and can be released when the capacitor discharges.
Understanding capacitor energy is critical for power supply design (hold-up time), flash photography (energy per flash), defibrillators (energy delivered to patient), and safety (discharge energy in high-voltage circuits). Capacitors charged to high voltages can store dangerous amounts of energy — even small capacitors at high voltage can deliver lethal shocks.
Capacitor Energy Formula
Energy Stored:
E = ½ × C × V²
Alternative Forms:
E = ½ × Q × V (using charge)
E = Q² / (2 × C) (using charge and capacitance)
Where:
E = energy in joules (J)
C = capacitance in farads (F)
V = voltage in volts (V)
Q = charge in coulombs (C) = C × V
Power During Discharge:
P = E / t (average power over discharge time)
Peak power depends on discharge circuit resistance
Example Calculation
Calculate the energy stored in a 100µF capacitor charged to 12V:
C = 100 µF = 100 × 10⁻⁶ F = 0.0001 F
V = 12V
E = ½ × 0.0001 × 12² = ½ × 0.0001 × 144
E = 0.0072 J = 7.2 mJ
Charge stored: Q = CV = 0.0001 × 12 = 0.0012 C = 1.2 mC
If discharged through 100Ω:
Time constant τ = RC = 100 × 0.0001 = 0.01s = 10ms
Peak current = V/R = 12/100 = 120mA
Peak power = V²/R = 144/100 = 1.44W
Capacitor Energy Reference Table
| Capacitance | Voltage | Energy Stored |
|---|---|---|
| 1 pF | 5V | 12.5 pJ |
| 100 pF | 5V | 1.25 nJ |
| 1 nF | 12V | 72 nJ |
| 100 nF | 12V | 7.2 µJ |
| 1 µF | 12V | 72 µJ |
| 10 µF | 25V | 3.125 mJ |
| 100 µF | 12V | 7.2 mJ |
| 100 µF | 50V | 125 mJ |
| 1000 µF | 25V | 312.5 mJ |
| 4700 µF | 50V | 5.875 J |
| 1 F | 2.7V | 3.645 J |
| 3000 F | 2.7V | 10.935 kJ |
Frequently Asked Questions
Why is capacitor energy proportional to V²?
As a capacitor charges, each additional increment of charge requires more work because it must be pushed against the increasing voltage. The energy is the integral of V×dQ from 0 to Q, which gives ½QV = ½CV². This means doubling the voltage stores four times the energy — making voltage the dominant factor in energy storage.
How much energy is dangerous?
As little as 1 joule can cause a painful shock, and 10-50 joules across the chest can cause cardiac fibrillation. A 400V, 470µF capacitor (common in power supplies) stores 37.6J — potentially lethal. Always discharge capacitors before working on equipment. Even "low voltage" supercapacitors can deliver dangerous currents due to their extremely low internal resistance.
How long can a capacitor hold its charge?
The discharge time depends on leakage resistance. Film capacitors can hold charge for hours to days. Electrolytic capacitors have higher leakage and may discharge in minutes to hours. The self-discharge time constant is τ = R_leakage × C. A 1000µF electrolytic with 1MΩ leakage has τ = 1000 seconds (~17 minutes to drop to 37% of initial voltage).
Can capacitors replace batteries?
Supercapacitors bridge the gap between capacitors and batteries. They offer much higher power density (fast charge/discharge) but lower energy density than batteries. A 3000F supercapacitor at 2.7V stores about 3 Wh — compared to a AA battery at about 3 Wh. Supercapacitors excel in applications needing rapid charge/discharge cycles: regenerative braking, UPS hold-up, and burst power delivery.
What is the energy density of different capacitor types?
Ceramic capacitors: 0.01-0.1 J/cm³. Film capacitors: 0.1-1 J/cm³. Electrolytic capacitors: 0.5-5 J/cm³. Supercapacitors: 10-30 J/cm³. For comparison, lithium batteries achieve 1000-2500 J/cm³. Capacitors trade energy density for power density — they can deliver their energy much faster than batteries.
How do I calculate hold-up time for a power supply?
Hold-up time is how long a capacitor can power a load after input power is lost. t = C × (V_max² - V_min²) / (2 × P_load). For example, a 1000µF cap from 400V to 300V powering 100W: t = 0.001 × (160000-90000) / (2×100) = 0.35 seconds. This determines the capacitor size needed in UPS systems and power supplies.
Safety Warning
Charged capacitors can be extremely dangerous. High-voltage capacitors in power supplies, camera flashes, defibrillators, and industrial equipment can retain lethal charge for extended periods after power is removed. Always verify capacitors are discharged before handling. Use a discharge resistor (not a screwdriver — the spark can damage the capacitor and create shrapnel). Capacitors rated above 50V should be treated with extreme caution.
Practical Applications
- Camera flash: 300µF at 300V = 13.5J released in milliseconds for intense light
- Power supply hold-up: Bulk capacitors maintain output during brief input interruptions
- Defibrillators: 32µF at 5000V = 400J delivered to restart the heart
- Regenerative braking: Supercapacitors capture kinetic energy during braking
- Pulsed lasers: Capacitor banks provide high peak power for laser excitation
- Welding: Capacitor discharge welding uses stored energy for precise spot welds