Capacitor Energy Calculator
Calculate capacitor energy with our free science calculator. Uses standard scientific formulas with unit conversions and explanations.
Formula
E = 0.5 C V^2
Where E is the energy stored in joules, C is the capacitance in farads, and V is the voltage across the capacitor in volts. Equivalent forms include E = Q^2/(2C) and E = QV/2, where Q is the charge in coulombs.
Worked Examples
Example 1: Camera Flash Capacitor Energy
Problem: A camera flash uses a 220uF capacitor charged to 330V. How much energy is stored?
Solution: E = 0.5 x C x V^2\nC = 220 x 10^-6 = 2.2 x 10^-4 F\nV = 330 V\nE = 0.5 x 2.2e-4 x 330^2\n= 0.5 x 2.2e-4 x 108900\n= 0.5 x 23.958 = 11.979 joules\nCharge Q = CV = 2.2e-4 x 330 = 0.0726 coulombs\nIf discharged in 1ms: Peak power = 11.979/0.001 = 11,979 W
Result: Energy: 11.98 J | Charge: 72.6 mC | Peak Power: ~12 kW (if 1ms flash duration)
Example 2: Supercapacitor Energy Storage
Problem: A 3000F supercapacitor rated at 2.7V. How much energy does it store in watt-hours?
Solution: E = 0.5 x C x V^2\n= 0.5 x 3000 x 2.7^2\n= 0.5 x 3000 x 7.29\n= 10,935 joules\nConvert to Wh: 10935 / 3600 = 3.038 Wh\nCharge Q = 3000 x 2.7 = 8100 coulombs\nThis is roughly equivalent to a small AA battery (about 3 Wh)
Result: Energy: 10,935 J = 3.04 Wh | Charge: 8,100 C | Comparable to an AA battery
Frequently Asked Questions
What is the energy stored in a capacitor and how is it calculated?
The energy stored in a capacitor equals one-half times the capacitance times the voltage squared: E = 0.5 CV^2. This energy is stored in the electric field between the capacitor plates. The energy can also be expressed as E = Q^2/(2C) or E = QV/2, where Q is the charge stored. The factor of one-half arises because as the capacitor charges, the average voltage during the charging process is V/2 (it starts at zero and ends at V). For example, a 1000 microfarad capacitor charged to 50 volts stores 0.5 x 0.001 x 2500 = 1.25 joules of energy. This energy can be released very quickly, making capacitors useful for applications requiring short bursts of high power.
How does capacitor energy compare to battery energy storage?
Capacitors store far less energy per unit mass or volume than batteries, but they can deliver that energy much more rapidly. A typical AA alkaline battery stores about 10,000 joules (roughly 2800 mAh at 1.5V), while even a large 10,000 microfarad capacitor at 50V stores only 12.5 joules. However, a capacitor can discharge its energy in microseconds, producing enormous instantaneous power. Supercapacitors (also called ultracapacitors) bridge the gap, storing 10-100 times more energy than conventional capacitors while maintaining faster charge/discharge rates than batteries. They typically store 5-10 Wh/kg compared to 100-250 Wh/kg for lithium-ion batteries but can sustain millions of charge cycles versus a few thousand for batteries.
What determines the maximum voltage rating of a capacitor?
The voltage rating is determined by the dielectric breakdown strength of the insulating material between the capacitor plates. If the applied voltage exceeds the rated voltage, the electric field across the dielectric can cause it to break down (become conductive), permanently damaging or destroying the capacitor. Different dielectric materials have different breakdown strengths: air breaks down at about 3 MV/m, paper at 10-15 MV/m, ceramic at 10-25 MV/m, and some polymers at 100+ MV/m. The voltage rating includes a safety margin below the actual breakdown voltage. In circuit design, it is standard practice to select a capacitor with a voltage rating at least 50 percent higher than the maximum expected circuit voltage to ensure reliability and long life.
How do different capacitor types compare in terms of energy density and applications?
Ceramic capacitors are small and inexpensive but limited to picofarads through low microfarads, used for decoupling and high-frequency filtering. Electrolytic capacitors (aluminum and tantalum) offer the highest capacitance per volume (up to thousands of microfarads) but are polarized and have limited voltage ratings and shorter lifespans. Film capacitors provide excellent stability, low losses, and high voltage ratings but are larger for a given capacitance. Supercapacitors achieve farad-level capacitance for energy storage applications but have low voltage ratings (typically 2.5-2.7V per cell). For energy storage applications, electrolytic capacitors provide the best joules per dollar, while supercapacitors provide the best joules per volume among capacitor technologies.
How do I calculate the peak current and power when a capacitor discharges?
When a capacitor discharges through a resistor, the initial peak current equals V/R (voltage divided by resistance). The current then decays exponentially as I(t) = (V/R) x e^(-t/RC). The instantaneous power is I^2 x R, which peaks at V^2/R at the instant of discharge. For a 1000 microfarad capacitor at 100V discharging through 1 ohm, the peak current is 100 amps and peak power is 10,000 watts (10 kW), though this lasts only milliseconds. If a capacitor is short-circuited (R approaches zero), the peak current and power are theoretically infinite, limited only by the internal resistance (ESR) of the capacitor and wiring. This is why short-circuiting capacitors is extremely dangerous and can cause explosions, fires, or severe burns.
Is my data stored or sent to a server?
No. All calculations run entirely in your browser using JavaScript. No data you enter is ever transmitted to any server or stored anywhere. Your inputs remain completely private.