Capacitor Energy Calculator – Stored Energy
Calculate the energy stored in a capacitor in joules using E = ½ × C × V² — instant results for electronics and electrical engineering.
Enter the capacitance in farads and the voltage across the capacitor to calculate stored energy (joules) and stored charge (coulombs).
Capacitor Energy Calculator – Stored Energy
Calculate the energy stored in a capacitor in joules using E = ½ × C × V² — instant results for electronics and electrical engineering.
About the Capacitor Energy Calculator
The energy stored in a capacitor is given by the formula E = ½ × C × V², where E is the energy in joules (J), C is the capacitance in farads (F), and V is the voltage across the capacitor in volts (V). This relationship arises from the work done to move charge onto the capacitor plates against the growing electric field: as each incremental charge dQ is moved, it must overcome a voltage V = Q/C, so the total work done is the integral of V dQ from 0 to Q_final, giving E = Q²/(2C) = ½CV².
The quadratic dependence on voltage is a critical design consideration: doubling the voltage across a capacitor quadruples the stored energy for the same capacitance. Conversely, doubling the capacitance while keeping the voltage constant only doubles the stored energy. This means that for high-energy storage applications (such as camera flashes, pulsed lasers, or defibrillators), using a higher voltage with a smaller capacitor is more volumetrically efficient than using a large capacitor at low voltage — though the higher voltage imposes stricter safety and insulation requirements.
In power electronics, capacitor energy storage is used in a wide range of applications. DC link capacitors in variable-frequency drives store energy to smooth out ripple current drawn from the rectifier and to provide instantaneous current during switching transients. Energy storage banks made from large electrolytic capacitors or supercapacitors are used in uninterruptible power supplies (UPS) and regenerative braking systems. The ability to charge quickly and discharge rapidly makes capacitors complementary to batteries, which have higher energy density but cannot sustain the high peak power required for pulsed applications.
Safety is a crucial concern with high-energy capacitors. A 1000 μF capacitor charged to 400 V (as found in many switch-mode power supplies) stores E = ½ × 0.001 × 400² = 80 J — comparable to the muzzle energy of a small firearm. Even after the supply is disconnected, the capacitor holds this charge and can deliver a lethal shock. Discharge resistors (bleeders) are used to safely dissipate stored energy; the discharge time constant τ = R × C must be short enough to discharge the capacitor in a reasonable time while not being so short that the resistor itself becomes a fire hazard.
Supercapacitors (also called ultracapacitors or electrochemical double-layer capacitors) can store 100–1000 farads at low voltages (2.5–2.7 V per cell). A 500 F supercapacitor charged to 2.5 V stores E = ½ × 500 × 2.5² = 1562.5 J ≈ 0.43 Wh. While small compared to a lithium-ion battery (150–300 Wh/kg), supercapacitors can charge and discharge thousands of times faster and last for millions of cycles, making them ideal for peak power buffering in hybrid vehicles, regenerative braking, and pulse applications.
Worked Examples
Three capacitor energy calculations across different applications from electronics to power systems.
| Capacitor Values | Stored Energy | Application Notes |
|---|---|---|
| C = 100 μF = 1×10⁻⁴ F, V = 12 V | E = ½ × 1×10⁻⁴ × 144 = 7.2 × 10⁻³ J = 7.2 mJ | Small DC supply filter capacitor. Energy is modest; this capacitor is primarily for ripple filtering, not energy storage. |
| C = 1000 μF = 0.001 F, V = 400 V | E = ½ × 0.001 × 160,000 = 80 J | DC link capacitor in a switch-mode power supply. 80 J is potentially lethal — always discharge before servicing. |
| C = 500 F (supercapacitor), V = 2.5 V | E = ½ × 500 × 6.25 = 1562.5 J ≈ 0.434 Wh | Supercapacitor energy storage. Low voltage but enormous capacitance gives useful energy for short-duration backup power. |
How to Use the Capacitor Energy Calculator
- Enter the capacitance in farads (F). Convert from common units as needed: 1 μF = 1×10⁻⁶ F, 1 mF = 1×10⁻³ F, 1 nF = 1×10⁻⁹ F.
- Enter the voltage across the capacitor in volts (V). This is the charged voltage, not the rated working voltage.
- Click Calculate to see the stored energy in joules (J) and the stored charge in coulombs (C). The energy result is highlighted.
- To find the voltage needed for a target energy level, rearrange: V = √(2E/C). To find the required capacitance: C = 2E/V².
- Click Reset to clear the fields and perform a new calculation.
Frequently Asked Questions
What is the formula for capacitor energy storage?
The stored energy is E = ½ × C × V², where C is capacitance in farads and V is voltage in volts. The result E is in joules. The same energy can also be expressed as E = Q²/(2C) = ½QV, where Q = CV is the stored charge in coulombs. All three forms are equivalent and useful in different calculation contexts.
Why does energy scale with V² and not just V?
As charge accumulates on a capacitor, each new increment of charge must be pushed against an increasing opposing voltage. The work to add a small charge dQ is V × dQ = (Q/C) × dQ. Integrating from 0 to the final charge Q gives E = Q²/(2C) = ½CV². The quadratic dependence means that doubling the voltage quadruples the stored energy, making high-voltage storage much more energy-dense per unit of capacitance.
How does capacitor energy compare to battery energy?
Capacitors store much less energy per kilogram than batteries. A typical electrolytic capacitor stores 0.01–0.1 Wh/kg, while a lithium-ion battery stores 150–300 Wh/kg — roughly 3,000 to 10,000 times more energy per unit mass. However, capacitors can deliver their energy in microseconds, accept full charge in seconds, and survive millions of cycles. Supercapacitors bridge the gap at 1–10 Wh/kg but with faster charge/discharge and much longer cycle life than batteries.
Is all the charged energy recoverable?
Ideally yes — in a lossless circuit, all stored energy E = ½CV² can be recovered. In practice, some energy is dissipated in the equivalent series resistance (ESR) of the capacitor and any external resistance during discharge. When charging through a series resistor, exactly 50% of the supplied energy is dissipated in the resistor regardless of R value; the other 50% is stored. When discharging into a resistive load, the energy stored in the capacitor is fully delivered to the load (minus ESR losses).
What is the energy stored in capacitors in series or parallel?
For capacitors in parallel charged to the same voltage V: Ctotal = C1 + C2 + …, so total energy = ½ × Ctotal × V². For capacitors in series charged to the same total voltage V: 1/Ctotal = 1/C1 + 1/C2 + …, and total energy = ½ × Ctotal × V². In both cases the formula E = ½CV² applies to the equivalent capacitance. Note that in series, each capacitor has the same charge Q but different voltages, so individual energies are E_i = Q²/(2C_i).
Why are large capacitors dangerous even when disconnected from power?
A charged capacitor maintains its stored energy (E = ½CV²) after the power source is removed. For large capacitors at high voltage — such as those in CRT televisions, microwave ovens, welding equipment, and power supplies — the stored energy can be tens to hundreds of joules, and the peak discharge current can be thousands of amperes. This is lethal. Always use a discharge resistor (bleeder) to safely drain large capacitors before any service work, and verify with a multimeter that the voltage is at a safe level.