Capacitor Energy Calculator: Joules in Seconds

This calculator works out the energy stored in a capacitor from its capacitance and the voltage across it, giving you the result in joules. You enter the capacitance in farads (or a subunit) and the voltage in volts, and it returns the stored energy using the standard formula E = ½·C·V². It also shows the charge in coulombs, which is useful when you need to know how much charge the capacitor delivers.

Formula

The energy stored in a capacitor is given by:

E = ½ · C · V²

where:

• E = stored energy in joules (J)
• C = capacitance in farads (F)
• V = voltage across the capacitor in volts (V)

The same energy can be expressed using charge (Q) and capacitance: E = Q² / (2C) and also E = ½·Q·V. The calculator uses the first form because capacitance and voltage are the values you typically know from the component label or circuit design.

Worked Example

Using the default values from the calculator above – capacitance 1000 µF and voltage 12 V – the calculation goes like this:

1. Convert capacitance to farads: 1000 µF = 1000 × 10�� F = 0.001 F
2. Square the voltage: 12² = 144
3. Multiply: 0.5 × 0.001 × 144 = 0.072

The stored energy is 0.072 joules. The charge is 0.001 × 12 = 0.012 coulombs. If you discharge this capacitor through a circuit, that 0.072 J is the total energy available (minus losses in the discharge path).

Why the Voltage Is Squared

The V² term is not an approximation – it comes from the physics of moving charge against a changing electric field. As you charge a capacitor, the voltage rises linearly with charge (V = Q/C). The work done to add a small increment of charge dQ is V·dQ, and integrating that from 0 to Q gives ½·Q·V, which equals ½·C·V². That means doubling the voltage quadruples the stored energy, which is why high-voltage capacitors can be surprisingly dangerous even with small capacitance.

Common Units and Conversions

Capacitance is usually given in microfarads (µF), nanofarads (nF), or picofarads (pF) rather than whole farads, because 1 F is a very large value. The calculator accepts any subunit as long as you convert to farads in the formula – the default already does this. Common conversions:

• 1 F = 1,000,000 µF
• 1 µF = 1,000 nF = 1,000,000 pF
• Energy in joules can be converted to watt-seconds (1 J = 1 W·s) or to millijoules (1 J = 1000 mJ)

If you have capacitance in pF and voltage in kV, convert both to base units first: pF to F (divide by 10¹²) and kV to V (multiply by 1000).

How to Read the Result

The energy value tells you how much work the capacitor can do if fully discharged through an ideal load. In real circuits, some energy is lost as heat in the resistance of the discharge path, so the usable energy is always less. For a capacitor used as a backup power source, the energy in joules divided by the discharge time in seconds gives the average power in watts – but the actual power varies as the voltage drops. For a camera flash, the capacitor discharges in milliseconds, delivering a high peak power despite a modest total energy.

Common Mistakes and Misconceptions

• Using peak voltage instead of working voltage: The energy formula uses the actual voltage across the capacitor, not the rated maximum. If you charge a 25 V capacitor to only 10 V, use 10 V in the formula.
• Forgetting to convert units: Mixing µF with V without converting to F gives results off by a factor of 10â�¶. Always convert capacitance to farads first.
• Confusing energy with power: Energy is stored; power is the rate of energy transfer. A capacitor can store 1 J but deliver it in 1 ms, giving 1000 W peak – that is why photoflash capacitors are rated in joules, not watts.
• Assuming the formula includes ESR or leakage: The ideal formula ignores internal resistance and leakage current. Real capacitors lose energy through these, especially at high frequencies or long storage times.

Limits of the Formula

The equation E = ½·C·V² assumes a linear, ideal capacitor with constant capacitance regardless of voltage. Most ceramic and film capacitors approximate this well, but supercapacitors (electric double-layer capacitors) can have capacitance that varies with voltage, so the formula gives only an estimate. Electrolytic capacitors also have a tolerance of ±20% or more, so the actual stored energy can differ from the calculated value by that margin. The formula also assumes the capacitor is fully charged to the specified voltage; in a real circuit with resistance, charging to 99% of the supply voltage takes about 5 time constants (5·R·C).

Safety Considerations

Even small capacitors can deliver a painful or dangerous shock if charged to a high voltage. A 100 µF capacitor at 50 V stores 0.125 J, which is enough to cause a muscle spasm; a 1000 µF capacitor at 100 V stores 5 J, which can be lethal under the right conditions. Always discharge capacitors with a properly rated resistor before handling, and verify that the voltage reads zero with a multimeter. Electrolytic capacitors must be connected with the correct polarity – reversing the voltage can cause them to vent or explode. High-voltage capacitors (above 50 V) should be shorted with a discharge tool and kept shorted during storage.

Frequently Asked Questions

How many joules does a 1 farad capacitor at 5 volts store?

Using E = ½·C·V², a 1 F capacitor at 5 V stores ½ × 1 × 25 = 12.5 joules. That is enough to power a 1 W LED for 12.5 seconds, assuming no losses in the discharge circuit.

What is the difference between energy and charge in a capacitor?

Charge (Q = C·V) is the total amount of electricity stored, measured in coulombs. Energy (E = ½·C·V²) is the work that charge can do, measured in joules. A capacitor with more charge does not necessarily have more energy if the voltage is lower – energy depends on both charge and voltage.

Why do camera flash capacitors store so much energy but are small?

Camera flash capacitors are typically 100–300 µF charged to 300–400 V. At 300 V and 200 µF, the energy is ½ × 200×10�� × 300² = 9 J. The high voltage allows a small capacitance to store a useful amount of energy, and the flash tube discharges it in about 1 ms, producing a bright burst of light.

Can I use this calculator for supercapacitors?

Yes, but the result is approximate. Supercapacitors often have capacitance that decreases as the voltage rises, so the linear formula overestimates the stored energy near the rated voltage. For rough sizing, the formula is fine; for precise energy management, consult the manufacturer’s discharge curves.

What happens if I double the voltage across a capacitor?

The stored energy quadruples, because energy scales with the square of voltage. Doubling the voltage also doubles the charge. This is why overvoltage is dangerous – a small increase in voltage leads to a disproportionate increase in stored energy.

How do I convert joules to watt-hours for battery comparison?

Divide joules by 3600 to get watt-hours (1 Wh = 3600 J). For example, 0.072 J from the worked example is 0.072 / 3600 = 0.00002 Wh, which is negligible compared to a typical AA battery (about 2–3 Wh). Capacitors store far less energy than batteries of similar size, but they can deliver it much faster.

Related

• Antigen Frequency Calculator: How It Works
• Antenna Wavelength Calculator: Find Length
• Annular Velocity Calculator: Formula & Guide

in c | Capacitance | µF | 1000 | 0
in v | Voltage | V | 12 | 0
out e | Stored energy | J | 0.5 * c * 1e-6 * v * v | 4
out q | Charge | C | c * 1e-6 * v | 6