Capacitor Energy Calculator - Joules, Wh and Discharge Time

Energy in a capacitor, from a 100 nF chip to a 3,000 F cell

Double the voltage on a capacitor and it holds four times the energy; double the capacitance and it holds only twice as much. That square is the whole story behind W = C × V² ÷ 2, and this calculator runs it in three directions: energy from capacitance and voltage, the capacitance a target energy needs, or the voltage it needs. A 100 µF part at 400 V stores 8 J and a charge of 40 mC. Add a discharge resistor and you also get the RC time constant, the peak current and how long the voltage takes to drop to a level you choose.

Quick start. Pick "Energy", type 470 with the unit µF and 400 volts. The result is 37.6 J. Now add 100 kΩ as the discharge resistance and 50 V as the target: the time constant is 47 s and the voltage needs 97.734 s, a little over a minute and a half, to fall to 50 V.

The form, one box at a time

What to find - the stored energy, the capacitance for an energy you need, or the voltage for an energy you need. The boxes below change with the choice.
Capacitance and its unit - the value printed on the part, in pF, nF, µF, mF or F. A supercapacitor is typed as 3000 with F, not as three billion microfarads.
Voltage - the voltage the capacitor is actually charged to, which is usually below the rating on the case.
Energy and its unit - only when you are solving for capacitance or voltage, in µJ, mJ, J, kJ or Wh.
Discharge resistance (optional) - a bleeder resistor or the load the capacitor feeds, in Ω, kΩ or MΩ.
Target voltage (optional) - the level you want to reach, for example 50 V before touching a board, or the lowest voltage a circuit still runs on.
Read the result - the answer, the four quantities side by side, a lift comparison for scale and, with a resistance, a discharge table from 1 to 5 time constants.

Eight capacitors and what they hold

All results below came straight from the calculator, with the inputs listed so you can repeat each one.

Example 1: a photoflash capacitor

Situation: Energy, 330 µF at 330 V

Result: 17.969 J and 108.9 mC, enough to lift 1 lb by 13.3 ft.

Charged only to 300 V, the same part holds 14.85 J. Ten percent less voltage costs about 17% of the energy.

Example 2: the bulk capacitor in a power supply

Situation: Energy, 470 µF at 400 V, bleeder 100 kΩ, target 50 V

Result: 37.6 J, time constant 47 s, 97.734 s to reach 50 V, with 587.5 mJ still left at that point.

The bleeder draws only 4 mA and dissipates 1.6 W at the start, which is why the wait is long.

Example 3: a 3,000 F supercapacitor cell

Situation: Energy, 3000 F at 2.7 V

Result: 10.935 kJ, which is 3.0375 Wh, and a charge of 8,100 C.

A huge capacitance at a low voltage: the cell holds more energy than every other example on this page put together, yet the square law keeps it at about three watt-hours.

Example 4: a defibrillator-sized pulse

Situation: Voltage, 200 J in 150 µF

Result: the capacitor has to be charged to 1.633 kV, holding 244.95 mC.

Biphasic external defibrillators typically deliver 120 to 200 J, and older monophasic units 360 J. The capacitance here is an illustration, not a device specification.

Example 5: a 1 F car audio capacitor

Situation: Energy, 1 F at 14.4 V

Result: 103.68 J, or 28.8 mWh.

A 1,000 W amplifier would drain that in about a tenth of a second, and only if the voltage were allowed to fall all the way to zero.

Example 6: sizing a capacitor for one joule

Situation: Capacitance, 1 J at 50 V

Result: 800 µF. The next standard value up is 1,000 µF, which at 50 V gives 1.25 J.

Example 7: hold-up time for a 5 V board

Situation: Energy, 1000 µF at 5 V, load 100 Ω, target 3.3 V

Result: 12.5 mJ stored, time constant 100 ms, and 41.552 ms before the rail sags to 3.3 V.

At that moment 5.445 mJ is still in the capacitor, but the circuit can no longer use it.

Example 8: a 100 nF decoupling capacitor

Situation: Energy, 100 nF at 5 V

Result: 1.25 µJ and 500 nC. Tiny, but it sits millimeters from the chip and delivers it in nanoseconds.

Stored energy for five capacitors at four voltages

Each cell is W = C × V² ÷ 2. Reading across a row shows the voltage effect, reading down a column shows the capacitance effect. A cell only means something if the part is rated for that voltage.

Capacitance	5 V	16 V	50 V	400 V
100 nF	1.25 µJ	12.8 µJ	125 µJ	8 mJ
10 µF	125 µJ	1.28 mJ	12.5 mJ	800 mJ
100 µF	1.25 mJ	12.8 mJ	125 mJ	8 J
1,000 µF	12.5 mJ	128 mJ	1.25 J	80 J
10,000 µF	125 mJ	1.28 J	12.5 J	800 J

Why the voltage matters more than the microfarads

Charging a capacitor means pushing charge onto a plate that already repels it. The first microcoulomb goes on almost for free, the last one against the full voltage, so the average push is half the final voltage and the work adds up to half of Q × V. With Q = C × V that becomes C × V² ÷ 2. The table follows one 470 µF capacitor rated 450 V as the voltage rises.

Charged to	Energy	Compared with 100 V
100 V	2.35 J	1×
200 V	9.4 J	4×
300 V	21.15 J	9×
400 V	37.6 J	16×
450 V (rating)	47.588 J	20.25×

The same logic works backward. Take energy out and the voltage falls with the square root of what is left, so a capacitor that has given away three quarters of its energy still sits at half its starting voltage. That is why a supply capacitor can read a scary number on a meter long after the device looks dead.

Waiting for a capacitor to discharge

Through a fixed resistance the voltage decays as V₀ × e^-t/RC. After one time constant (R × C) 36.79% of the voltage is left, after five only 0.67%. To reach a particular voltage directly, the time is RC × ln(V₀ ÷ V_target): from 400 V to 50 V that is RC × ln 8, about 2.08 time constants. The energy drops twice as fast as the voltage in these terms, because it goes with the square.

50 V is a useful target because the OSHA work-practice rules for electrical safety treat live parts under 50 V to ground as not requiring de-energizing. Measure before touching anyway: a bleeder resistor can fail open, and then the calculated time means nothing.

Things people ask before they charge one

What is the formula for the energy stored in a capacitor?

W = C × V² ÷ 2, with C in farads, V in volts and W in joules. Two equivalent forms are W = Q × V ÷ 2 and W = Q² ÷ (2C). For 100 µF at 400 V all three give 8 J.

How do I convert joules to watt-hours?

Divide by 3,600, because one watt-hour is one watt for 3,600 seconds. The calculator adds the Wh figure next to the joules once it reaches 1 mWh, so a 3,000 F cell at 2.7 V shows 10.935 kJ = 3.0375 Wh.

Can I use all of the stored energy?

Rarely. Most circuits stop working below some minimum voltage, and whatever is left below it stays in the capacitor. In Example 7 the board quits at 3.3 V with 5.445 mJ of the original 12.5 mJ unused, so only 56.4% was delivered.

What resistor should I use to discharge a capacitor?

Pick the time you are willing to wait, then check the peak current V ÷ R and the peak power V² ÷ R the result shows. A 10 kΩ resistor across 470 µF at 400 V brings it to 50 V in under ten seconds, but it starts at 16 W, so a small quarter-watt part is the wrong choice.

Why is the charge shown in coulombs?

Charge Q = C × V is the amount of electricity moved onto the plates. It helps when you compare with a battery rating: 1 mAh equals 3.6 C, so the 8,100 C in a 3,000 F cell at 2.7 V corresponds to about 2,250 mAh, delivered while the voltage slides from 2.7 V toward zero.

Does a real capacitor hold exactly this much?

The formula is exact for an ideal part. Real ones carry a tolerance, often ±20% for aluminum electrolytics, lose some charge through leakage, and waste part of a fast discharge in their equivalent series resistance. Treat the answer as the nominal figure.

What happens if I connect capacitors in series to get a higher voltage?

Two equal capacitors in series have half the capacitance and can take twice the voltage, so the stored energy at the doubled voltage is twice that of one part. Supercapacitor banks need balancing resistors or a balancing circuit, because unequal leakage lets one cell drift above its rating.

Is 8 J a lot?

In mechanical terms it would lift a 1 kg mass by 815.77 mm, about knee height, which is why the result adds that comparison. Released in a few milliseconds through a short circuit, though, the same energy can pit a screwdriver tip or weld contacts, so the delivery time matters as much as the number.

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