Capacitor Energy Calculator

E = ½ × C × V², and Q = C × V

C

The capacitance, in microfarads by default, with nanofarads, picofarads and farads in the same box. It appears to the first power, so doubling the capacitance doubles the energy at the same voltage. The farad is an enormous unit — one farad charged to one volt holds one coulomb, which is more charge than most circuits ever move — which is why the everyday values are microfarads and picofarads and why the box offers them first. It is the electrical twin of mass in the kinetic energy expression: both resist a change in the rate of something

V

The voltage across the capacitor, in volts, and it is squared, which is what makes this page worth a calculator. Twice the voltage is four times the energy and three times the voltage is nine times, so the numbers grow much faster than intuition suggests: a capacitor bank at 400 V holds sixteen times what the same bank holds at 100 V. The value entered here is treated as a magnitude, and a negative voltage is rejected, because energy storage does not care about polarity even though a capacitor in an AC circuit spends half of every cycle charged the other way

E

The energy stored, in joules, with a second row in millijoules for the small end of the range. One joule is one watt for one second, which is the comparison that makes the number concrete: 4.5 J is a 4.5 watt load for a second, and 80 J — a 1000 µF capacitor at 400 V — is 80 watts for a second. The same ½ appears in the kinetic energy expression ½mv² and in the elastic potential energy ½kx², and the three are the same piece of mathematics with the conjugate pair of quantities swapped in

Q

The charge the capacitor holds, in coulombs, which is capacitance times voltage. It is reported but never asked for, because charge is not something you measure on a capacitor directly and it is not a quantity you would design to: you choose a capacitance and a voltage rating and the charge follows. It matters when you go on to how long the capacitor can supply a current, since a coulomb per second is an amp, and it is the electrical counterpart of momentum in the mechanical analogue, where p = mv pairs with Q = CV

Use it when a capacitor is being used as an energy store rather than as a filter: sizing the bank behind a camera flash, working out what a capacitor can hold up during a power interruption, checking how much energy is available for a pulse, or estimating how hot a capacitor will get when it is discharged into a low resistance. Use it too before working on anything with a large capacitor in it, since the number this page returns is the energy that will come out of it if something shorts the terminals. It is the wrong tool for a filter or a decoupling capacitor, where the energy stored is incidental and the capacitance is chosen for impedance rather than for storage.

1. The defaults: 100 µF charged to 300 V

   1. Capacitance 100 µF (0.0001 F), voltage 300 V
   2. E = ½ × 0.0001 × 300² = ½ × 0.0001 × 90000
   3. E = 4.5 J, which is the same as 4500 mJ
   4. Charge is 0.0001 × 300 = 0.03 C

   4.5 joules is a familiar amount of energy in an unfamiliar place. It is what a 4.5 W lamp uses in one second, and it is enough to make a visible spark and a sharp crack if the capacitor is shorted with a screwdriver. The number worth carrying away is the 300 squared: because the voltage is squared and the capacitance is not, choosing a capacitor rated at twice the voltage holds four times the energy for the same capacitance, not twice. That is why the voltage rating of a capacitor is the specification to read first.

2. The energy in a power supply filter: 1000 µF at 400 V

   1. Capacitance 1000 µF (0.001 F), voltage 400 V
   2. E = ½ × 0.001 × 400² = ½ × 0.001 × 160000
   3. E = 80 J
   4. Charge is 0.001 × 400 = 0.4 C

   This is the capacitor sitting on the high-voltage side of a switched-mode power supply, and 80 J is the number that makes it dangerous. Released in a millisecond — which is what happens when something shorts a charged capacitor — 80 J is a burst of 80 kW; released through your hand it is a burn and a muscle contraction. This capacitor holds its charge with the mains unplugged, because nothing in the circuit discharges it, which is why a board that has been off for an hour can still be live. The safety rule that follows is mechanical rather than electrical: bleeder resistor, or a meter check with the supply off, every time.

3. A real one from a 12 V rail: 4700 µF at 16 V

   1. Capacitance 4700 µF (0.0047 F), voltage 16 V
   2. E = ½ × 0.0047 × 16² = ½ × 0.0047 × 256
   3. E = 0.602 J, which is 601.6 mJ
   4. Charge is 0.0047 × 16 = 0.0752 C

   This is the biggest capacitor most people actually have on a bench — a 4700 µF electrolytic rated 16 V, the sort that sits on the output of a 12 V supply — and it stores six tenths of a joule, which is harmless. Comparing it with the previous example is the whole lesson of the page: the capacitance only went up by a factor of 4.7 while the energy went up by a factor of 133, because the voltage went from 16 to 400. Voltage is where the energy lives, which is why high-voltage capacitors are the ones with the warning labels and why this figure is 0.602 rather than a round 0.6.

4. The small end: 1 µF at 10 V

   1. Capacitance 1 µF (0.000001 F), voltage 10 V
   2. E = ½ × 0.000001 × 100 = 0.00005 J
   3. The joule row shows 0.000, because three decimal places cannot resolve fifty millionths of a joule
   4. The millijoule row shows 0.05, and the charge is 0.00001 C

   The first output reads 0.000 J and that is not a failure: the energy really is fifty millionths of a joule, and the joule row is scaled for a range that starts around one joule. This is why the page carries a second row in millijoules rather than one row with more decimals — a single row would either lose the small end or clutter the large end. For small capacitors the millijoule row is the one to read, and for a 100 µF at 300 V it is the joule row. Nothing is wrong with the zero; it is the unit that is too large.

5. An uncharged capacitor: 100 µF at 0 V

   1. Capacitance 100 µF, voltage 0 V
   2. E = ½ × 0.0001 × 0 = 0 J
   3. All three outputs are exactly zero, not an error
   4. The page accepts this rather than refusing it

   Zero volts is a real reading — a capacitor that has been discharged, or one sitting in a circuit that is off — so the page returns three zeros rather than an error message. It is worth seeing once, because it establishes that zero is a legitimate answer here and not a sign that something failed to compute. The one value that is refused is a negative voltage, and that is a different kind of rejection: not because the physics forbids it, since a capacitor in an AC circuit is reverse-charged every half cycle, but because a stored energy is a magnitude and a negative entry means the wrong number was typed.

The formula gives the energy stored in an ideal capacitor, and a real one does not return all of it: dielectric absorption, equivalent series resistance and leakage all take a share, so the energy you can actually get back is somewhat less. The calculation says nothing about how fast the energy can be delivered, which is limited by the series resistance and by how much current the rest of the circuit can take. It also says nothing about the voltage rating, which is a hard limit: exceed it and the capacitor fails, often shorted, sometimes violently.

Why is the voltage squared in the energy formula?

Because charging a capacitor is not a matter of pushing a fixed amount of charge at a fixed voltage. The first bit of charge goes in at nearly zero volts, and each later bit goes in at a higher voltage because the charge already there has raised the potential. The work done is the average voltage times the charge, and the average of a straight line from 0 to V is V/2 — which is where the half in ½CV² comes from, and where the square comes from once you substitute Q = CV. The practical consequence is the one to remember: doubling the volts quadruples the joules.

Is it dangerous to touch a charged capacitor?

It can be, and it depends entirely on the energy rather than on the voltage or the capacitance alone. The figure this page returns is the energy that will come out of the capacitor if something completes the circuit, and 80 J — which is a 1000 µF capacitor at 400 V, an ordinary part in a power supply — is enough to cause a serious injury. Energy stored in a capacitor does not disappear when the supply is switched off, and in most circuits nothing discharges it, so the rule is to fit a bleeder resistor or to measure the voltage with the equipment isolated before touching anything.

What is a millijoule and why is there a second row?

A millijoule is a thousandth of a joule, and the second row exists because the range of real capacitors spans about six orders of magnitude in energy. A 1 µF capacitor at 10 V holds 0.00005 J, which rounds to 0.000 in the joule row — the number is correct but the unit is too coarse to show it — while a 1000 µF at 400 V holds 80 J, where the millijoule row reads 80000 and is equally awkward. Rather than print a single row with an unusable number of decimals, the page gives the same quantity twice at two scales, and you read whichever row has a sensible number in it.

Does the capacitance or the voltage matter more?

The voltage, by a wide margin, because it is squared and the capacitance is not. Going from 16 V to 400 V multiplies the energy by 625 at the same capacitance, while going from 100 µF to 1000 µF multiplies it by ten. In the examples on this page a 4700 µF capacitor at 16 V holds 0.602 J and a 1000 µF at 400 V holds 80 J: the second one has less than a quarter of the capacitance and more than a hundred times the energy. It is also why high-voltage capacitors are the dangerous ones, and why the voltage rating is the first specification to check.

How long can a capacitor power my circuit?

That is a different calculation and this page does not do it, but the charge row is the bridge. A coulomb per second is one amp, so the 0.4 C held by a 1000 µF capacitor at 400 V is 0.4 amp-seconds of charge: at a steady 0.1 A it is about four seconds if the voltage were allowed to fall all the way to zero. In practice the answer is much shorter, because the circuit stops working when the voltage drops below its minimum and the discharge follows an exponential curve rather than a straight line. Hold-up time calculations need the load's minimum voltage as well.

Can the energy be negative?

No, and the page will not accept a negative voltage. Energy storage depends on the square of the voltage, so a capacitor charged to −300 V holds exactly the same 4.5 J as one charged to +300 V, and a negative answer would be meaningless. The guard is there because a capacitor in an AC circuit really is reverse-charged every half cycle, so a reader who has measured −300 V has measured something real — but the number this page wants is the magnitude, 300, and entering the signed value is a data-entry slip rather than a physical case.

• OpenStax College Physics 2e, §19.7 Energy Stored in Capacitors — the derivation of E = ½CV² from the work done charging a capacitor, and the millijoule-scale examples used here — OpenStax
• Capacitor — the farad defined through charge and voltage, the practical submultiples, and the energy-storage and safety sections — Wikipedia
• IEC 60384-14 and the capacitor safety literature on stored charge — why a capacitor retains its energy after the supply is removed, and the bleeder-resistor practice for high-voltage equipment — Wikipedia