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PHYSICS

Joules to Volts Calculator — energy, charge and voltage from V = J / C

Convert between energy, electric charge and voltage using the definition of the volt, solving for whichever of the three you do not already have, in any common unit.

The relation is a definition rather than an approximation: one volt is exactly one joule per coulomb. Any two of the three quantities determine the third.
An amp-hour is 3,600 coulombs, because a coulomb is an amp-second. That single conversion is what lets a battery capacity in milliamp-hours turn into an energy figure once you supply a voltage.
The box for the quantity you are solving for is ignored on input and filled in with the answer, so you can leave whatever is in it.
Voltage
 
 
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Energy in joules
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Energy in watt-hours
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Charge in coulombs
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Charge in milliamp-hours
Tip: a battery rated in milliamp-hours is telling you about charge, not energy. Two cells with the same milliamp-hour figure but different voltages store different amounts of energy, and this relation is exactly what separates them.
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The joules to volts calculator above applies the definition of the volt. One volt is one joule per coulomb: the energy given to, or taken from, each unit of charge as it moves through a potential difference. Rearranged, energy is voltage times charge, and charge is energy divided by voltage. The tool solves for whichever of the three you are missing and shows the other two in several common units.

Arb Digital publishes free engineering calculators that make the boundaries between similar tools explicit. This page relates three quantities to each other. It is deliberately different from a unit converter, which changes one quantity into different units of the same quantity, and different from the single-particle case, where the numbers are so small that a dedicated unit exists for them.

What This Joules to Volts Calculator Does

Voltage is not a property of a place; it is a ratio. It describes how much energy each coulomb of charge picks up or gives up between two points. That is why a nine-volt battery and a nine-volt mains adapter produce identical voltages while storing wildly different amounts of energy: they differ in how much charge they can deliver, not in the energy per unit of charge.

Working out any one of the three from the other two is a common practical need. Battery capacity is published as charge, in milliamp-hours, while what you usually want is energy, in watt-hours. Capacitor energy problems run the other way. Electroplating and electrochemistry work in coulombs or faradays and need an energy figure for the power supply.

The hero figure is whichever quantity you asked for. The grid always shows the full picture: energy in joules and watt-hours, charge in coulombs and milliamp-hours, so the conversions are visible rather than hidden.

How to Use It

  1. Choose what you are solving for. The box for that quantity is ignored on input and filled with the answer, so its current contents do not matter.
  2. Enter the two you know, with their units. Each quantity has its own unit selector, so you can mix watt-hours with milliamp-hours without converting first.
  3. Read the grid for the full picture. It restates both energy and charge in two units each, which is usually where a battery or capacitor question actually gets answered.
  4. Watch for amp-hours against coulombs. The factor is 3,600, and forgetting it is the most common error in this calculation by a wide margin.
  5. Treat the voltage as an average. A real battery's voltage falls as it discharges, so a single figure gives an approximation rather than an exact energy.

The Formula: How Energy, Charge and Voltage Relate

The defining relation is V = E ÷ Q, with energy in joules, charge in coulombs and the result in volts. Rearranged, E = V × Q and Q = E ÷ V. There is no approximation and no material property involved; this is how the volt is defined in the International System of Units.

The unit conversions matter more than the arithmetic. A coulomb is an ampere-second, so an amp-hour is 3,600 C and a milliamp-hour is 3.6 C. A watt-hour is 3,600 J for the same reason. The elementary charge is 1.602176634 × 10−19 C exactly, a fixed value in the revised SI, published by NIST as the CODATA elementary charge. A faraday, the charge on a mole of electrons, is 96,485.33 C.

Work the defaults through by hand. With 7,500 J delivered by 500 C of charge, the voltage is 7,500 ÷ 500 = 15 V. Reversing it, 500 C at 15 V stores 7,500 J, which is 7,500 ÷ 3,600 = 2.083 Wh, and 500 C is 500 ÷ 3.6 = 138.9 mAh.

A more familiar example: a 2,000 mAh phone cell at a nominal 3.7 V. The charge is 2,000 × 3.6 = 7,200 C, so the energy is 7,200 × 3.7 = 26,640 J. Dividing by 3,600 gives 7.4 Wh, which matches the shortcut of multiplying 2 Ah by 3.7 V directly. The OpenStax University Physics section on Ohm's law sets out the circuit relationships these quantities sit inside.

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Why Milliamp-Hours Are a Misleading Way to Compare Batteries

Battery capacity is almost always advertised in milliamp-hours, and milliamp-hours measure charge. Charge alone tells you nothing about stored energy until you also know the voltage, and that is exactly why the figure misleads.

A 10,000 mAh power bank built around lithium cells at 3.7 V stores 37 Wh. The same 10,000 mAh rating on a pack presented at 5 V output would be 50 Wh. Manufacturers frequently quote the cell-level milliamp-hours while the device delivers at a different voltage, so a straight comparison of the headline numbers compares quantities that are not the same thing.

This is also why airline restrictions and shipping rules are written in watt-hours rather than milliamp-hours. Watt-hours are energy, and energy is what determines how much damage a failure can do. Converting a capacity to watt-hours before comparing anything is the single most useful habit here.

The Same Relation in Capacitors and Electrochemistry

A capacitor stores charge Q = CV, but its stored energy is E = ½CV², not QV. The factor of a half catches people out constantly, and it has a real cause: as a capacitor charges, its voltage rises from zero to the final value, so the average voltage each coulomb experienced was half the final one. Multiplying total charge by final voltage overstates the energy by a factor of two. That correction does not apply to a battery, whose voltage stays roughly constant over most of its discharge, which is why the simple product works there.

Electrochemistry uses the same relation from the other direction. Faraday's laws relate the charge passed through a cell to the mass of material deposited or dissolved, and one faraday, 96,485 C, corresponds to one mole of electrons. Multiplying that charge by the cell voltage gives the electrical energy consumed, which is how the energy cost of electroplating or aluminium smelting is calculated.

Where This Sits Next to the Other Electrical Tools

Four pages here look similar and answer different questions. This one relates energy, charge and voltage to each other and solves for any of the three. The energy converter converts one energy figure into different energy units — joules to calories to kilowatt-hours — without any charge or voltage involved. The electric charge converter does the same for charge alone, moving between coulombs, milliamp-hours and elementary charges. Neither of those relates the two quantities.

The electron volt calculator is the closest neighbour and the most important boundary to state. It handles the single-particle case: an accelerating voltage applied to a charged particle, giving the energy per particle in electron-volts and joules, with a charge state and a particle count. It is the tool for beam physics, spectroscopy and semiconductor band energies. This page is the bulk case, where charge is measured in coulombs rather than elementary charges and energy in joules rather than electron-volts. The relation is identical; the scale and the audience are not.

For the circuit that surrounds all of this, the Ohm's law calculator covers voltage, current and resistance, the capacitor energy calculator handles the half-CV-squared case directly, the Joule heating calculator covers energy dissipated as heat, and the voltage drop calculator deals with losses along a conductor.

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Common Mistakes to Avoid

  • Treating milliamp-hours as energy — they measure charge, and two packs with the same rating at different voltages store different amounts of energy.
  • Forgetting the factor of 3,600 — an amp-hour is 3,600 coulombs and a watt-hour is 3,600 joules, and dropping it produces an answer wrong by three orders of magnitude.
  • Using Q times V for a capacitor — capacitor energy is half of that, because the voltage rose from zero as it charged rather than staying constant.
  • Assuming a battery holds one voltage — terminal voltage falls through the discharge, so the nominal figure gives an approximate energy rather than an exact one.
  • Mixing elementary charges with coulombs — the two differ by about 6.24 times ten to the eighteenth, and the single-particle scale belongs on the electron volt page.

Related Free Tools From Arb Digital

Use the energy converter when you only need to change energy units, and the electric charge converter for charge units alone. For the single-particle scale, the electron volt calculator handles accelerating voltages and energies in electron-volts. The capacitor energy calculator covers stored energy in a capacitor with the correct factor of a half, the Ohm's law calculator and voltage divider calculator cover the surrounding circuit, and the Joule heating calculator and voltage drop calculator deal with energy lost along the way. Everything Arb Digital publishes is on the free online tools hub.

Frequently Asked Questions

How do you convert joules to volts?

Divide the energy in joules by the charge in coulombs. One volt is defined as one joule per coulomb, so the conversion needs a charge figure as well as an energy figure. Joules alone cannot be converted to volts any more than metres alone can be converted to metres per second.

How many coulombs are in a milliamp-hour?

Exactly 3.6. A coulomb is an ampere-second, so an amp-hour is 3,600 coulombs and a milliamp-hour is a thousandth of that. This single conversion is what turns a battery capacity rating into a charge figure that can be multiplied by a voltage to give energy.

How do I turn a battery's mAh rating into watt-hours?

Multiply the capacity in amp-hours by the nominal voltage. A 2,000 milliamp-hour cell at 3.7 volts is 2 amp-hours times 3.7 volts, which is 7.4 watt-hours or 26,640 joules. Watt-hours are the fair way to compare batteries, because milliamp-hours measure charge and say nothing about voltage.

Why is capacitor energy half of charge times voltage?

Because the voltage rises from zero to its final value as the capacitor charges, so the average voltage experienced by the charge that flowed in was half the final voltage. Integrating over the charging process gives one half C V squared. A battery holds a roughly constant voltage throughout, so the simple product applies there instead.

What is the difference between this and the electron volt calculator?

Scale and audience. This page works in coulombs and joules and answers bulk questions about batteries, capacitors and electrochemistry. The electron volt calculator works with individual charged particles, taking an accelerating voltage and a charge state and returning energy per particle in electron-volts. The underlying relation is the same one.

Does the voltage stay constant while a battery discharges?

No. Terminal voltage falls through the discharge, steeply at first, then across a long plateau, then sharply at the end. Using the nominal voltage gives a good approximation of total energy, which is what manufacturers do when they print a watt-hour figure, but the instantaneous voltage varies throughout.

What is a faraday and when would I use it?

It is the charge carried by one mole of electrons, 96,485 coulombs. It appears in electrochemistry, where Faraday's laws relate the charge passed through a cell to the mass of substance deposited or dissolved. Multiplying that charge by the cell voltage gives the electrical energy consumed, which is how the energy cost of electroplating or smelting is worked out.

This tool is provided for educational and general engineering use. It applies the SI definition of the volt with ideal unit conversions and assumes a single constant voltage; real batteries vary in terminal voltage through discharge, and real systems lose energy to internal resistance and conversion inefficiency. Use manufacturer data for any design decision.

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