Joulesto Volts Calculator
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Joulesto Volts Calculator
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Formula: Voltage (V) = Energy (J) / Charge (C)
Worked example — Voltage = 12 V (12,000 mV)
Formula
Voltage (V) = Energy (J) / Charge (C)
Voltage in volts equals energy in joules divided by electric charge in coulombs. This derives from the definition of the volt: one volt is the potential difference that transfers one joule of energy per coulomb of charge passing through it. Additional relationships include Power = Energy / Time and Current = Voltage / Resistance.
Worked Examples
Example 1: Battery Voltage Calculation
Problem:A battery transfers 240 joules of energy while delivering 20 coulombs of charge. What is the voltage?
Solution:Voltage = Joules / Coulombs V = 240 / 20 = 12 V Millivolts = 12,000 mV Watt-hours = 240 / 3600 = 0.0667 Wh
Result:Voltage = 12 V (12,000 mV)
Example 2: Capacitor Energy to Voltage
Problem:A capacitor stores 50 joules with a charge of 0.5 coulombs. Find the voltage across it.
Solution:Voltage = Joules / Coulombs V = 50 / 0.5 = 100 V kV = 0.1 kV Calories = 50 / 4.184 = 11.95 cal
Result:Voltage = 100 V (0.1 kV)
Frequently Asked Questions
How do you convert joules to volts?
Joules cannot be directly converted to volts because they measure different quantities: joules measure energy and volts measure electric potential. However, if you know the electric charge in coulombs, you can calculate voltage using the formula V = J / C, where V is volts, J is joules, and C is coulombs. This relationship comes from the definition of a volt: one volt is the potential difference that imparts one joule of energy per coulomb of charge. You always need at least two quantities to find the third.
What is the relationship between joules, volts, and coulombs?
These three quantities are linked by the fundamental equation E = V times Q, where E is energy in joules, V is voltage in volts, and Q is charge in coulombs. One joule equals one volt times one coulomb. A 12-volt car battery delivering 100 coulombs of charge transfers 1200 joules of energy. This relationship is essential in circuit design, battery sizing, and understanding how electrical energy is stored and delivered in capacitors and other components.
How are joules related to watts and electrical power?
A watt is defined as one joule per second. Power (watts) equals energy (joules) divided by time (seconds). In electrical circuits, power also equals voltage times current (P = V x I) and voltage squared divided by resistance (P = V^2 / R). If a 100-watt light bulb runs for 60 seconds, it consumes 6000 joules of energy. Understanding these relationships is fundamental for sizing power supplies, calculating energy costs, and designing electrical systems.
What are typical joule values in everyday electrical devices?
A single AA battery stores about 9360 joules (1.5V x 2600mAh x 3600s/h divided by 1000). A fully charged smartphone battery holds roughly 40,000-50,000 joules. A car battery stores about 2.6 million joules. A lightning bolt delivers approximately 1 billion joules. A defibrillator shock uses 200-360 joules. One kilowatt-hour, the standard unit on electricity bills, equals 3.6 million joules. These numbers illustrate the vast range of energy scales in electrical systems.
How do you find the voltage across a capacitor from its stored energy?
A capacitor stores energy as E = 0.5 x C x V squared, so the voltage is V = square root of (2E / C). With 2 joules stored in a 0.01 farad capacitor, V = sqrt(4 / 0.01) = sqrt(400) = 20 volts. Note this differs from the simple V = J / C relationship used for a moving charge, because the capacitor voltage rises as charge accumulates and the stored energy is the integral of that rising voltage.
What is an electronvolt and how does it relate to the joule?
An electronvolt is the kinetic energy an electron gains when accelerated through a potential difference of one volt, so it is simply the elementary charge multiplied by one volt. Since the elementary charge is 1.602176634 x 10^-19 coulombs, one electronvolt equals 1.602176634 x 10^-19 joules exactly. Physicists use it in atomic, semiconductor and particle work because joules are inconveniently huge at that scale; the silicon band gap is about 1.1 eV.
How many coulombs does an ampere-hour of battery capacity represent?
An ampere-hour is one ampere flowing for 3600 seconds, and one ampere is one coulomb per second, so 1 Ah equals 3600 coulombs. This is what lets you turn a battery rating into energy: a 12 V, 5 Ah pack carries 5 x 3600 = 18,000 coulombs, and multiplying by the 12 V terminal voltage gives 216,000 joules, or 60 watt-hours. Real usable energy is lower because terminal voltage sags as the cell discharges.
What is the difference between electromotive force and potential difference?
Electromotive force (EMF) is the energy per coulomb that a source such as a battery or generator supplies to the circuit, while potential difference is the energy per coulomb delivered to a particular component. Both are measured in volts. They differ because a real source has internal resistance, so terminal voltage equals EMF minus the current multiplied by that internal resistance. This is why a battery reads lower under heavy load than at open circuit.
How do you calculate current from the energy delivered over a period of time?
Divide the energy in joules by the elapsed time in seconds to get power in watts, then divide that power by the voltage to get current in amperes. If a circuit transfers 600 J in 10 seconds at 12 volts, power is 60 W and current is 60 / 12 = 5 A. This chain of relationships, P = E / t and I = P / V, is how energy figures are turned into the conductor sizing and fuse ratings an installation actually needs.
What separates charge in coulombs from current in amperes?
Charge is a quantity and current is a rate. One ampere is one coulomb passing a point each second, so charge equals current multiplied by time. Mixing them is the most frequent error on this page: entering 2 amperes where the calculator expects 2 coulombs treats a one-second flow as if it were the whole transfer. If you only know current, multiply it by the duration in seconds first, then divide the joules by that result.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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