Joule Heating Calculator
Calculate the heat a steady current releases in a resistance over time, E = I²Rt, in joules from current in amperes, resistance in ohms and duration in seconds.
Joule Heating Calculator
Background.
Every conductor carrying a current turns some of the electrical energy into heat. In a kettle element or a toaster that is the whole point. In a cable, a circuit-board trace or the internal resistance of a battery it is waste, and sometimes a hazard. James Prescott Joule measured the relationship in the early 1840s, and it is strikingly simple: the heat released is proportional to the resistance, to the time, and to the square of the current.
That square is what makes the law matter in practice. Double the current through the same resistance and the heat does not double, it quadruples. That is why electricity grids move power at high voltage and low current, why a cable that runs warm at 10 A runs far hotter at 20 A, and why a fuse, a deliberately thin strip of metal, melts within moments when a fault drives the current far past its rating.
The quantity worked out here is energy in joules, E = I²Rt, with current I in amperes, resistance R in ohms and time t in seconds. Divide the answer by the time and you have the heating power in watts; divide it by 3,600 and you have watt-hours, the unit printed on most battery packs.
What the number is not is a temperature. How hot something gets depends on its mass, its material and how fast it sheds heat to its surroundings, and none of those enter the equation. Current and resistance are also held fixed for the whole period, although metal conductors become more resistive as they warm. For alternating current, use the RMS value, which is defined as the steady current that would give the same heating.
What is joule heating calculator?
Joule heating, also called resistive or ohmic heating, is the conversion of electrical energy into heat when a current flows through a material that has resistance. The energy released over a period is E = I²Rt, in joules, and the rate at which it is released, the heating power, is I²R in watts. Every component with resistance produces it, from a heater element designed to glow to a long extension lead that is supposed to stay cool.
How to use this calculator.
- Enter the current in amperes. For an AC circuit, use the RMS current, which is the value that ratings and ordinary meter readings quote.
- Enter the resistance in ohms. For a cable, multiply the length by the manufacturer's resistance per metre, and remember a two-wire circuit carries the current out and back, so count both conductors.
- Enter the duration in seconds: 60 for a minute, 3,600 for an hour.
- Read the heat in joules. Divide by the duration for the heating power in watts, or by 3,600 for watt-hours.
The formula.
E = I² × R × t. The current I, in amperes, is squared; R is the resistance in ohms and t the time in seconds. One ampere squared times one ohm is one watt, and one watt sustained for one second is one joule, so the answer comes out in joules with no conversion constant. Ohm's law gives two equivalent forms: with V the voltage across the resistance, E = V × I × t = V² × t ÷ R, and all three agree whenever V = I × R. Current dominates the result. A 10% rise in current raises the heat by 21%, while a 10% rise in resistance or time raises it by only 10%. All three inputs must be positive, and they are treated as constant for the whole period; heat loss, temperature and any drift in resistance are outside the calculation.
A worked example.
A current of 4 A flows through a 12 Ω resistor for 90 seconds. Squaring the current gives 4² = 16. Multiplying by the resistance gives 16 × 12 = 192 W of heating power, and multiplying by the time gives 192 × 90 = 17,280 J, or 17.28 kJ, which is 4.8 Wh. The voltage across the resistor is 4 × 12 = 48 V, and 48 × 4 × 90 gives the same 17,280 J. If every joule went into a litre of water, about 1 kg, it would warm it by roughly 17,280 ÷ 4,186 ≈ 4.1 °C; in practice some would escape to the air first. Raise the current to 8 A and the energy becomes 8² × 12 × 90 = 69,120 J, four times as much, not twice.
Frequently asked questions.
Why does doubling the current quadruple the heat?
How do I turn the joules into a temperature rise?
Why do power lines run at such high voltages?
What current should I enter for an AC circuit?
How does a fuse use Joule heating?
Does a wire's resistance change as it heats up?
References& sources.
- [1]OpenStax, University Physics Volume 2, section 10.5, Electrical Energy and Power.
- [2]NIST Special Publication 811, SI derived units for power and energy.
- [3]BIPM, The International System of Units (SI Brochure), 9th ed., version 3.01, coherent derived units and quantity equations.
How this page was produced
- Published by
- Quanta Calculator
- Overseen by
- Daniel, Quanta Calculator Researcher
- Primary sources
- 3 cited below
- Method
- E = I²Rt
- Published
- Last verified
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