Joule’s law of heating states that the heat produced in a current-carrying conductor is directly proportional to the square of the current, to the resistance of the conductor and to the time for which the current flows. In symbols, H = I²Rt, where H is heat in joules, I is current in amperes, R is resistance in ohms and t is time in seconds. James Prescott Joule established it experimentally in 1841, and it explains every electric heater, iron, geyser and fuse you have ever used.
The three parts of Joule’s law of heating
The law is usually stated as three separate proportionalities, each with the other two quantities held constant. Questions that say “state Joule’s law of heating” expect all three.
- H ∝ I² when R and t are constant. Heat is proportional to the square of the current. Double the current and the heat becomes four times, not twice.
- H ∝ R when I and t are constant. Heat is proportional to the resistance of the conductor.
- H ∝ t when I and R are constant. Heat is proportional to the time the current flows.
Combining the three gives H ∝ I²Rt, and in SI units the constant of proportionality is 1, so:
H = I²Rt joules
The squared term is what makes the law feel counter-intuitive at first. A 10 per cent rise in current raises the heat by 21 per cent, because 1.1² = 1.21. This is why overloaded wiring heats up so quickly.
Derivation of H = I²Rt
The derivation is short and is worth writing out in full in an exam answer.
When a charge Q moves through a potential difference V, the work done by the source is:
W = QV
Current is charge per unit time, so Q = It. Substituting:
W = VIt
In a purely resistive conductor, none of this energy is stored or converted into mechanical work, so all of it appears as heat. Therefore H = W = VIt.
Now apply Ohm’s law, V = IR, and replace V:
H = (IR) × I × t = I²Rt
That is Joule’s law of heating. Note the two assumptions it carries: the conductor obeys Ohm’s law and its resistance stays effectively constant over the period considered, and the whole of the electrical energy is converted into heat rather than into motion or light output in a motor or a lamp.
SI units and related quantities
| Quantity | Symbol | SI unit | Note |
|---|---|---|---|
| Heat produced | H | joule (J) | 1 calorie = 4.18 J, so H in calories = 0.24 I²Rt |
| Current | I | ampere (A) | Squared in the formula |
| Resistance | R | ohm (Ω) | |
| Time | t | second (s) | Convert minutes and hours before substituting |
| Power | P = H/t = I²R | watt (W) | 1 W = 1 J/s |
| Commercial energy | – | kilowatt hour (kWh) | 1 kWh = 1 unit = 3.6 × 10⁶ J |
The three forms of the formula and when to use each
All three forms are algebraically equivalent for one resistor obeying Ohm’s law, but they behave very differently when you have more than one resistor in a circuit. Picking the wrong one is the single most common error on this topic.
| Form | Use it when | Result |
|---|---|---|
| H = I²Rt | The current is the quantity common to the components, that is, a series circuit | Heat is proportional to R. The largest resistance gets hottest |
| H = V²t / R | The voltage is the quantity common to the components, that is, a parallel circuit or a device across the mains | Heat is proportional to 1/R. The smallest resistance gets hottest |
| H = VIt | You know the voltage across the element and the current through it | Always valid for that element; the safest general form |
Why the two extreme cases disagree: in a series circuit the same current I passes through every resistor, so H = I²Rt makes the bigger resistor produce more heat. In a parallel circuit each branch has the same voltage V but different currents, so H = V²t/R makes the smaller resistor draw more current and produce more heat. The numerical example below proves it with figures.
Practical consequence: a heater element of low resistance connected across 230 V draws a big current and gives a lot of heat, while the thick connecting wire, with very small resistance but carrying the same current in series, stays comparatively cool only because its resistance is tiny. The element and the wire are in series, so the element, being the larger resistance, takes almost the entire heat.
Solved numerical examples
Example 1: heat produced in a coil
A heating coil of resistance 20 Ω carries a current of 3 A for 2 minutes. Find the heat produced.
Given: R = 20 Ω, I = 3 A, t = 2 min = 120 s.
H = I²Rt = (3)² × 20 × 120 = 9 × 20 × 120
H = 9 × 2400 = 21,600 J = 21.6 kJ
The power of the coil is P = I²R = 9 × 20 = 180 W, and 180 W × 120 s = 21,600 J, which confirms the answer.
Example 2: an electric iron on the mains
An electric iron draws 5 A from a 220 V supply and is used for 30 minutes. Find the heat produced, its resistance, and the energy used in units.
Given: V = 220 V, I = 5 A, t = 30 min = 1800 s.
Using H = VIt = 220 × 5 × 1800 = 1100 × 1800
H = 1.98 × 10⁶ J = 1980 kJ
Resistance from Ohm’s law: R = V / I = 220 / 5 = 44 Ω.
Cross-check with the squared form: H = I²Rt = 25 × 44 × 1800 = 1100 × 1800 = 1.98 × 10⁶ J. The two forms agree, as they must.
Energy in commercial units: P = VI = 220 × 5 = 1100 W = 1.1 kW, for 0.5 h, so energy = 1.1 × 0.5 = 0.55 kWh. At a tariff of Rs 8 per unit that half hour of ironing costs about Rs 4.40.
Example 3: series against parallel, the same two resistors
Resistors of 4 Ω and 6 Ω are connected across a 12 V supply for 60 s. Which one gets hotter (a) in series, (b) in parallel?
(a) In series. Total resistance = 4 + 6 = 10 Ω. Current I = V / R = 12 / 10 = 1.2 A, the same through both.
- Heat in the 4 Ω resistor: H = I²Rt = (1.2)² × 4 × 60 = 1.44 × 240 = 345.6 J
- Heat in the 6 Ω resistor: H = (1.2)² × 6 × 60 = 1.44 × 360 = 518.4 J
The 6 Ω resistor, the larger one, gets hotter. Total heat = 864 J, which matches I²Rtotalt = 1.44 × 10 × 60 = 864 J.
(b) In parallel. Each resistor now has the full 12 V across it.
- Heat in the 4 Ω resistor: H = V²t / R = (144 × 60) / 4 = 8640 / 4 = 2160 J
- Heat in the 6 Ω resistor: H = (144 × 60) / 6 = 8640 / 6 = 1440 J
Now the 4 Ω resistor, the smaller one, gets hotter. Same two resistors, same supply, opposite answer. The connection decides which quantity is common, and that decides which form of the formula to use.
Applications of Joule heating
- Electric heater and room heater. A nichrome coil of high resistivity and high melting point (about 1400 °C) glows red hot without oxidising away. Nichrome is used rather than copper precisely because its resistance is high.
- Electric iron. The same heating element, clamped to a heavy metal sole plate that spreads the heat, with a bimetallic thermostat cutting the supply at the set temperature.
- Electric geyser and immersion rod. A sheathed element transfers Joule heat to the surrounding water.
- Electric kettle and toaster. Short heating time because the element sits directly in or next to what is being heated.
- Incandescent bulb. A thin tungsten filament, melting point 3422 °C, is heated to roughly 2500 °C so that it emits visible light. Only a small fraction of the input energy leaves as light; the rest is Joule heat, which is why these bulbs have been replaced by LEDs.
- Electric fuse. A deliberate weak link, described in the next section.
- Soldering iron, electric oven, hair dryer and induction-free hotplates, all built on the same I²Rt effect.
- Industrial use: resistance welding, where a large current through the joint melts the metal locally, and electric arc and resistance furnaces.
The electric fuse and why its wire has a low melting point
A fuse is a short piece of wire connected in series with the live wire of a circuit, deliberately made so that it melts and breaks the circuit before the appliance or the wiring is damaged.
Its design follows straight from H = I²Rt:
- Low melting point. The fuse wire is an alloy of tin and lead, which melts at roughly 200 °C, far below copper’s 1085 °C. The fuse must give way while the rest of the circuit is still intact, so it has to be the part that melts first.
- Higher resistance and small cross-section. The fuse wire is thin, so its resistance is higher than that of the household wiring around it. In series the current is the same everywhere, so by H = I²Rt the highest-resistance element produces the most heat, and that element is the fuse.
- Current rating. Because heat rises with the square of current, a modest overload heats the wire sharply. A fuse rated slightly above the normal working current carries it indefinitely but melts within seconds on a short circuit. Common domestic ratings in India are 5 A for lighting circuits and 15 A for heavy-load points serving geysers and air conditioners.
- The fuse is placed in the live wire, not the neutral, so that when it blows the appliance is genuinely disconnected from the supply potential.
A blown fuse should be replaced with one of the same rating. Substituting a thicker wire or a nail restores the circuit but removes the protection completely, because the heat it needs to melt will never be reached before the cable behind the wall overheats.
Unwanted Joule heating and its disadvantages
The same effect that runs a heater wastes energy everywhere else.
- Transmission line losses. Power lost in a line is I²R. To deliver the same power P = VI, raising V lowers I, and because loss depends on I², halving the current cuts the loss to one quarter. This is the entire reason electricity is transmitted at 220 kV, 400 kV or 765 kV and stepped down near the consumer.
- Copper losses in transformers and motors. Current in the windings produces heat that has to be carried away by oil, air or fans. It lowers efficiency and, more seriously, ages the insulation. A rough working rule used in machine design is that every 10 °C of extra winding temperature roughly halves insulation life.
- Cable and switchgear heating, which is why conductors carry an ampacity rating and why loose or corroded connections, having extra contact resistance, are a fire risk.
- Electronic devices. Heat sinks, thermal paste and fans exist only to remove Joule heat from processors and power semiconductors.
- Battery heating due to internal resistance, which wastes charge and shortens battery life.
Joule heating is not the Joule-Thomson effect
Two different things share the name. Joule’s law of heating, H = I²Rt, is about electrical energy turning into heat in a resistor. The Joule-Thomson coefficient belongs to thermodynamics and describes the temperature change of a real gas when it is throttled through a valve at constant enthalpy, which is the principle behind gas liquefaction and refrigeration. They are unrelated apart from being named after the same physicist.
One more distinction worth keeping straight: Joule heating is irreversible and always produces heat whichever way the current flows. The Peltier effect at a junction of two different conductors is reversible and can heat or cool depending on current direction.
FAQs
State Joule’s law of heating.
The heat produced in a conductor is directly proportional to the square of the current through it, to the resistance of the conductor, and to the time for which the current flows. Combining the three gives H = I²Rt, with H in joules, I in amperes, R in ohms and t in seconds.
What is the formula for Joule’s law of heating?
H = I²Rt. Using Ohm’s law it can also be written as H = VIt or H = V²t/R. Use H = I²Rt when the current is common to the elements, as in a series circuit, and H = V²t/R when the voltage is common, as in a parallel circuit.
Why is heat proportional to the square of the current?
Because two current-dependent quantities multiply together. The work done is W = VIt, and by Ohm’s law V itself equals IR, so substituting gives one factor of I from the voltage and one from the current, producing I²Rt. In practice it means a 10 per cent overload raises the heating by about 21 per cent.
Which wire gets hotter in a series circuit, the thick one or the thin one?
The thin one. A thinner wire of the same material and length has a larger resistance, and in series every part carries the same current, so by H = I²Rt the larger resistance produces more heat. This is exactly how a fuse wire is designed to fail first.
What are the applications of Joule’s law of heating?
Electric heaters, irons, geysers, kettles, toasters, soldering irons, incandescent bulbs and electric fuses all depend on it, as does resistance welding in industry. The same effect appears as unwanted loss in transmission lines, transformer and motor windings, and electronic components.