Example 1
A lab resistor: 2 A flows through 10 Ω. The square of the current is 4 A², so the loss is 40 W and the voltage drop is 20 V.
P = 40 W
Joule Power Loss Calculator
Enter current I in amperes and resistance R in ohms. The calculator finds the loss from the Joule–Lenz law, P = I²·R, and reports it in watts. For 2 A and 10 Ω the power is 40 W, and at 4 A with the same resistance it is 160 W. The model assumes fixed R: it does not compute resistance from ρ·l/S or a temperature rise.
Result
Enter current and resistance. Loss power, the square of the current, and the heat over 60 s will appear here.
For the heat released over a chosen time, use the Q = I²R·t calculator; for power from voltage and current, use the P = U·I calculator.
How it works
I2 A
I² · R4 · 10 Ω
P40 W
Joule heating loss, the heat a current releases in a resistance per unit time, is the power this calculator computes: P = I²R. Textbooks call the relation Joule’s law, and in many European courses the Joule–Lenz law. The SI unit of power is the watt: 1 W = 1 J/s.
The calculator uses the two numbers you enter: current I in amperes and resistance R in ohms. The result is the product of the squared current and the resistance. For 2 A and 10 Ω the square of the current is 4 A², and the product with the resistance is 40 W. The voltage drop on an ohmic resistor is then U = IR = 20 V, and 20 V × 2 A is the same 40 W.
What the result panel shows. Besides the power in watts, and in kilowatts when the value is large, the calculator reports the square of the current, the voltage drop U = IR, the heat over 1 s (numerically equal to the power, because 1 W = 1 J/s), and the heat over 60 s, Q = P·60. For 40 W that is 2,400 J. Time is not a form field; 1 s and 60 s only illustrate the product Q = P·t. The chart draws the same parabola: power at the resistance you entered, against current.
Limits of the model. Resistance is constant, and the current is the value you enter: steady or RMS. The calculator does not compute wire resistance from R = ρl/S, does not read an appliance nameplate, and does not turn power into a temperature rise. A temperature needs mass and specific heat, or a thermal resistance, and this calculator does not use those inputs. A negative resistance is rejected. Zero resistance gives zero power.
Neighboring pages compute other quantities. The heat Q = I²Rt is energy over a time you enter there. The power P = UI starts from voltage and current. On a purely resistive element the two powers agree, because U = IR. The unit switch in the header does not change amperes, ohms, or watts.
The reference example on this page: 2 A and 10 Ω → 40 W, 20 V, and 2,400 J over 60 s. The button “Fill this example” puts the example’s numbers into the fields and refreshes the panel and the chart. The fields start empty.
Formula
P = I² · R
U = I · R on an ohmic resistance; Q = P · t = I² · R · t
[P] = W = J/s. This calculator finds P. The voltage U and the heat over 1 s and over 60 s are consequences of that same product. Time t is not a form field.
Model: Joule–Lenz law (Joule, 1841; Lenz, 1842). Order of computation: P = I · I · R, rejecting a negative R.
Theory, derivation, and domain
Definition of Joule heating loss and the watt
Joule heating loss is the heat released in a conductor per unit time while current flows through it. For an ohmic resistance the definition reduces to the product P = I²R. Power is a scalar: the direction of the current does not change it, because the current is squared.
The SI unit of power is the watt, 1 W = 1 J/s = 1 kg·m²·s⁻³. The name honors James Watt. Engineering also uses the kilowatt (1,000 W) and the milliwatt (0.001 W). The ampere, the ohm, and the watt are tied so that 1 W = 1 V·A = 1 A²·Ω.
History: Joule, Lenz, and the heat of a current
James Prescott Joule put this heat on a quantitative footing in the early 1840s. In 1841, he published, in the Philosophical Magazine, a paper on the heat evolved in metallic conductors of electricity and in the cells of a battery during electrolysis. He immersed a wire in a known mass of water, passed a current for a set time, and measured the temperature rise. The heat rose with the square of the current and with the resistance of the wire.
Heinrich Friedrich Emil Lenz reached a similar relation independently in 1842. Many European textbooks therefore call it the Joule–Lenz law. This calculator computes the power from that law. It does not rebuild Joule’s calorimeter, and it does not say how many degrees the water would warm.
Deriving P = I²R from Ohm’s law
Step 1. For an ohmic resistance, Ohm’s law says the voltage across the element is the product of current and resistance: U = IR. Step 2. The electric power delivered to the element is P = UI. Step 3. Substituting the voltage gives P = (IR) · I = I²R. For 2 A and 10 Ω the voltage is 20 V, and the power is 20 V × 2 A = 40 W, the same as 4 A² × 10 Ω.
Step 4. When the current and the resistance stay constant, the energy over a time t is the product Q = P·t = I²R·t. This calculator has no time field. The panel shows Q for 1 s and for 60 s as two illustrations of that same product. Any other time is covered by the separate Joule-heat calculator.
Assumptions: ohmic resistance, steady current, no reactance
The formula assumes a resistance that stays constant during the calculation. The current is the value you enter. For direct current that is the steady current. For a sine wave the average heating power is the square of the RMS current times the resistance, so the field should receive the RMS value, not the amplitude. The calculator does not divide the entered number by √2.
The model does not include the reactance of an inductor or a capacitor. Reactive power moves between the source and the element and does not become heat in this formula. A negative resistance, such as a tunnel diode, is outside the model: R must be zero or positive.
The square of the current, and where the thermal model stops
Because the current is squared, twice the current at the same resistance gives four times the power. 2 A on 10 Ω is 40 W, and 4 A on 10 Ω is 160 W. Half the current leaves one quarter of the loss. That is why conductor cross-sections and fuses are chosen for current, not for voltage alone.
The result is a power, not a temperature. A temperature rise also needs mass and specific heat, Q = mcΔT, or a thermal resistance to the surroundings. This calculator does not use those quantities. Tungsten and nichrome increase resistance as they heat. At the fixed current you enter here, a higher hot resistance increases the power. A resistance measured cold will not describe hot operation.
Units, heat over time, and the neighboring formulas
Display conversions: 1 kW = 1,000 W, 1 mW = 0.001 W, 1 W · 1 s = 1 J, and 1 W · 60 s = 60 J. For 40 W one minute gives 2,400 J, or 2.4 kJ. The ampere, the ohm, and the watt stay in SI regardless of the unit switch in the header.
Neighboring formulas compute something else. P = UI starts from voltage and current. Q = I²Rt starts from current, resistance, and time and returns joules. R = ρl/S computes the resistance of a wire from the material, the length, and the cross-section. On this page R is already an input, and the result is power.
Sources
J. P. Joule, On the Heat evolved by Metallic Conductors of Electricity, and in the Cells of a Battery during Electrolysis, Philosophical Magazine, 1841. H. F. E. Lenz, independent measurements in 1842, which is why textbooks speak of the Joule–Lenz law. D. Halliday, R. Resnick, J. Walker, Fundamentals of Physics, the chapter on current and resistance. H. D. Young, R. A. Freedman, University Physics, the chapter on current, resistance, and electromotive force. BIPM, The International System of Units (SI), 9th edition (watt, ohm, ampere). NIST Special Publication 811. This page is for educational purposes.
How to use
- Enter current I in amperes, for example 2. The current enters the formula as a square.
- Enter resistance R in ohms, for example 10. Resistance cannot be negative.
- Click Calculate or wait for the live result. For 2 A and 10 Ω the calculator shows 40 W.
- Check the panel: the square of the current, the voltage drop U = IR, and the heat over 1 s and over 60 s.
- Clear removes the inputs and hides the panel. The unit switch in the header does not change amperes or ohms.
Symbol glossary
P is the loss power, I is the current, and R is the resistance. For 2 A and 10 Ω the result is 40 W.
- P
- Loss power [W]. I²R. For 2 A and 10 Ω this is 40 W.
- I
- Current [A]. It enters the formula as a square.
- R
- Resistance [Ω]. Zero is allowed; a negative number is not.
- I²
- Square of the current [A²]. At 2 A it is 4.
- U
- Voltage drop on an ohmic resistance, U = IR [V]. At 2 A and 10 Ω it is 20 V.
- Q
- Heat Q = P·t [J]. The panel shows t = 1 s and t = 60 s. Another time is covered by the Joule-heat calculator.
- W
- Watt, 1 J/s. 1,000 W = 1 kW, and 0.001 W = 1 mW.
Worked examples
Example 2
A kettle-heater model near 230 V: 8.7 A and 26.45 Ω give 2002 W. That is the scale of a typical kettle, not a measurement of one appliance.
P = 2002 W
Example 3
A USB lead at a 2 A charging current and a conductor resistance of 0.2 Ω loses 0.8 W as heat in the cable. The voltage drop on that resistance is 0.4 V.
P = 0.8 W
Example 4
A car bulb treated as ohmic: 4.58 A through a hot filament of 2.62 Ω gives 54.96 W. The voltage across the filament is close to 12 V.
P = 54.96 W
Example 5
A series resistor for an LED: 0.02 A and 150 Ω give 0.06 W. The voltage drop on the resistor is 3 V.
P = 0.06 W
Example 6
A thicker conductor: 5 A on 0.5 Ω gives 12.5 W. At the same current a thinner conductor, with a higher resistance, releases more heat.
P = 12.5 W
Example 7
A resistive heater: 6 A and 3 Ω release 108 W. The voltage drop is 18 V.
P = 108 W
Example 8
A thin heating wire: 0.5 A and 100 Ω give 25 W. The voltage on the wire is 50 V.
P = 25 W
Example 9
The same 10 Ω as the lab example, but 4 A instead of 2 A. The power is 160 W, four times 40 W, because the current was squared.
P = 160 W
Example 10
An ideal short of 0 Ω: even with a current of 4 A, the Joule loss is 0 W. A real contact has a small but nonzero resistance, so it still releases a little heat.
P = 0 W
Example 11
A 1 kΩ resistor at 0.1 A releases 10 W. An element at that power needs a heat sink or a large body.
P = 10 W
Example 12
One ampere on 50 Ω gives 50 W. The voltage drop is 50 V, and the heat over one minute is 3,000 J.
P = 50 W
Common questions
What is Joule heating loss, and what does this calculator compute?
Joule heating loss is the heat released in a resistance per unit time: P = I²R. The calculator squares the current you enter, multiplies by the resistance you enter, and reports the result in watts. For 2 A and 10 Ω the result is 40 W. The panel also shows the square of the current, the voltage drop U = IR, and the heat over 1 s and over 60 s.
Which units should I enter for current and resistance?
Current in amperes [A] and resistance in ohms [Ω]. The result is in watts [W], which are joules per second. A comma and a period are the same decimal separator. The unit switch in the header does not convert amperes, ohms, or watts, because this formula stays in SI.
Why does the power rise with the square of the current?
Because current stands to the second power in P = I²R. Doubling the current at a fixed resistance quadruples the power: 2 A on 10 Ω is 40 W, and 4 A on 10 Ω is 160 W. Half the current leaves one quarter of the loss.
How is this result different from the heat Q = I²Rt?
This calculator computes power, in watts. Joule heat is energy over a chosen time, in joules: Q = P·t. Time is not a field here. The panel illustrates only two fixed intervals, 1 s and 60 s. For 40 W that is 40 J and 2,400 J. Any other time is covered by the separate Joule-heat calculator.
How is P = I²R different from P = UI?
Here the power comes from current and resistance. On the P = UI page the power comes from voltage and current. For a purely resistive element the two numbers agree, because Ohm’s law gives U = IR, so UI = I²R. For 2 A and 10 Ω the voltage is 20 V, and both paths give 40 W.
What happens with a negative current, zero resistance, or empty fields?
The square of the current is never negative, so −2 A on 10 Ω gives the same 40 W as +2 A. The sign remains only in the voltage drop U = IR. A resistance of 0 Ω gives 0 W, because zero resistance releases no Joule heat. A negative resistance is rejected. When both fields are empty, the prompt stays without an error message. One number without the other does not produce a result.
Is the result the nameplate power of a whole appliance?
No. The result is the heating power in the resistance you enter, at the current you enter. A kettle or bulb nameplate covers the whole circuit and the hot resistance. This calculator does not read a nameplate.
Will the calculator find wire resistance from length and area?
No. Resistance R is an input here. The relation R = ρl/S, from resistivity, length, and cross-section, is covered by the separate wire-resistance calculator. That result can then be entered in the resistance field.
What does the formula assume, and what does it leave out?
Resistance is constant, and the current is a steady value or an RMS value. The calculator does not convert a sine-wave amplitude into RMS, does not compute reactive power in an inductor or a capacitor, and does not compute a temperature rise. A temperature needs mass and specific heat, or a thermal resistance, and this calculator does not use those inputs.
What does the chart under the result show?
The chart draws loss power in watts against current, at the resistance just entered. The curve is a parabola, because power rises with the square of the current. The current axis runs from zero to twice the absolute value of the current you entered. At zero current the sample scale runs to 4 A, so the parabola is still visible.
For alternating current, may I enter the RMS value?
Yes. The average heating power of a sine wave is the square of the RMS current times the resistance, so the current field should receive the RMS value, not the amplitude. The amplitude is √2 times larger, and using it instead of RMS doubles the computed power. The calculator does not make that conversion itself.
Why can a hot wire heat differently from its cold resistance?
This model treats resistance as constant. Tungsten and nichrome increase resistance as they warm up. At the fixed current you enter here, a higher resistance increases the power. If you measure the cold resistance while current is already flowing in the hot wire, the result will not describe the real power.
How do I calculate Joule power loss with P = I²·R?
Square the current and multiply by the resistance. For 2 A and 10 Ω: P = 2²·10 = 40 W. The result is in watts.
What is P at 2 A and 10 Ω?
40 W. The voltage drop U = I·R is then 20 V, and the heat over 60 s is 2400 J.
What is P at 4 A and 10 Ω?
160 W. Doubling the current at fixed R multiplies the power by four, because I enters the formula squared.
Is the result the temperature of the wire?
No. P is power in watts. A temperature rise needs mass, specific heat, or a thermal resistance — those fields are not on this page.
Does the calculator compute 2·I²R or 3·I²R cable losses?
No. The result is for one resistance R at a known I. For a single-phase or three-phase line you must sum the loaded conductors separately.
Sources
The formula, the unit, and the history below come from textbooks and from metrology documents.
- Wikipedia — Joule heating
- Polish Wikipedia — Joule’s law
- BIPM — The International System of Units (SI), 9th edition
- NIST SP 811 — Guide for the Use of the International System of Units
- NIST SP 330 — The International System of Units
- J. P. Joule, Philosophical Magazine, 1841 — heat in metallic conductors.
- D. Halliday, R. Resnick, J. Walker, Fundamentals of Physics — current and resistance.
- H. D. Young, R. A. Freedman, University Physics — current, resistance, and electromotive force.
Page updated November 3, 2026.