Free Power Dissipation Calculator (P = V·I)
Calculate electrical power dissipation from any two of voltage, current, and resistance. This tool applies the three forms of Joule's law — P = V·I, P = I²·R, and P = V²/R — to return power in watts and solve for the quantity you didn't enter.
Enter any two of voltage, current, and resistance — the power and the missing quantity are calculated.
P = V × I = I²R = V²/R. For resistive DC loads; use amperes (A), not milliamps. One watt = one joule per second.
Quick answer
Electrical power dissipation is found with Joule's law: P = V·I, which is equivalent to P = I²·R and P = V²/R, where P is power in watts (W), V is voltage in volts (V), I is current in amperes (A), and R is resistance in ohms (Ω). For example, 12 V across a component carrying 2 A dissipates P = 12 × 2 = 24 W, and its resistance is R = V/I = 6 Ω. Enter any two of voltage, current, and resistance and the calculator returns the power and the third quantity.
Formula & method
Power dissipation (three equivalent forms)
P = V · I = I² · R = V² / R
- P — power dissipated, in watts (W)
- V — voltage across the component, in volts (V)
- I — current through the component, in amperes (A)
- R — resistance, in ohms (Ω)
All three forms give the same answer for a resistive (Ohmic) load because V = I·R. Use P = V·I when you know voltage and current, P = I²·R when you know current and resistance, and P = V²/R when you know voltage and resistance. One watt equals one joule per second (1 W = 1 J/s).
Solving for the missing quantity
R = V / I • V = I · R • I = V / R
- R — resistance from voltage and current, in ohms (Ω)
- V — voltage from current and resistance, in volts (V)
- I — current from voltage and resistance, in amperes (A)
These are the Ohm's-law rearrangements the calculator uses to fill in whichever of V, I, or R you left blank before computing power. Division by zero is guarded: a missing value cannot be recovered if the divisor is zero.
Examples
- Input
- V = 12 V, I = 2 A
- Result
- P = 24 W, R = 6 Ω
- Why
- With voltage and current given, use P = V·I = 12 × 2 = 24 watts. The resistance follows from Ohm's law, R = V / I = 12 / 2 = 6 Ω. A 12 V supply pushing 2 A through a 6 Ω load therefore dissipates 24 W as heat.
- Input
- V = 12 V, R = 4 Ω
- Result
- P = 36 W, I = 3 A
- Why
- Use P = V² / R = 12² / 4 = 144 / 4 = 36 watts. The current is I = V / R = 12 / 4 = 3 A. As a check, P = V·I = 12 × 3 = 36 W, which matches — the three forms always agree.
- Input
- I = 0.5 A, R = 220 Ω
- Result
- P = 55 W, V = 110 V
- Why
- Use P = I²·R = 0.5² × 220 = 0.25 × 220 = 55 watts. The voltage across the resistor is V = I·R = 0.5 × 220 = 110 V. A resistor dissipating 55 W needs a power rating well above that with a safety margin, so a 100 W heatsinked resistor would be appropriate.
- Input
- V = 2 V, I = 0.02 A
- Result
- P = 0.04 W (40 mW), R = 100 Ω
- Why
- Use P = V·I = 2 × 0.02 = 0.04 watts, i.e. 40 milliwatts. The effective resistance at this operating point is R = V / I = 2 / 0.02 = 100 Ω. Typical indicator LEDs run at roughly 20 mA and 2 V, so tens of milliwatts is expected.
- Input
- I = 1 A then I = 2 A, both at R = 10 Ω
- Result
- 10 W → 40 W (4× the power)
- Why
- At 1 A: P = I²·R = 1² × 10 = 10 W. At 2 A: P = I²·R = 2² × 10 = 4 × 10 = 40 W. Doubling the current quadrupled the dissipated power (40 ÷ 10 = 4) because current is squared in P = I²·R — the key reason wiring overheats when load current rises.
When to use this tool
- Sizing a resistor's power (wattage) rating so it does not overheat — compute the dissipation from the operating voltage or current, then choose a rating above it with margin.
- Estimating the heat a component, heating element, or load releases, since dissipated electrical power equals the rate of heat generation in watts (1 W = 1 J/s).
- Solving introductory electronics and physics problems that give any two of voltage, current, and resistance and ask for power and the third quantity.
- Checking how a change in load current or supply voltage affects heating — for example showing that doubling current quadruples I²R losses in a cable.
- Sanity-checking a DC power-supply or battery budget by confirming that the total dissipation across components matches the source power V·I.
Common mistakes
- Mixing milliamps with amperes. Power comes out in watts only when current is in amperes and voltage in volts. A 20 mA LED current must be entered as 0.02 A, not 20 — using 20 would overstate the power by a factor of a thousand.
- Forgetting to square the current or voltage. In P = I²·R you square the current first, then multiply by resistance; in P = V²/R you square the voltage. Multiplying only once gives a wildly wrong (usually far smaller) answer.
- Confusing the power rating with the dissipated power. A resistor's wattage rating is the maximum it can shed safely, not what it actually dissipates. Always size the rating above the calculated dissipation, typically with a 2× safety margin.
- Applying P = V²/R or I²·R to a non-resistive load. These two forms assume an Ohmic resistor where V = I·R. For reactive loads (motors, capacitors, inductors on AC) you must use real power P = V·I·cosφ instead, where φ is the phase angle.
- Dividing by zero. Resistance R = V/I is undefined when current is zero, and the V²/R and I²·R forms break down at R = 0; the calculator guards these cases and shows a dash rather than a misleading number.
Frequently asked questions
What is the formula for power dissipation?
Power dissipation is given by Joule's law, P = V·I, where P is power in watts, V is voltage in volts, and I is current in amperes. For a resistor this is equivalent to P = I²·R and P = V²/R, where R is resistance in ohms, because Ohm's law (V = I·R) links the three quantities. You pick whichever form matches the values you know.
Which power formula should I use — V·I, I²R, or V²/R?
Use the one that matches the values you already have. If you know voltage and current, use P = V·I. If you know current and resistance, use P = I²·R. If you know voltage and resistance, use P = V²/R. For a purely resistive load all three give the same answer; this calculator first fills in the missing quantity with Ohm's law, then computes power, so you only need any two inputs.
How do I calculate the power dissipated by a resistor?
Find the resistor's current or the voltage across it, then apply P = I²·R or P = V²/R. For example, 0.5 A through a 220 Ω resistor dissipates P = I²·R = 0.25 × 220 = 55 W. Always pick a resistor whose power rating exceeds this dissipation, usually by a factor of about two, so it runs cool and reliably.
What units does this power dissipation calculator use?
It uses SI units: voltage in volts (V), current in amperes (A), resistance in ohms (Ω), and the resulting power in watts (W). One watt is one joule per second. If your data is in milliamps or millivolts, convert first — divide milliamps by 1000 to get amperes (20 mA = 0.02 A) — or the power will be off by a factor of a thousand.
Does this work for AC circuits and motors?
It computes the dissipation of a resistive (Ohmic) DC load, and it also gives the correct real power for AC across a pure resistor. For reactive AC loads such as motors, transformers, or capacitive and inductive loads, you must use real power P = V·I·cosφ, where cosφ is the power factor; using V·I alone there gives apparent power (volt-amperes), which overstates the heat dissipated.
Why does doubling the current quadruple the power?
Because current is squared in P = I²·R. If the resistance stays the same and you double the current, then I² becomes four times larger, so the dissipated power increases by a factor of four. This is why cables and connectors heat up so quickly as load current rises, and why engineers work hard to keep currents (and therefore I²R losses) low in power distribution.
Sources & references
- Wikipedia — Joule heating (P = I²R)
- Wikipedia — Electric power
- HyperPhysics — Ohm's Law and Electric Power
External references open in a new tab. We are independent and not affiliated with these organizations.
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Provided “as is” for general information only — results may be inaccurate, so verify before you rely on them. No warranty; use at your own risk.
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