Ohm's law calculator

Give any two of voltage, current, resistance and power, and this works out the other two — including the power the resistor has to dissipate, which is the figure that decides whether it survives.

Voltage

12 V

Current

2 A

Resistance

6 Ω

Power

24 W

V = I × R · P = V × I = I²R = V²/R

Ohm's law is one equation with three letters, and the fourth quantity — power — is the one that actually burns things. A resistor picked correctly for resistance and wrongly for wattage works perfectly until it discolours the board around it. Give this page any two of the four and it fills in the rest, so you can see the dissipation before you order the part rather than after.

How it is calculated

V = I × R · P = V × I = I²R = V²/R

The ohm is defined so that one volt across one ohm drives one ampere, and the watt so that one volt at one ampere delivers one joule per second. Every form above follows from substituting one into the other, which is why any two quantities are enough and a third can only agree or contradict.

Source: NIST SP 811 (2008), Appendix B.8 — ohm (Ω), the SI unit of resistance: 1 Ω = 1 V/A

Questions people ask

Why does it ask for exactly two values?
Because two fix the circuit and a third has nothing left to add. If you enter 12 V, 2 A and 99 Ω, the first two already say the resistance is 6 Ω — the page cannot know which of your numbers is the wrong one, so it asks you to remove one rather than silently picking a winner.
Which resistor wattage should I buy?
Take the power figure here and give it room. Common practice is to fit a resistor rated at roughly twice the dissipation, because a part run at its limit gets hot enough to drift in value and to cook whatever sits next to it. A quarter-watt resistor dissipating 0.24 W is a mistake that works on the bench and fails in a warm enclosure.
Does this work for mains or for AC?
Only for the simple case. With alternating current the useful figures are RMS rather than peak, and anything with a coil or a capacitor has impedance rather than plain resistance, so current and voltage stop being in step. For a resistive load on AC, entering RMS values gives the right answer; for anything else this is the wrong tool.
Why is my LED calculation wrong here?
Because a diode is not a resistor: its voltage barely moves while the current climbs, so Ohm's law does not describe it. The way to use this page for an LED is to subtract the forward voltage from the supply first, then enter the remainder with the current you want — what you are sizing is the series resistor, not the diode.
What do the units mean at these sizes?
Everything here is in base units: volts, amperes, ohms and watts. A 220 Ω resistor on 5 V draws 0.0227 A — that is 22.7 mA — and dissipates 0.114 W, comfortably inside a quarter-watt part. Milliamps are thousandths of an amp and kilohms are thousands of ohms, so shift the decimal point rather than looking for a separate field.

Found a problem, or want more?

A number that disagrees with its source is a defect, not a rounding preference.

What did you enter, what did the tool show, and what did you expect instead? If you have a source that disagrees with ours, a link to it is the most useful thing you can send.

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