Ohm's Law Calculator
Enter any two of voltage, current, resistance and power; the other two are computed, with the formulas used shown alongside.
Volts (V)
Amps (A)
Ohms (Ω). 4k7 and 1M are understood.
Watts (W)
5 V22.7 mA220114 mWI = V / RP = V² / RThe Ohm’s law wheel
Each quadrant is one quantity, and the three formulas inside it are the three ways to get that quantity from two of the others. It looks like twelve things to memorise. It is two — V = I × R and P = V × I — and the other ten are those two rearranged.
- V Volts
- I Amps
- R Ohms
- P Watts
The relation, and the three others that come with it
Ohm’s law is one equation: the current through a conductor is proportional to the voltage across it, with resistance as the constant of proportionality. V = I × R. Everything else on the wheel chart is that equation rearranged, or Joule’s P = V × I rearranged, or the two substituted into each other.
That is worth saying plainly because the twelve-formula chart makes it look like twelve things to memorise. It is two, and the rest is algebra you can redo in your head when you need it.
Worked examples
A resistor on a supply
220 Ω across 5 V. I = V/R = 5 / 220 = 22.7 mA, andP = V²/R = 25 / 220 = 114 mW. A quarter-watt part handles that with room to spare; an 0603 surface-mount part rated 100 mW does not.
From a power rating backwards
A 60 W bulb on a 120 V supply. I = P/V = 60 / 120 = 0.5 A andR = V²/P = 14400 / 60 = 240 Ω. Measure a cold incandescent bulb and you will find far less than 240 Ω — the filament’s resistance climbs by a factor of ten as it heats, which is why they fail at switch-on.
Sizing for a current limit
You want 10 mA from a 3.3 V rail. R = V/I = 3.3 / 0.01 = 330 Ω, dissipatingP = I²R = 0.0001 × 330 = 33 mW. 330 Ω is an E24 value, so this one lands on a real part without compromise.
Where the law stops holding
Ohm’s law describes ohmic components, and plenty of the parts on your bench are not.
A diode’s current rises exponentially with voltage; dividing one by the other gives a number, but it is not a resistance you can reuse at another operating point. An LED does the same, which is why you put a resistor in series with it rather than computing an equivalent resistance for the LED itself. A filament lamp changes resistance by an order of magnitude between cold and hot. A thermistor is built to change with temperature. A switching regulator draws less current as its input voltage rises, which makes its apparent resistance negative.
For those, the arithmetic here still applies at a single operating point — the instantaneous V, I and P are whatever they are — but the resistance it reports is only valid at that point.
AC, and why this calculator is DC
With alternating current, resistance generalises to impedance, which has a magnitude and a phase. A capacitor or inductor stores energy and returns it instead of dissipating it, so P = V × Ioverstates the real power by the power factor. Using these formulas on an AC circuit with reactive elements gives apparent power in volt-amps, not watts.
For a purely resistive AC load the formulas hold as written, provided the voltage and current are RMS values rather than peaks. Mains at 230 V RMS peaks at 325 V; using the peak would overstate power by a factor of two.
Reading the result
The tool shows which two formulas it used, because the same four numbers can be reached several ways and knowing which one produced them tells you which input an error came from. If the power looks wrong and the resistance is right, the current is the input to re-check.
Frequently asked questions
What are the twelve Ohm’s law formulas?
They are six pairs, one for each combination of two known quantities. V and I give R = V/I and P = V×I. V and R give I = V/R and P = V²/R. V and P give I = P/V and R = V²/P. I and R give V = I×R and P = I²×R. I and P give V = P/I and R = P/I². R and P give V = √(P×R) and I = √(P/R). The wheel chart you have seen is those twelve arranged in a circle.
Is P = V×I part of Ohm’s law?
Strictly no. Ohm’s law is V = I×R and nothing more. The power relations are Joule’s law, and they hold for things that do not obey Ohm’s law at all. They get solved together because that is how the question arrives — you know two of the four and want the rest — not because they are the same law.
How do I convert watts to amps?
You need a second quantity. With the voltage, I = P/V: a 60 W bulb on 120 V draws 0.5 A. With the resistance, I = √(P/R). Watts alone do not give amps, because the same power can be a small current at a high voltage or a large one at a low voltage — which is the whole argument for transmitting power at high voltage.
What resistor do I need for an LED?
Subtract the LED’s forward voltage from the supply, then divide by the current you want. A red LED at about 2 V on a 5 V rail at 20 mA needs (5 − 2) / 0.02 = 150 Ω, and the nearest part up is 150 Ω or 220 Ω. The dedicated LED resistor calculator on this site does that subtraction for you and picks a standard value.
Why does this refuse a resistance of zero?
Because the answer would be infinite current, which is not a measurement — it means the thing has been shorted and the question is wrong. Returning Infinity would put something on the screen that looks like a result. The same applies to a current of zero with a voltage: no current through a component tells you nothing about its resistance.
What power rating should the resistor have?
At least double the computed dissipation, and more if it sits in still air or beside something hot. A quarter-watt part running at 0.24 W is a part running at its limit for its whole life. The figure this tool gives is what the resistor must shed continuously; derate from there rather than toward it.
The code behind this
The arithmetic on this page is published as the ohmslaw module — the same functions you are using here, with the same tests. MIT, no dependencies.
Read the sourceOn npmnpm i electronics-math
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