Ohm's Law Calculator

Ohm's Law and the Power Formula

There are four quantities and two relationships between them, which is what makes any two of the four enough to find the rest. Everything below is those two relationships rearranged — there is nothing else to learn.

Ohm's law

The current through a resistance is proportional to the voltage across it. Double the voltage and you double the current; double the resistance and you halve it:

The proportionality is the claim being made, and it is worth knowing it is a property of the component rather than a law of nature. A resistor obeys it closely over a wide range. A diode, a lamp filament and a thermistor do not, which is exactly why they are useful — and why this calculator describes a resistive load and not every two-terminal part.

The power law

Power is the rate at which energy is turned into heat. It is the voltage across the component times the current through it:

Substituting Ohm’s law into that gives the two forms that matter most in practice, because they are the ones with a square in them:

That square is the single most useful thing on this page. Doubling the current through a resistor does not double the heat, it quadruples it — so a part that runs warm at 100 mA is not merely twice as warm at 200 mA, it is dissipating four times the power. Most burnt resistors are a linear intuition applied to a squared relationship.

The six ways round

Which pair you were given decides which rearrangement you need. There are six, and the calculator above picks the right one from whichever two fields you fill:

• V and I → R = V/I, and P = V × I
• V and R → I = V/R, and P = V²/R
• V and P → I = P/V, and R = V²/P
• I and R → V = I × R, and P = I² × R
• I and P → V = P/I, and R = P/I²
• R and P → V = √(P × R), and I = √(P/R)

Only the last pair needs a square root, and that is a consequence of the squares above: knowing the power and the resistance but neither the voltage nor the current means undoing a square to get back to either one.

Worked example

The current is the headline answer, and the 6 W is the one that decides whether the part survives — a quarter-watt resistor here fails within seconds. Note how little the current tells you about that: 0.5 A sounds modest, and the heat it makes in 24 Ω is twenty-four times what a quarter-watt part can shed.

Choosing a Resistor With Ohm's Law

Working out the resistance is usually the easy half. Three more numbers decide whether the part you order actually works.

Power rating, and why half is the usual answer

A resistor’s power rating is what it can dissipate in free air at around 25 °C without exceeding its maximum temperature. Real boards are hotter than that, and a part run at its rating runs hot enough to shorten its life and to drift in value. The common working rule is to pick a rating at least twice the calculated dissipation.

So the 6 W in the worked example wants a 10 W or larger part, not a 6 W one — and certainly not the quarter-watt resistor that a schematic symbol looks like.

The tolerance you are buying

The resistance you calculated is a target, not a promise. A ±5% part at 24 Ω is anything from 22.8 Ω to 25.2 Ω, which moves the current by the same 5%. The standard-values table above the theory shows the nearest orderable value and what it can actually measure once the tolerance is allowed for.

Working voltage

Every resistor has a maximum working voltage as well as a power rating, and on small surface-mount parts it can be as low as 50 V. It is easy to satisfy the power rating and exceed the voltage rating at the same time, which is why high-voltage dividers are built from several resistors in series.

Where the Squared Term Catches People

Three everyday situations where the intuition that power scales with voltage — rather than with voltage squared — gives the wrong answer by a factor of four.

• Running a 12 V heater from 24 V does not double its output. It quadruples it, and the element usually fails.
• Halving a series resistor to get more LED current also quadruples that resistor’s dissipation for the same supply, because both the current through it and the current squared have changed.
• A supply sagging from 5 V to 4.5 V — down only 10% — drops the power in a fixed resistive load by 19%, not 10%.

The general form is worth internalising: a change of factor k in voltage or current across a fixed resistance changes the power by k².

What This Calculator Assumes

Stated plainly, because a number that arrives without its assumptions is a number you cannot check:

• DC, or a steady value. For AC, V and I have to be the same kind of measurement and the load has to be resistive; with a reactive load, power is no longer simply V × I.
• A resistive, ohmic load. A diode, an LED, a filament lamp or a motor does not have one resistance, so a single V/I answer describes only the operating point you measured it at.
• All four quantities positive. This solves a passive load, not a source — the sign convention that tells a battery from a resistor is not modelled.
• No wiring losses. The voltage you enter is the voltage across the component. Over a long cable run, some of the supply is dropped before it gets there.
• A constant resistance. Real resistors drift with temperature, and the resistance that produced your heat calculation is not quite the resistance you end up with.

Energy, Not Just Power

Power is a rate; energy is a rate multiplied by time, and it is energy that a meter bills and a battery stores. The conversion is simple and constantly confused with power:

A 6 W load left on for a day is 144 Wh, or 0.144 kWh — a fraction of a unit of electricity. The same 6 W in a resistor rated for 6 W is a component running at its limit continuously. Identical power, entirely different questions, which is why "how much does it cost to run" and "will this part survive" need separate sums.

Common mistakes

• Treating power as proportional to voltage. It goes with the square: double the volts across a fixed resistor and the heat is four times, not twice.
• Choosing a resistor rated at exactly the calculated power. Aim for at least twice it, because the rating assumes free air at around 25 °C and your board is warmer.
• Mixing prefixes before dividing. 12 V ÷ 24 mΩ is 500 A, not 0.5 A — convert to base units first, or let the unit dropdowns do it.
• Applying Ohm’s law to a diode or an LED as though it had a single resistance. It has a forward voltage instead, and the current is set by whatever else is in the loop.
• Using the supply voltage rather than the voltage across the component. The rest of the circuit has dropped some of it before it arrives.
• Entering three of the four values. Two fix all four; a third either repeats what the first two already said or contradicts them.

Frequently asked questions

What is Ohm's law?

It states that the voltage across a resistance equals the current through it times that resistance: V = I × R. Rearranged, I = V/R and R = V/I. It holds closely for resistors and only approximately, or not at all, for diodes, lamps and motors.

How do you calculate current from voltage and resistance?

Divide the voltage by the resistance: I = V/R. Twelve volts across 24 ohms gives 0.5 A. Convert both to base units first — volts and ohms — or a stray prefix moves the answer by a factor of a thousand.

How do you find watts from volts and amps?

Multiply them: P = V × I. If you have the resistance instead of one of the two, use P = V²/R or P = I² × R. All three give the same answer for a resistive load.

How do you work out resistance from power?

It depends which other quantity you have. With voltage, R = V²/P. With current, R = P/I². With neither, power and resistance alone still fix everything: V = √(P × R) and I = √(P/R).

What power rating should the resistor be?

At least twice the calculated dissipation. Ratings assume free air at roughly 25 °C, and a part run at its limit gets hot enough to drift in value and to age quickly. A 6 W dissipation wants a 10 W or larger part.

Why do I only need two values to use Ohm's law?

Because four quantities are tied together by two independent relationships — V = I × R and P = V × I. Two equations plus two known values leave two unknowns, which is exactly solvable. A third value adds no information and may contradict the first two.

Does Ohm's law work for AC?

For a purely resistive load, yes, provided the voltage and current are the same kind of measurement — RMS with RMS. Once the load has capacitance or inductance, current and voltage fall out of step and the power is no longer simply V × I.