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AESTECHNO
V = R I

Ohm's Law and Power Calculator

Enter any two of the four quantities (voltage V, current I, resistance R, power P) and the calculator solves the other two using Ohm's law and the Joule power wheel. Example: 12 V and 2 A give R = V / I = 6 ohm and P = V x I = 24 W.

Inputs

Fill in exactly two quantities; leave the other two blank.

Circuit type

In AC (RMS values), R is read as the impedance magnitude |Z| and P as real (active) power in W.

Between 0 and 1 (1 = resistive load). Links real power P to apparent power S = V × I: P = S × cos phi.

V volts V = I · RV = P / IV = √(P · R) I amps I = V / RI = P / VI = √(P / R) R ohms R = V / IR = V² / PR = P / I² P watts P = V · IP = I² · RP = V² / R

Ohm's law / power wheel

Result

24 W

Breakdown

V 12 V · I 2 A · R 6 Ω · P 24 W

Formula R = V / I · P = V × I

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Enter any two of V, I, R, P and we solve the other two. DC: purely resistive load. AC (single phase, RMS): R is read as the impedance magnitude |Z| and P as real power, with apparent power S = V × I = P / cos phi.

TOOL // OHM'S LAW

This calculator gets you a first pass at sizing, but a real power study accounts for derating, thermals and margins. Let's talk it through in a free 30-minute audit with our design team.

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Frequently asked questions

FAQ

Which quantities do I need to use the calculator?
Any two different quantities out of four: voltage V (volts), current I (amperes), resistance R (ohms) and power P (watts). From that pair, the tool solves the two missing quantities using the matching pair of power-wheel equations (12 formulas in total) and shows the exact pair of formulas it applied under the result.
Which formula is used for each known pair?
Known (V, I): R = V / I and P = V x I. Known (V, R): I = V / R and P = V x V / R. Known (V, P): I = P / V and R = V x V / P. Known (I, R): V = I x R and P = I x I x R. Known (I, P): V = P / I and R = P / (I x I). Known (R, P): V = sqrt(P x R) and I = sqrt(P / R). The calculator shows the pair it applied under the result.
Why are some combinations rejected?
For physical reasons. Solving R with zero current (I = 0) is a division by zero: that is an open circuit, undefined. Any division by a zero known quantity is blocked. A negative resistance, power or current is non-physical and rejected. The square-root branch (the R, P pair) requires a strictly positive resistance.
How are metric prefixes (mA, kohm, mW) handled?
Each field has its own unit selector: voltage (mV, V, kV), current (uA, mA, A), resistance (mohm, ohm, kohm, Mohm), power (uW, mW, W, kW). Everything is converted to SI base units before computing (u = 1e-6, m = 1e-3, k = 1e3, M = 1e6), then the result is shown with a readable prefix: for example 0.0227 A becomes 22.73 mA and 1500 ohm becomes 1.5 kohm.
Is the calculation valid for AC?
The core V/I/R/P solve assumes a purely resistive (ohmic) load in DC. In AC mode (single phase, RMS values), R is read as the impedance magnitude |Z| and the P field as real power: P = V x I x cos phi (W), with apparent power S = V x I (VA). If you enter P, we derive S = P / cos phi before solving. Three-phase (factor sqrt(3) on line quantities) and non-linear loads are out of scope.
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An electronics design house based in Montpellier, France, AESTECHNO brings 10+ years of design experience and a 100% success rate on CE/FCC certifications, across dozens of campaigns run over more than 10 years of practice. Formulas based on Ohm's law and the Joule power wheel.