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Ohm's Law Calculator

Solve V, I, R, P from any two known values. V=IR, P=VI, P=V²/R, P=I²R — pick which pair you have, get the other two instantly.

Known pair

V

12 V

I

0.12 A

R

100 Ω

P

1.44 W

Formula: I = V / R, P = V² / R

All four values are computed in base SI units (V, A, Ω, W). Display uses the unit prefix you picked. Highlighted outputs are the computed pair.

How to use

  1. Pick which two values you know (the highlighted pair).
  2. Enter their magnitudes and units.
  3. Read the computed pair and the formula used. Copy the full V/I/R/P summary.

About this tool

Show

Ohm's Law (V = I × R) combined with the power equation (P = V × I) lets you solve any electrical circuit value from any two others. This calculator handles all six combinations: V+I, V+R, V+P, I+R, I+P, R+P. Pick the pair you know, enter values with sensible unit prefixes (mV, A, kΩ, mW, etc), and the other two appear with the formula used. Useful for resistor sizing, LED current limiting, power supply calculations, and basic electronics homework.

The law describes an ohmic component, which is a resistor and not much else. An LED is a diode: its voltage barely moves while its current climbs steeply, so there is no resistance value to plug in and connecting one directly across a supply destroys it. The standard approach is a series resistor sized from the voltage left over, (supply − forward voltage) / desired current, and it is that subtraction rather than the raw supply voltage that goes into the calculation. Motors, batteries and semiconductor junctions are all non-ohmic in the same way.

The power figure is the one that decides whether a part survives. A quarter-watt resistor dissipating 0.4 W does not fail immediately; it runs hot, drifts out of tolerance and dies later, which makes it a fault that appears weeks after the build. Derating to about half the rated dissipation is the usual habit.

Resistance is also not a constant across conditions. A resistor's value shifts with temperature by its stated coefficient, and a filament or a fuse changes by a large factor between cold and hot, which is why an incandescent lamp draws a surge at switch-on that the cold-resistance calculation predicts and the hot one does not.

Everything here is direct current. With alternating current the resistance term becomes impedance, phase separates real power from apparent power, and a power factor enters the arithmetic, none of which this models.

Frequently asked questions

When does Ohm's Law not apply?

Non-linear / non-ohmic devices: diodes, LEDs, transistors, incandescent bulbs (resistance changes with temperature). For those, use the device's I-V curve or datasheet rather than V = IR.

Calculate dropping resistor for an LED?

Use R + P known. R = (V_supply − V_LED) / I_LED. For a 3.2V/20mA LED from 5V: R = (5 − 3.2) / 0.020 = 90 Ω. Power dissipated in the resistor = (V_supply − V_LED) × I_LED = 36 mW, so a 1/4 W resistor is fine.

Is this AC or DC?

DC, or AC with purely resistive loads. AC circuits with capacitors/inductors involve impedance (Z) and phase — that's a separate calculation with complex numbers.

Why is one of my fields disabled?

It's part of the 'unknown' pair. The calculator works from two knowns to two unknowns. Switch the radio buttons at top to change which pair you're solving for.

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