Quick answer: Current equals voltage divided by resistance: 12 V across 6 Ω drives 12 ÷ 6 = 2 A, and the power is 12 × 2 = 24 W.
How it's calculated
Ohm's law says the voltage across a resistor is the current through it times its resistance: V = I × R. Electric power is voltage times current: P = V × I.
Putting the two together gives every other form: I = V ÷ R, R = V ÷ I, P = I² × R, P = V² ÷ R, I = √(P ÷ R) and V = √(P × R). That is why any two values fix the other two.
Units: volts (V), amps (A), ohms (Ω) and watts (W). Enter milliamps as amps (250 mA = 0.25 A) and kilo-ohms as ohms (4.7 kΩ = 4,700 Ω).
Worked examples
- 12 V, 6 Ω: I = 12 ÷ 6 = 2 A, P = 12 × 2 = 24 W
- A 2,000 W heater on 230 V: I = 2000 ÷ 230 = 8.696 A, R = 230² ÷ 2000 = 26.45 Ω
- 0.5 A through 100 Ω: V = 0.5 × 100 = 50 V, P = 0.5² × 100 = 25 W
Ohm's law formulas
| Find | From V and I | From V and R | From I and R | From P and another |
|---|---|---|---|---|
| V | — | — | I × R | P ÷ I or √(P × R) |
| I | — | V ÷ R | — | P ÷ V or √(P ÷ R) |
| R | V ÷ I | — | — | V² ÷ P or P ÷ I² |
| P | V × I | V² ÷ R | I² × R | — |
Frequently asked questions
What is Ohm's law?
V = I × R: the voltage across a conductor equals the current through it multiplied by its resistance, as long as its temperature stays the same.
How many amps does a 2,000 W appliance draw on 230 V?
I = P ÷ V = 2000 ÷ 230 ≈ 8.7 A. On 127 V it would be about 15.7 A.
Does Ohm's law work for AC?
For resistive loads, yes, using RMS voltage and current. For motors, transformers and electronics you need impedance and power factor, which this calculator does not include.
What resistor do I need for an LED?
Take the LED's voltage away from the supply voltage and divide by the current. A 2 V LED at 20 mA on a 5 V supply needs (5 − 2) ÷ 0.02 = 150 Ω.
How much power does a resistor turn into heat?
Current squared times resistance. 2 A through 6 Ω is 24 W, so you'd want a resistor rated comfortably above that.
What happens if I double the voltage?
The current doubles too, and the power goes up four times. That's why a slightly higher voltage can cook a part fast.