Fill in any two of voltage, current, resistance and power, leave the other two blank, and get the missing values. Unit prefixes are handled for you (mV, kV, mA, µA, kΩ, MΩ, mW, kW).
Worked example
These are example values. Change any entry above and press Calculate to run your own numbers.
Voltage (V)
24 V
Current (I)
0.5 A 500 mA
Resistance (R)
48 Ω
Power (P)
12 W
Ohm's law applies to resistive loads, and to DC. On AC, reactive loads such as motors and coils also depend on impedance and power factor.
How it works
Ohm's law links voltage, current and resistance, and the power law links them to power. Between them, any two of the four quantities fix the other two for a resistive load.
The formulas
Voltage: V = I × R, V = P ÷ I, V = √(P × R)
Current: I = V ÷ R, I = P ÷ V, I = √(P ÷ R)
Resistance: R = V ÷ I, R = V² ÷ P, R = P ÷ I²
Power: P = V × I, P = I² × R, P = V² ÷ R
Resistors in series: R = R₁ + R₂ + … in parallel: 1 ÷ R = 1 ÷ R₁ + 1 ÷ R₂ + …
Frequently asked questions
What is Ohm's law?
Ohm's law says the current through a resistor is proportional to the voltage across it: I = V ÷ R, or equivalently V = I × R. Twelve volts across a 48 Ω resistor passes 0.25 A.
How do I find power if I know voltage and resistance?
Use P = V² ÷ R. A 24 V supply across a 48 Ω load dissipates 24² ÷ 48 = 12 W. If you know the current instead, use P = I² × R, or P = V × I when you have both voltage and current.
Why must I fill in exactly two values?
Two known quantities fix the other two. With one value the problem has many answers, and with three or four you could enter values that contradict each other, so the calculator asks for exactly two and solves for the rest.
How do I size a resistor for an LED?
Subtract the LED forward voltage from the supply voltage to get the voltage the resistor must drop, then divide by the LED current: R = (V supply − V LED) ÷ I. Enter that voltage and the current here for the resistance and the power the resistor dissipates, and choose a resistor rated comfortably above that power.
Does it work for AC?
For resistive AC loads such as heaters, with RMS values, yes. Inductive and capacitive loads such as motors and transformers depend on impedance and power factor as well as resistance, so Ohm's law with a single resistance does not describe them completely.
