Ohm's law calculator
Voltage, current or resistance from the known pair, with power dissipation.
Fill in the fields and the result will appear here automatically.
Find the missing voltage, current or resistance for an ohmic load, plus dissipated power. Choose the known pair explicitly: U/I, U/R or I/R. Power is an output, not an input mode. Values are nonnegative magnitudes; the model does not describe current direction, a diode’s nonlinear characteristic or resistance changing with temperature.
How it works
Formula and logic
U = IR; I = U/R when R > 0; R = U/I when I > 0. P = UI = I²R = U²/R for positive R. For DC this is resistive power; for a purely resistive AC load use RMS voltage and current.
Example
12 V and 2 A give R = 12/2 = 6.00 Ω and P = 12 × 2 = 24.00 W. 5 V across 250 Ω give I = 0.020 A and P = 0.10 W. U = 0 with R = 100 Ω gives I = 0 and P = 0.
Fields and units
- Known pair — list option
- Voltage — V
- Current — A
- Resistance — Ω
How to use
- — Choose the actual known pair; the third input marked as computed is not used.
- — Enter volts, amperes and ohms. Divide milliamperes by 1000: 20 mA = 0.020 A.
- — Compare calculated dissipation with the component rating under its cooling conditions; this tool does not choose a part rating.
Method and limitations
- Calculation method
- Formula and logic
- Data or methodology source
- OpenStax: Ohm’s law and the limits of ohmic behavior OpenStax: AC power and a purely resistive load
- Limitation
- Ideal ohmic load with constant resistance. This calculation does not establish circuit, cooling or component-rating safety.
FAQ
Can power be supplied instead of voltage?
No. The supported pairs are U/I, U/R and I/R; power is calculated after recovering the third quantity.
When does Ohm’s law accept zero?
U = 0 with R > 0 gives zero current. I = 0 with known R also gives U = P = 0. Solving R from zero I or I from zero R requires division by zero, so these modes cannot provide a result.
Does this describe a motor or diode?
Not in general. Reactive loads need impedance and power factor; a diode does not have constant ohmic resistance U/I.
What happens to power if resistance doubles at the same voltage?
With ideal constant voltage U, I = U/R and P = U²/R. Doubling R halves both current and power. At 12 V and 6 Ω, power is 24 W; at 12 Ω it is 12 W. Constant current is different: P = I²R.