Ohm's Law Calculator
Solve for voltage, current, resistance or power from any two known values.
V = I × R · P = V × I = I² × R = V² / R
DC and purely resistive loads. AC with reactance needs impedance and power factor, neither of which this models. Values you type are never overruled — clear a field to have it recomputed.
How it works
Type any two of voltage (V), current (A), resistance (Ω) or power (W). Leave the two you want solved blank.
V = I × R gives the third quantity and P = V × I gives the fourth, whichever two you started from.
The solver substitutes repeatedly rather than matching your pair against a fixed list of formulas, so all six possible pairs resolve — including the two that need a square root.
All four values are shown together, the two you entered beside the two derived from them, so the answer can be checked against the input rather than replacing it.
Ohm's law and DC power
Ohm's law fixes the relationship between the three electrical quantities; the power identity adds the fourth. Substituting one into the other gives the two forms that avoid a quantity you do not have — P = I²R when voltage is unknown, P = V²/R when current is.
V = I × R P = V × I = I² × R = V² / R
Use cases
LED series resistor
A red LED drops about 2 V at 20 mA, so on a 5 V rail the resistor absorbs the other 3 V. Enter 3 V and 0.02 A: 150 Ω, dissipating 0.06 W — a margin any through-hole part clears by an order of magnitude.
Resistor power rating
A 100 Ω resistor across 12 V passes 0.12 A and dissipates 1.44 W. That last figure is what picks the part: a 1/4 W resistor in that position runs hot enough to drift and then fail, and neither the voltage nor the current says so on its own.
Current draw before sizing a supply
Enter the rail voltage and a measured or datasheet resistance to get what one branch pulls, then add the branches up. A 5 V supply feeding four 50 Ω loads is 0.4 A, so a 500 mA brick has nothing left for inrush.
Checking a measurement against the maths
The meter reads 4.86 V across a part and 0.51 A through it. Enter both: 9.53 Ω and 2.48 W. If the part is marked 10 Ω, the gap is lead resistance or a sagging supply rather than a mystery.
Battery runtime, roughly
A load drawing 0.35 A at 3.7 V is 1.3 W. A 2000 mAh cell at that voltage holds about 7.4 Wh, which puts the ceiling a little under six hours before any converter loss is counted.
Voltage divider sanity check
Two resistors in series across a rail carry the same current. Solve for that current from the rail voltage and the total resistance, then pair it with either resistor to get the voltage across it — the divider ratio falls out without a second formula.
Questions
- Can it solve for two unknowns at once?
- That is the normal case. Any two of the four determine the other two, including the pairs that need a square root: power with resistance gives I = √(P/R), and power with current gives V = P/I.
- Does it handle AC?
- Only purely resistive AC, and only if you enter RMS values — the results are then RMS as well. Anything reactive, meaning motors, transformers and capacitive-input supplies, needs impedance and power factor, and this models neither.
- What units does it expect?
- Base SI units: volts, amps, ohms, watts. There is no prefix parsing, so 20 mA is 0.02, 4.7 kΩ is 4700 and 100 mW is 0.1.
- What if I enter three values that disagree?
- It fills blanks and checks nothing. Type V = 5, I = 1 and R = 10 and all three come back exactly as typed, with P = 5 W derived from the first two. Clear a field to have that field recomputed.
- Why do some negative values return nothing?
- Where the solve needs a square root — power together with resistance, or power together with current — a negative input makes the root imaginary. The result stays empty rather than showing a fabricated number.
- Are the numbers sent anywhere?
- No. The arithmetic is a few lines of JavaScript running in the page. Nothing is uploaded and nothing is stored.
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