About Ohm's Law Calculator — Volts, Amps, Ohms & Watts
The Ohm's law calculator solves the two definitional relationships of a resistive circuit — Ohm's law V = I·R and the power law P = V·I — from whichever two quantities you actually know. Pick the known pair (voltage and resistance, current and power, or any of the six combinations), enter the two values, and the tool returns all four: volts, amps, ohms, and watts. Because P = V·I combines with V = I·R, every remaining quantity has a closed-form answer: P = V²/R, P = I²·R, R = V²/P, V = √(P·R), and I = √(P/R).
These formulas describe DC circuits and AC circuits with purely resistive loads — heaters, incandescent lamps, resistors. For AC loads with a power factor below 1 (motors, drives, transformers), real power is P = V·I·PF and the current for a given kW is higher than the resistive formula suggests; size those with the three-phase power calculator. For wire sizing and voltage drop over a run of conductor, follow the related tools.
How It Works
- Select the pair of quantities you know. The two input fields switch to match — for example 'Voltage & resistance' asks for volts and ohms.
- Enter the two known values. Every field is validated: values must be positive and finite, and a zero divisor (such as solving R = V/I with 0 A) is rejected with a clear message instead of returning infinity.
- Read all four quantities in the results card. The two you entered are echoed back and the other two are computed via Ohm's law and the power law.
- Cross-check intuition: doubling voltage across a fixed resistance doubles the current but quadruples the power (P = V²/R) — the reason a 240 V element rated from 120 V service draws four times the watts.
Worked Example
A 120 V lamp is rated 60 W — what current does it draw and what is its hot filament resistance? Known pair: voltage and power. Current I = P/V = 60/120 = 0.5 A, and resistance R = V²/P = 120²/60 = 14,400/60 = 240 Ω. The tool echoes V = 120 V and P = 60 W and computes I = 0.5 A and R = 240 Ω — the second row of the examples table. Reverse the check with the (I, R) pair: 0.5 A through 240 Ω gives V = 0.5 × 240 = 120 V and P = 0.5² × 240 = 60 W, closing the loop.
Ohm's law examples: solving V, I, R and P from two knowns
Four everyday examples solved through the calculator, one from each common known pair. Each row starts from the two bold-named knowns and fills in the other two quantities via V = I·R and P = V·I — resistive-load basis, so the power figures assume no power factor.
| Known values | Volts (V) | Amps (A) | Resistance (Ω) | Power (W) |
|---|---|---|---|---|
| 12 V across 6 Ω (automotive load) | 12 | 2 | 6 | 24 |
| 120 V lamp rated 60 W | 120 | 0.5 | 240 | 60 |
| 5 A through 10 Ω | 50 | 5 | 10 | 250 |
| 230 V heater rated 2,300 W | 230 | 10 | 23 | 2,300 |
Formulas
- Ohm's law
V = I × R- Power law
P = V × I- Derived forms (combining the two)
P = V²/R = I²·R; R = V²/P = P/I²; V = √(P·R); I = √(P/R)
Standards & References
- Ohm's law (V = I·R) and the electric power law (P = V·I) — definitional physics relationships; no empirical constants involved
- DC / purely resistive AC basis: for reactive AC loads real power is P = V·I·PF — use the three-phase power calculator for motors and transformers
Frequently Asked Questions
How do I calculate watts from volts and amps?
Multiply them: P = V × I. A 120 V circuit carrying 15 A delivers 120 × 15 = 1,800 W. This is exact for DC and for resistive AC loads (heaters, incandescent lamps); for motors and electronics with a power factor below 1, the same V × I gives apparent power in VA and the real watts are V × I × PF.
How do I find resistance from voltage and power?
Use the derived form R = V²/P. A 2,300 W heater on 230 V has R = 230²/2,300 = 23 Ω — the last row of the examples table. The same element moved to 120 V would draw only P = V²/R = 120²/23 ≈ 626 W, which is why heating elements are voltage-specific.
Why does the calculator reject 0 A when solving from voltage and current?
Because R = V/I divides by the current: at exactly 0 A the resistance is undefined (an open circuit is better modeled as infinite resistance, not computed from Ohm's law). The tool returns a clear error instead of infinity — every output is guaranteed finite.
Does Ohm's law work for AC circuits?
For purely resistive AC loads, yes — use RMS voltage and current and the formulas hold as written. For reactive loads (motors, transformers, LED drivers) the current lags or leads the voltage, real power becomes P = V·I·PF, and impedance replaces resistance. Size those circuits with the three-phase power calculator, which includes the power factor.
What is I²R loss and why does it matter?
P = I²·R is the heat dissipated in any conductor carrying current — the reason wire gauge is chosen by amps, not watts. Because the loss grows with the square of current, halving the current (by doubling the voltage for the same power) cuts conductor heating to a quarter; that is the entire rationale for high-voltage transmission and for checking voltage drop on long runs.