Volts to Watts Calculator - Power from Current or Resistance

Volts do not become watts until a load draws current. With current, W = V x A. With resistance, W = V^2 / R. For AC loads, include power factor when current is involved.

Formula at a glance

  • with current: W = V x A x PF
  • with resistance: W = V^2 / R
  • three-phase: W = 1.732 x V x A x PF

Field note: The voltage printed on the outlet does not tell you the watts. The load decides how much current it takes.

Calculator Tool

Convert voltage to power

V
A
Result

Formulas

With IP = V × I
With RP = V² / R

Quick Reference

120V @ Watts
1 A 120 W
5 A 600 W
10 A 1200 W
15 A 1800 W

How to use the Volts to Watts Calculator

Use this as a fast electrical check, then compare the result with the nameplate, measured voltage and power factor. The formula is clean. Real panels, motors and UPS loads usually have one extra wrinkle.

Worked example

Example: 120 V at 5 A is 600 W for a resistive load. At PF 0.9, real power is 540 W.

Practical checks before you trust the number

  • Use resistance formulas for heaters and simple resistive loads.
  • Use PF for motors, LED drivers and transformers.
  • Use a power meter when the waveform is messy.

Common mistake

The voltage printed on the outlet does not tell you the watts. The load decides how much current it takes.

Sources and references

Related calculators

Frequently Asked Questions

Watts = volts × amps for DC. For single-phase AC, W = V × A × PF. For three-phase, W = √3 × V × A × PF. Example: 230 V × 5 A × PF 0.9 single-phase = 1035 W. The voltage alone doesn't determine power — you need the current too. Always confirm the load type and use the appropriate formula. We size every panel circuit with this as the starting equation.

Single-phase: W = V × A × PF. Three-phase: W = √3 × V × A × PF. The PF accounts for how much of the apparent power (VA) is actually doing useful work. For purely resistive loads PF = 1, so W = V × A. For motors, capacitors, and ballasts, PF is less than 1. Memorize the √3 factor (1.732) for three-phase work — that's where most calculation slip-ups happen.

On a resistive load (PF = 1): W = 120 × 5 = 600 W. With a small inductive load at PF 0.9: W = 120 × 5 × 0.9 = 540 W. So at 5 A you're somewhere between 540 and 600 W depending on PF. The current is the same in both cases, but the energy delivered to do real work is different. This catches many juniors when they assume amps directly equal watts in AC.

Use P = V² ÷ R when only voltage and resistance are known. Example: 230 V across a 100 Ω heater element → P = 230² ÷ 100 = 529 W. This is handy for resistive heaters and incandescent lamps where R is a fixed property. For inductive or capacitive AC loads, use impedance Z instead of R, and apply PF separately. Memorize all three Ohm's law power forms (P = VI, P = I²R, P = V²/R).

Power factor reduces the real power delivered. W = V × A × PF, so even at the same V and A, lower PF means lower W. Example: 230 V × 10 A = 2300 VA, but at PF 0.7 only 1610 W of real power. The remaining 690 VAR is reactive — it flows back and forth without doing work. Improving PF with capacitor banks reduces line current and saves on cable losses.

Not directly. You need at least two of the three (V, A, R) to find power. With voltage and resistance: P = V² ÷ R. With voltage and current: P = V × A × PF. Voltage by itself only tells you the potential difference, not the power. So always ask: what's the load resistance or what's the current? Otherwise, the answer cannot be determined.

Yes, you can estimate device power from a known supply voltage and measured or rated current. Example: a fan rated 230 V, 0.4 A, PF 0.95 → W = 230 × 0.4 × 0.95 ≈ 87 W. This matches typical ceiling fan ratings. For more accuracy, use a clamp meter and a true-RMS power meter that reports W, VA, and PF directly. Calculator estimates are good for sizing, not for energy auditing.