Watts to Volts Calculator
Find voltage from real power (watts) using either current (amps) for DC and AC (with power factor) or resistance (ohms) for resistive DC via V = √(P × R). Matches our Amps to Volts and Watts to Amps assumptions. All calculations run in your browser.
Same rearrangement as P = V × I: V = P ÷ I (DC) and AC forms with PF — mirrors our Amps to Volts power mode with watts entered first.
Real power in watts (not VA).
DC: current through the load. AC: RMS line current.
Table of Contents
Why watts alone do not determine volts
Power in watts is only one piece. To solve for voltage you also need either current (with DC vs AC rules) or, for a purely resistive model, resistance.
With P and I: V = P ÷ I (DC), V = P ÷ (I × PF) (single-phase RMS), VL-L = P ÷ (√3 × I × PF) (balanced three-phase, line-to-line).
With P and R on a resistive branch: P = V² ÷ R ⇒ V = √(P × R).
This tool complements Amps to Volts (same current-mode math, watts-first UI) and Watts to Amps / Volts to Amps for a consistent set of electrical calculators.
Formulas
From power and current
DC
V (V) = P (W) ÷ I (A)
From P = V × I. Power factor does not apply.
AC single-phase
VRMS (V) = P (W) ÷ (I (A) × PF)
Real power P, RMS current I, PF between 0 and 1.
AC three-phase (balanced)
VL-L (V) = P (W) ÷ (√3 × I (A) × PF)
Line-to-line RMS voltage. √3 ≈ 1.732.
Resistive DC (power and resistance)
V (V) = √(P (W) × R (Ω))
From P = V² ÷ R. Use only when dissipated power in R is the quantity you entered as P.
Quick Reference Table
| Method | Circuit | Power | I or R | PF | Voltage | Example |
|---|---|---|---|---|---|---|
| P + I | DC | 120 W | 10 A | — | 12 V | Small DC load |
| P + I | AC 1φ | 2,736 W | 12 A | 0.95 | 240 V | RMS result |
| P + I | AC 3φ | 6,462 W | 10 A | 0.9 | 415 V | VL-L |
| P + R | Resistive | 100 W | 25 Ω | — | 50 V | √(100 × 25) |
FAQ
The current mode uses the same equations; this calculator puts watts first in the form for people who think in power → voltage.
When P is the power dissipated in a resistor R and you want the DC voltage across it. It is not a general AC substitute for motor or inverter loads.
Enter power in watts (multiply kW by 1000). Example: 2.5 kW → 2500.
Balanced real power P = √3 × VL-L × I × PF ⇒ VL-L = P ÷ (√3 × I × PF).