Volts to Amps Calculator
Find current (amps) from voltage using either real power (watts) for DC/AC circuits or resistance (ohms) with Ohm’s law (I = V ÷ R). Power mode supports single- and three-phase AC with power factor. Same assumptions as our Watts to Amps and Amps to Volts tools.
Uses P = V × I rearranged: I = P ÷ V (and AC forms with PF). Same math as our Watts to Amps Calculator.
Volts across the load.
Real power in watts (not VA).
Table of Contents
Why volts alone do not fix current
Voltage by itself does not determine amps. You need another relationship — usually real power in watts (with AC power factor and phase count as appropriate) or resistance for Ohm’s law.
With P and V, current follows the same forms as Watts to Amps: I = P ÷ V (DC), I = P ÷ (V × PF) (single-phase RMS), I = P ÷ (√3 × VL-L × PF) (balanced three-phase, line-to-line).
With V and R, I = V ÷ R — the inverse of using V = I × R in our Amps to Volts calculator’s Ohm mode.
Pick the mode that matches your knowns. All four electrical tools (Amps ↔ Watts, Watts → Amps, Amps → Volts, Volts → Amps) share consistent RMS and line-to-line conventions.
Formulas
From power and voltage
DC
I (A) = P (W) ÷ V (V)
Rearrangement of P = V × I. Power factor does not apply.
AC single-phase
I (A) = P (W) ÷ (VRMS × PF)
Real power P, RMS voltage V, power factor PF (0–1).
AC three-phase (balanced)
I (A) = P (W) ÷ (√3 × VL-L × PF)
Use line-to-line RMS voltage. √3 ≈ 1.732. Same form as our Watts to Amps three-phase mode.
Ohm’s law
I (A) = V (V) ÷ R (Ω)
Equivalent to V = I × R solved for current. Resistive/DC idealization only.
Quick Reference Table
| Method | Circuit | Voltage | P or R | PF | Current | Example |
|---|---|---|---|---|---|---|
| Power | DC | 12 V | 120 W | — | 10 A | Small DC load |
| Power | AC 1φ | 240 V | 2,736 W | 0.95 | 12 A | RMS supply |
| Power | AC 3φ | 415 V | 6,462 W | 0.9 | 10 A | VL-L (AU industrial) |
| Ohm’s law | — | 100 V | 50 Ω | — | 2 A | Resistor |
FAQ
In power mode, yes — same formulas. This page emphasizes “I have volts and watts” and adds an Ohm’s law mode (**I = V ÷ R**).
No — you also need power (watts) (with AC details) or resistance (ohms), or another complete circuit relationship.
When you know voltage across and the resistance of a resistive branch (or equivalent R), and want current through it.
Balanced real power: P = √3 × VL-L × I × PF. Solving for I gives P ÷ (√3 × VL-L × PF).