Amps to Volts Calculator - Voltage from Power, Current or Resistance

Amps alone cannot tell you voltage. You need either resistance, watts, VA or kW. With resistance, V = I x R. With power, V = W / A for DC or V = W / (A x PF) for single-phase AC.

Formula at a glance

  • Ohm law: V = I x R
  • DC power: V = W / A
  • single-phase AC: V = W / (A x PF)

Field note: The trap is asking "how many volts is 10 amps" with no other information. That question has no answer. A phone charger and a welder can both involve 10 A in different parts of the circuit.

Calculator Tool

Convert current to voltage

A
W
Result

Formulas

With PV = P / I
With RV = I × R

Ohm's Law

V = I × R
I = V / R
R = V / I

The foundation of electrical calculations.

How to use the Amps to Volts 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: a 5 A load through 24 ohms needs 120 V. A 1,000 W heater drawing 8.33 A is also on roughly 120 V.

Practical checks before you trust the number

  • If the load is AC and not resistive, include power factor.
  • For three-phase voltage from kW, divide by 1.732 x amps x PF.
  • When troubleshooting, measured current and measured resistance often beat nameplate guesses.

Common mistake

The trap is asking "how many volts is 10 amps" with no other information. That question has no answer. A phone charger and a welder can both involve 10 A in different parts of the circuit.

Sources and references

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Frequently Asked Questions

Volts = watts ÷ amps. Example: a 1000 W appliance drawing 10 A needs V = 1000 ÷ 10 = 100 V. For AC with PF: V = W ÷ (A × PF) for single-phase. This calculation is mostly useful when you're back-checking a power supply rating against a known load and current. In normal residential work, voltage is fixed by the utility (230 V or 415 V), so this formula is more relevant for control-circuit and DC design.

Ohm's law: V = I × R. Example: 2 A through a 50 Ω resistor produces V = 100 V. This is the foundation for voltage divider design and for calculating shunt voltages in measurement circuits. In AC analysis replace R with impedance Z. Always make sure your resistor wattage rating exceeds I²R; for the example above, P = 2² × 50 = 200 W, which most discrete resistors cannot handle without a heatsink.

V = W ÷ A (DC or PF = 1). For 1000 W at 10 A, V = 100 V. With PF 0.8 (single-phase AC): V = 1000 ÷ (10 × 0.8) = 125 V. So load type matters when reverse-calculating voltage. In real residential design we don't pick voltage this way — utility supply is fixed — but for DC equipment specs and battery system planning the formula tells us what supply rating the device needs.

Current alone tells you flow rate, not the driving voltage. The same 5 A can flow through a low-voltage 10 V circuit (50 W) or a 230 V circuit (1150 W). To know the voltage, you need either the power rating or the resistance. From Ohm's law: V = I × R. Always specify both current and one of voltage or resistance when describing a circuit, otherwise the description is incomplete.

DC: V = I × R or V = P ÷ I. Example: a 12 V battery driving 3 A through a coil means the coil resistance is 4 Ω. Or, if a load draws 6 A and consumes 72 W, V = 72 ÷ 6 = 12 V. Both work. For DC there's no power factor to worry about, which simplifies battery, solar, and EV system calculations significantly.

AC single-phase: V = W ÷ (I × PF). Example: 1500 W motor drawing 8 A at PF 0.75 → V = 1500 ÷ (8 × 0.75) = 250 V. For three-phase: V_line = W ÷ (√3 × I × PF). PF is essential — using PF = 1 for an inductive load gives a wrong voltage, often misleading you about supply requirements during commissioning.

Yes. If a piece of imported equipment is rated in current and power, divide W by A (and PF if AC) to confirm whether it matches your local supply voltage. Example: a US-built 110 V appliance drawing 9 A at 1000 W cannot run directly on 230 V without a transformer. Spotting voltage mismatches early saves expensive equipment damage and prevents fire risk during installation.