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Ohm’s Law and Power Calculator

Solve voltage, current, resistance and power from any two of them — and real and apparent power for single- or three-phase AC loads with a power factor.

Method reviewed October 26, 2026. Version 1.0.0. How it works

Your inputs

Result

Voltage
240V
Current
10A
Resistance
24Ω
Power
2,400W

240 V, 10 A, 24 Ω, 2,400 W.

Heat equivalent
8,189BTU/h
Show the working
  1. V = I × R and P = V × I
    240 V = 10 A × 24 Ω; P = 2,400 W

The relationships

  • Voltage = current × resistance (V = I × R).
  • Power = voltage × current (P = V × I), so also P = I² × R and P = V² ÷ R.

These hold exactly for resistive loads such as heating elements. For motors and other AC loads, real power also depends on the power factor: P = V × I × PF for single-phase, and P = √3 × V × I × PF for three-phase with line-to-line voltage.

What this does not do

Choosing wire sizes, breakers and overload protection depends on the equipment nameplate, conductor type, installation conditions and the electrical code adopted where you are. This calculator does the arithmetic only; it does not size circuits.

Worked example

These are the example values the calculator opens with, and the result the engine calculates for them.

Inputs

Circuit
Resistive (DC, or AC with no power factor)
I know
Voltage and current
Voltage
240 V
Current
10 A

Result

Voltage
240 V
Current
10 A
Resistance
24 Ω
Power
2,400 W
Heat equivalent
8,189 BTU/h

240 V, 10 A, 24 Ω, 2,400 W.

Assumptions

  • Resistive circuits: V = IR, P = VI. AC: real power = V × I × PF (× √3 for three-phase line-to-line voltage).

Limitations

  • For wire, breaker and overcurrent sizing, follow the equipment nameplate and the electrical code that applies; this tool does not size conductors or protection.

Frequently asked questions

How many watts is 1 amp at 240 volts?

For a resistive load, 240 watts. For a motor, multiply by its power factor as well.

What is the difference between kW and kVA?

kVA is apparent power (volts × amps); kW is the real power that does work. kW = kVA × power factor.

Why √3 for three-phase?

With balanced three-phase loads and line-to-line voltage, total power is √3 (about 1.732) × V × I × PF.

Sources

  1. University Physics Volume 2, §9.4 Ohm’s Law — OpenStax, Rice University (CC BY 4.0). V = IR; electrical power P = IV.
  2. Determining Electric Motor Load and Efficiency — U.S. DOE. Equation 1: three-phase input kW = V × I × PF × √3 ÷ 1,000 (worked example 469.7 V, 37 A, PF 0.763 → 22.9 kW); Equation 2: full-load input kW = hp × 0.7457 ÷ efficiency.

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