1 watt is 1 joule per second. In circuit terms, 1 W = 1 V·A.

What is power factor and reactive power?

Power factor is the ratio of real power delivered to apparent power supplied, PF = P/S where S = V times I. Reactive power is the gap between them, Q equals the square root of S squared minus P squared, measured in VAR.

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Electrical Power Calculator

Q: Can I use this for AC circuits?

For a purely resistive AC load, treat RMS values as DC and use P = V times I. When voltage and current are out of phase, use the P = V times I times PF method, which accounts for the gap between real and apparent power.

Solve for power (W), voltage (V), current (A), resistance (Ω), or power factor using P = V·I, P = I²R, P = V²/R, Ohm's Law (V = I·R), or P = V·I·PF for AC circuits — with unit conversion, a method-specific diagram, and a plain-language read on what the number actually means.

Background

Electrical power tells you how fast electrical energy is used or delivered. For simple DC circuits, power links voltage and current: P = V·I. Using Ohm's Law V = I·R, you can also write P = I²R and P = V²/R — the same relationship, rearranged for whichever two quantities you already know. Those three formulas assume voltage and current stay in phase (a purely resistive load) — for AC circuits where they don't, use P = V·I·PF, which accounts for the gap between real and apparent power using the power factor (PF = cos φ).

Set up your calculation

Step 1 — Which formula fits your circuit?

Pick the relationship that matches what you're given.

Use Ohm's Law when you want V, I, or R directly from the other two. Use a P-form when power is involved. Use P = V·I·PF for AC circuits where voltage and current are out of phase.

Step 2 — Solve for

Options that aren't solvable with your chosen method are disabled automatically.

Step 3 — Enter your values

The field you're solving for is grayed out — fill in the rest.

Power (P)

Voltage (V)

Current (I)

Resistance (R)

Must be > 0 when used in division or square roots.

Power Factor (PF = cos φ)

A decimal between 0 and 1. This model treats PF as a magnitude only — it doesn't distinguish leading vs. lagging.

Learning options

Show step-by-step solution

Show visual breakdown

Quick picks (click to calculate)

Result

No results yet. Enter values and click Calculate.

How to use this calculator

Pick a method: P = V·I, P = I²R, P = V²/R, V = I·R, or P = V·I·PF.
Choose what you want to solve for (P, V, I, R, or PF) — the input fields adjust to match.
Enter the other required values in whatever units you have; everything converts internally.
Click Calculate to see the result, unit conversions, a method-specific diagram, and a callout on what the number actually means.

How this calculator works

All inputs are converted to consistent base units first — watts, volts, amps, and ohms — so mixing kΩ with mA is never a problem.

The selected variable is solved algebraically from the chosen equation, using whichever two (or three, for the AC method) quantities you supplied.

Ohm's Law connects the three P-forms: substituting V = I·R into P = V·I gives both P = I²R and P = V²/R, so all three describe the same physics from different starting points.

For the AC power-factor method, apparent power S = V·I (in VA), reactive power Q = √(S²−P²) (in VAR), and the phase angle φ = cos⁻¹(PF) are all computed alongside the real power, so you can see the full power triangle.

Common unit equivalents are shown automatically (W ↔ kW ↔ hp, Ω ↔ kΩ ↔ MΩ, and so on) and a plain-language callout grounds the result against a familiar comparison.

Formulas & Equations Used

P = V·I

P = I²·R

P = V² / R

V = I·R

P = V·I·PF, where PF = cos φ = P/S and S = V·I is the apparent power (VA)

Q = √(S² − P²) — reactive power (VAR)

Identity: 1 W = 1 V·A

Example Problems & Step-by-Step Solutions

Example 1 — P = V·I

A charger outputs 5 V at 2 A.

Step: P = 5 × 2 = 10 W.

Result: 10 W.

Example 2 — P = V²/R

A device runs at 12 V with 6 Ω resistance.

Step: P = 12² / 6 = 144 / 6.

Result: 24 W.

Example 3 — P = I²R

A current of 0.5 A flows through a 10 Ω resistor.

Step: P = (0.5)² × 10 = 0.25 × 10.

Result: 2.5 W.

Example 4 — P = V·I·PF

An AC motor draws 5 A at 120 V with PF 0.9.

Step: S = 120×5 = 600 VA; P = 600 × 0.9.

Result: 540 W (Q ≈ 261.5 VAR).

Frequently Asked Questions

What is 1 watt?

1 watt means 1 joule per second. In circuit terms, 1 W = 1 V·A.

Can I use this for AC circuits?

For a purely resistive AC load (voltage and current in phase), treat RMS values as you would DC and use P = V·I. When voltage and current are out of phase, use P = V·I·PF instead — it accounts for the gap between real and apparent power, though it doesn't compute the phase angle's sign (leading vs. lagging) directly.

What is power factor, and what is reactive power?

Power factor (PF = cos φ) is the ratio of real power actually delivered to the apparent power supplied. The gap between them is reactive power, Q = √(S²−P²), measured in VAR — power that sloshes back and forth without doing net work, typically from inductive or capacitive loads.

Why are there multiple formulas for power?

Because combining P = V·I with Ohm's Law (V = I·R) gives P = I²R and P = V²/R — the same relationship rearranged for whichever two quantities you already know.

What if I enter kΩ or mA?

That's fine — the calculator converts everything to base units (W, V, A, Ω) internally before solving.

Why does the calculator compare my result to everyday devices?

A raw number like "540 W" is hard to picture. Comparing it to something familiar — a hair dryer, a laptop charger, a household circuit breaker — gives you a quick sanity check on whether the number is even in the right ballpark.