Electrical Power Calculator

Electrical power (P) is the rate at which electrical energy is transferred or consumed, measured in watts (W) or kilowatts (kW). The fundamental formula is P = V × I, where V is voltage (volts) and I is current (amperes). Combined with Ohm's Law (V = I × R), this gives a family of formulas for any combination of P, V, I, and R. For AC circuits, true power also depends on the power factor (cos φ) — the ratio of real power to apparent power — because inductors and capacitors cause voltage and current to be out of phase. Three-phase systems multiply power by √3 for line voltage calculations.

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input Known Values

Enter any two values — the rest will be calculated.

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Real Power (P)

1,200 W

1.200 kW

Apparent Power (S)

1,200 VA

Voltage (V)

120 V

Current (A)

10 A

Resistance (Ω)

12 Ω

lightbulb Tips

  • P = V × I (DC) or V × I × PF (AC)
  • Three-phase power = √3 × V × I × PF ≈ 1.732×
  • Size breakers at 125% of continuous load (NEC)
  • kW = kVA × Power Factor

How to Use This Calculator

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Select Circuit Type

Choose DC, AC Single-Phase, or AC Three-Phase to match your circuit. For AC circuits, a power factor field will appear to account for reactive loads.

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Choose What to Solve For

Select whether you want to calculate Power (W), Voltage (V), Current (A), or Resistance (Ω). The input fields will update to ask for the two values you already know.

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Enter Known Values

Type in the known electrical values such as voltage and current. For AC circuits, enter the power factor (0–1) to get accurate real power results.

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Read Your Results

Instant results show real power (W), apparent power (VA), reactive power (VAR), and the solved unknown. Use these values to size wiring, breakers, or generators correctly.

The Formula

In DC circuits, P = V × I is exact. In AC circuits, voltage and current may be out of phase by angle φ, so real power P = V × I × cos(φ) where cos(φ) is the power factor. Apparent power (S, in VA) is the product V × I regardless of phase. Reactive power (Q, in VAR) represents energy stored and released by inductors/capacitors. These three form the power triangle: S² = P² + Q².

P = V × I = I² × R = V² / R | AC: P = V × I × PF | 3φ: P = √3 × V_L × I_L × PF

lightbulb Variables Explained

  • P Real power in watts (W) or kilowatts (kW)
  • V Voltage in volts (V)
  • I Current in amperes (A)
  • R Resistance in ohms (Ω)
  • PF Power factor (0–1) — ratio of real power to apparent power (AC only)
  • S Apparent power in volt-amperes (VA) = V × I
  • Q Reactive power in VAR = √(S² − P²)

tips_and_updates Pro Tips

1

A power factor below 0.9 on industrial equipment often incurs utility surcharges — consider power factor correction capacitors.

2

For safety, always size circuit breakers at 125% of the continuous load (NEC 210.20).

3

Three-phase systems carry √3 ≈ 1.732 times more power than single-phase at the same voltage and current.

4

kW × 1000 = W; kVA × 1000 = VA — watch units when mixing large appliances.

5

Efficiency = Output Power ÷ Input Power × 100%. A 90% efficient motor drawing 1000 W delivers only 900 W of mechanical work.

Electrical power — measured in watts — is the rate at which electrical energy is consumed or produced, governed by the fundamental relationship P = V × I (power equals voltage times current). Understanding these relationships is essential for sizing circuits, selecting wire gauges, choosing breakers, and estimating energy costs. A 1,500-watt space heater on a 120V circuit draws 12.5 amps — close to the 15A limit of a standard household circuit, which is why running a heater with other appliances on the same circuit often trips breakers. Our electrical power calculator converts between watts, volts, amps, and ohms using Ohm's law and the power formula, handles both AC and DC calculations (including power factor for AC circuits), and computes energy consumption in kilowatt-hours for cost estimation. Whether you are sizing a generator, checking circuit capacity, or calculating electricity costs, this tool provides instant answers with the underlying formulas displayed.

The power triangle: watts, volts, and amps

Four key relationships connect power (P in watts), voltage (V in volts), current (I in amps), and resistance (R in ohms): P = V × I, P = I²R, P = V²/R, and V = IR (Ohm's law). Knowing any two values lets you calculate the other two. A 60-watt light bulb on 120V draws I = 60/120 = 0.5 amps with resistance R = 120/0.5 = 240 ohms. For DC circuits, these formulas are straightforward. For AC circuits, the power factor (PF) affects the relationship: real power (watts) = V × I × PF. A motor drawing 10 amps at 240V with 0.85 power factor consumes 240 × 10 × 0.85 = 2,040 watts of real power, though it appears to draw 2,400 VA (volt-amps) from the utility.

Circuit capacity and breaker sizing

Standard US residential circuits: 15A at 120V provides 1,800 watts maximum capacity, 20A at 120V provides 2,400 watts, and 30A at 240V provides 7,200 watts. The NEC recommends loading circuits to no more than 80% of breaker rating for continuous loads (running more than 3 hours): a 15A circuit should not continuously draw more than 12A (1,440 watts). Common appliance loads: microwave 1,000-1,500W, hair dryer 1,200-1,800W, toaster 800-1,500W, refrigerator 100-400W (running), window AC 500-1,500W, electric dryer 4,000-5,500W (requires 30A 240V circuit), and electric range 8,000-12,000W (requires 40-50A 240V circuit). Exceeding circuit capacity causes breaker trips — a safety feature preventing wire overheating and fire.

Energy consumption and cost calculation

Energy (kilowatt-hours) = power (watts) × time (hours) / 1,000. Cost = kWh × electricity rate. A 1,500W space heater running 8 hours daily consumes 12 kWh/day, costing $1.44/day at $0.12/kWh ($43.20/month). Standby power (phantom loads) from electronics in standby mode consumes 5-10% of household electricity — a TV on standby draws 5-15W, a game console 10-25W, and a cable box 15-30W. Collectively, phantom loads cost the average household $100-200 annually. LED bulbs have revolutionized lighting costs: a 10W LED produces the same light as a 60W incandescent, saving $8-10/year per bulb at average usage. Replacing 30 incandescent bulbs saves $250-300 annually in electricity costs.

Frequently Asked Questions

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All formulas verified against official standards.