Electrical Power Calculator
Calculate electrical power (P = VI), voltage, current, and resistance. Includes AC apparent power, power factor, and three-phase power calculations.
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Enter your values above to see the results.
Tips & Notes
- ✓DC power P = V × I = I²R = V²/R. For AC resistive loads, use RMS values — the result is real power (watts). A 120 V AC circuit drawing 5 A through a pure resistor: P = 120 × 5 = 600 W real power.
- ✓Power factor (PF) accounts for reactive loads. Real power (W) = Apparent power (VA) × PF. An air conditioner rated 1,200 VA with PF = 0.85 draws 1,020 W of real power but 1,200 VA from the supply.
- ✓Three-phase power: P = √3 × V_line × I_line × PF for balanced loads. A 480 V three-phase motor drawing 25 A at PF = 0.90: P = 1.732 × 480 × 25 × 0.90 = 18,686 W ≈ 18.7 kW.
- ✓Power dissipated in resistors converts entirely to heat. A 10 Ω resistor carrying 2 A dissipates P = 4 × 10 = 40 W — this is why high-power components need heatsinks and forced air cooling.
- ✓Energy = Power × Time. 1 kWh = 1 kW running for 1 hour = 3,600,000 J. A 100 W device running 8 hours uses 0.8 kWh. At $0.15/kWh, cost = 0.8 × 0.15 = $0.12/day, $43.80/year.
Common Mistakes
- ✗Confusing VA (volt-amperes) with W (watts) for AC loads — they are equal only for purely resistive loads. Inductive loads (motors, transformers) have VA > W due to reactive power. Circuits must be sized for VA, not just W.
- ✗Using peak voltage instead of RMS for AC power calculations — power calculations require RMS values. Using 170 V peak instead of 120 V RMS for a resistive load overstates power by a factor of 2.
- ✗Forgetting power factor in motor sizing — a motor nameplate showing 5 kW at PF = 0.85 requires 5/0.85 = 5.88 kVA of apparent power. The circuit must be rated for this higher current demand.
- ✗Ignoring line losses in distribution systems — a cable carrying 20 A with 0.5 Ω total resistance dissipates P = 20² × 0.5 = 200 W. At high power levels, cable losses are a significant cost and heat management concern.
- ✗Treating power as always positive — reactive power is negative (capacitive) or positive (inductive). Only real power does useful work. Reactive power oscillates between source and load without net energy transfer.
Electrical Power Calculator Overview
Electrical power is the rate at which energy is transferred or consumed in a circuit. Understanding power — and the distinction between watts, volt-amperes, and volt-amperes reactive — is essential for safe circuit design, conductor sizing, equipment selection, and energy cost estimation.
DC power formulas:
P = V × I = I²R = V²/R | Units: watts (W), kilowatts (kW), megawatts (MW)
EX: 48 V DC system, 8 Ω load → I = 48/8 = 6 A → P = 48 × 6 = 288 W → Energy in 24 hr = 0.288 kW × 24 = 6.912 kWhAC power — real, reactive, and apparent:
S (VA) = V_RMS × I_RMS | P (W) = S × PF = S × cos(φ) | Q (VAR) = S × sin(φ) | PF = P/S
EX: Motor 230 V, 10 A, PF = 0.85 → S = 230 × 10 = 2,300 VA → P = 2,300 × 0.85 = 1,955 W real power → Q = 2,300 × sin(arccos(0.85)) = 2,300 × 0.527 = 1,212 VAR reactivePower conversion — key equivalences:
| Unit | Watts | kW | Horsepower | BTU/hr |
|---|---|---|---|---|
| 1 Watt | 1 W | 0.001 kW | 0.001341 hp | 3.412 BTU/hr |
| 1 Kilowatt | 1,000 W | 1 kW | 1.341 hp | 3,412 BTU/hr |
| 1 Horsepower | 745.7 W | 0.7457 kW | 1 hp | 2,545 BTU/hr |
| 1 BTU/hr | 0.293 W | 0.000293 kW | 0.000393 hp | 1 BTU/hr |
| Appliance | Power (W) | Daily Use | Monthly kWh | Monthly Cost ($0.15) |
|---|---|---|---|---|
| LED bulb | 10 W | 8 hr | 2.4 kWh | $0.36 |
| Laptop computer | 50 W | 8 hr | 12 kWh | $1.80 |
| Refrigerator | 150 W avg | 24 hr | 108 kWh | $16.20 |
| Electric water heater | 4,000 W | 2 hr | 240 kWh | $36.00 |
| Central AC (3-ton) | 3,500 W | 8 hr | 840 kWh | $126.00 |
| Electric vehicle (charging) | 7,200 W | 8 hr | 1,728 kWh | $259.20 |
Frequently Asked Questions
DC power: P = V × I, where P is in watts, V in volts, and I in amperes. Equivalently: P = I²R = V²/R. Example: 24 V DC supply powering a 4 Ω heater element → I = 24/4 = 6 A → P = 24 × 6 = 144 W. Or P = 6² × 4 = 144 W. For AC resistive loads, use RMS values: a 240 V RMS AC circuit powering a 960 Ω element → I = 240/960 = 0.25 A → P = 240 × 0.25 = 60 W.
Real power (W, watts): the power actually doing useful work — heating, lighting, mechanical work. Reactive power (VAR, volt-amperes reactive): power oscillating between source and reactive components (inductors, capacitors) — does no net work but loads the circuit. Apparent power (VA): the product of RMS voltage and RMS current regardless of phase — what determines conductor and transformer sizing. Power factor (PF) = Real power / Apparent power = cos(φ), where φ is the phase angle between voltage and current. For motors: PF ≈ 0.7-0.9. For purely resistive loads: PF = 1.0.
For balanced three-phase loads: Real power P = √3 × V_line × I_line × PF = 3 × V_phase × I_phase × PF. Apparent power S = √3 × V_line × I_line. Reactive power Q = √3 × V_line × I_line × sin(φ). Example: 480 V three-phase, 30 A line current, PF = 0.88 → P = 1.732 × 480 × 30 × 0.88 = 21,907 W ≈ 21.9 kW. The √3 factor comes from the 120° phase angle between the three phases.
1 kW = 1,000 W = 1.341 hp = 3,412 BTU/hr. 1 hp = 745.7 W = 0.7457 kW = 2,545 BTU/hr. 1 BTU/hr = 0.2931 W. Examples: a 100 hp electric motor = 74,570 W = 74.57 kW. A 3-ton air conditioner = 3 × 12,000 BTU/hr = 36,000 BTU/hr = 10,551 W ≈ 10.55 kW. A 750 W microwave draws: 750/120V = 6.25 A from a 120 V outlet.
Power factor (PF) = cos(φ) = Real power (W) / Apparent power (VA), ranging from 0 to 1. PF = 1.0 for purely resistive loads (heaters, incandescent bulbs). PF = 0.5-0.9 for inductive loads (motors, fluorescent ballasts, switching power supplies). Low PF means more current flows for the same real power delivered, which increases conductor losses, requires larger cables and transformers, and can incur utility penalties for industrial users. Capacitor banks are added to industrial facilities to correct poor power factor and reduce apparent power demand.
Energy (kWh) = Power (kW) × Time (hours). Cost = Energy (kWh) × Rate ($/kWh). Example: a 2.5 kW electric space heater running 6 hours/day for 30 days: Energy = 2.5 × 6 × 30 = 450 kWh. At $0.15/kWh: Cost = 450 × 0.15 = $67.50/month. For multiple appliances: sum their watt-hours (W × hours) for each, convert to kWh by dividing by 1,000, then multiply by your rate. Check your electricity bill for the exact rate per kWh — it varies by location from $0.08 to $0.35+.