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For motors and inductive loads, use AC with PF
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Sources & Methodology
Formulas verified against IEEE Std 100 (IEEE Standard Dictionary of Electrical and Electronics Terms) and NIST SI units reference.
NIST SP 811 — Guide to SI Units
Official NIST reference for SI electrical unit definitions: watt (W), volt (V), ampere (A), and ohm, plus conversion factors for horsepower and BTU/hr.
All About Circuits — Power Calculations
Reference for Ohm's Law power formulas (P=VI, P=V²/R, P=I²R), worked examples, and the relationship between real power, reactive power, and apparent power in AC circuits.
Methodology: DC / resistive AC: P = V × I. AC with power factor: P = V × I × PF. Solve for I: I = P / (V × PF). Solve for V: V = P / (I × PF). Conversions: 1 kW = 1000 W. 1 hp = 745.7 W (mechanical). 1 BTU/hr = 0.29307 W. Apparent power VA = V × I (no power factor). Reactive power VAR = √(VA² − W²).
⏱ Last reviewed: April 2026
How Is Electrical Power in Watts Calculated?
A watt is the SI unit of power, defined as one joule of energy transferred per second. In electrical circuits, power is the rate at which energy is delivered or consumed. The fundamental relationship connecting power, voltage, and current is the basis of all electrical engineering calculations.
The Core Power Formula
P = V × I I = P / V V = P / I
P = power (watts, W) • V = voltage (volts, V) • I = current (amps, A)
Example 1 — Find power: 120V circuit drawing 10A: P = 120 × 10 = 1200 W = 1.2 kW
Example 2 — Find amps: 1500W kettle on 120V: I = 1500 / 120 = 12.5 A
Example 3 — AC with PF: 230V, 8A motor at PF 0.85: P = 230 × 8 × 0.85 = 1564 W
Example 1 — Find power: 120V circuit drawing 10A: P = 120 × 10 = 1200 W = 1.2 kW
Example 2 — Find amps: 1500W kettle on 120V: I = 1500 / 120 = 12.5 A
Example 3 — AC with PF: 230V, 8A motor at PF 0.85: P = 230 × 8 × 0.85 = 1564 W
All Four Ohm’s Law Power Formulas
| Known | Solve For | Formula | Example (R=10Ω) |
|---|---|---|---|
| V and I | P | P = V × I | 120V, 12A → 1440W |
| V and R | P | P = V² / R | 120V, 10Ω → 1440W |
| I and R | P | P = I² × R | 12A, 10Ω → 1440W |
| P and V | I | I = P / V | 1440W, 120V → 12A |
| P and I | V | V = P / I | 1440W, 12A → 120V |
| P and R | V | V = √(P × R) | 1440W, 10Ω → 120V |
Common Appliance Wattages
| Appliance | Typical Watts | Amps (120V) | Amps (240V) |
|---|---|---|---|
| LED light bulb | 9 W | 0.08 A | 0.04 A |
| Laptop computer | 65 W | 0.54 A | 0.27 A |
| Refrigerator | 200 W | 1.67 A | 0.83 A |
| Microwave oven | 1200 W | 10.0 A | 5.0 A |
| Electric kettle | 2000 W | 16.7 A | 8.3 A |
| Electric shower | 9500 W | 79.2 A | 39.6 A |
| Electric car charger (L2) | 7200 W | 60 A | 30 A |
Power Conversions
1 kW = 1000 W 1 hp = 745.7 W 1 BTU/hr = 0.293 W
1 MW = 1,000,000 W • 1 kWh = energy of 1 kW for 1 hour = 3,600,000 J
To convert watts to amps: divide by voltage. To convert amps to watts: multiply by voltage.
To convert watts to amps: divide by voltage. To convert amps to watts: multiply by voltage.
💡 Doubling voltage halves the current for the same power. A 1500W appliance on 120V draws 12.5A. The same 1500W on 240V draws only 6.25A. This is why high-power appliances (ovens, dryers, EV chargers) use 240V circuits — the lower current allows smaller, cheaper wiring and reduced resistive losses in conductors (P_loss = I² × R, so halving current reduces heat losses by 75%).
Frequently Asked Questions
Watts = Volts × Amps (P = V × I). For example, 120V × 10A = 1200W = 1.2 kW. For AC circuits with inductive loads, multiply by power factor: P = V × I × PF. A motor at 240V, 5A, PF 0.90: P = 240 × 5 × 0.90 = 1080W.
1 amp at 120V = 120W. 1 amp at 240V = 240W. 1 amp at 12V = 12W. Power is directly proportional to both voltage and current. A 15A, 120V household circuit supplies a maximum of 1800W; NEC limits continuous loads to 80% = 1440W.
Amps = Watts / Volts (I = P / V). A 1500W appliance on 120V: I = 1500/120 = 12.5A. The same 1500W on 240V draws only 6.25A. This formula determines the correct wire gauge and breaker size. For AC loads with a power factor: I = P / (V × PF).
Watts (W) = real power (actual work done). Volt-amperes (VA) = apparent power (V × I with no phase correction). For resistive loads, W = VA. For inductive loads, VA > W. Power factor = W/VA. UPS units and transformers are rated in VA because they must handle the full current regardless of power factor.
Divide watts by 1000. 2500W = 2.5 kW. To calculate electricity cost: kWh = kW × hours. A 2.5 kW heater for 4 hours = 10 kWh. At $0.15/kWh = $1.50 cost. Monthly cost of a continuously running 100W device = 0.1 kW × 720 hours = 72 kWh = $10.80 at $0.15/kWh.
1 mechanical horsepower = 745.7 watts. 1 electrical hp = 746W. 1 metric hp (PS) = 735.5W. A 5 hp motor output = 3730W. Motor electrical input is higher: a 5 hp motor at 90% efficiency draws 3730/0.90 = 4144W = 4.14 kW from the supply.
A motor at 120V, 10A with PF 0.85 uses only 1020W of real power, not 1200W. The circuit must still carry 10A (1200VA), so wiring and breakers are sized for 10A. The 15% reactive portion does no useful work but heats conductors. Poor PF wastes electrical infrastructure capacity and may incur utility surcharges in industrial settings.
P = V²/R and P = I²×R. For a 10Ω resistor at 120V: P = 120²/10 = 1440W. For 12A through 10Ω: P = 144×10 = 1440W. The I²R formula is used to calculate heat losses in wiring: a 1Ω wire carrying 20A dissipates 400W as heat — which is why wire sizing is critical.
Three-phase power: P = V_line × I_line × 1.732 × PF. For 480V, 50A, PF 0.85: P = 480 × 50 × 1.732 × 0.85 = 35,362W = 35.4 kW. The factor 1.732 (square root of 3) accounts for the three phases each contributing to total power. Three-phase is more efficient for high-power loads than single-phase.
Typical wattages: LED bulb 7–15W, phone charger 5–20W, laptop 45–100W, refrigerator 100–400W, microwave 700–1500W, hair dryer 1200–2000W, kettle 1500–3000W, washing machine 500–2000W, air conditioner 750–3500W, electric oven 2000–5000W. Always check the appliance nameplate for exact wattage.
Calculate current: I = P / V. For a 2400W, 120V circuit: I = 20A. Per NEC, continuous loads must not exceed 80% of breaker rating, so the breaker must be at least 20/0.80 = 25A. Select the next standard size: 30A breaker. For 240V, 5000W oven: I = 20.8A, minimum breaker = 26A, use 30A breaker.
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