Electrical Power Calculator
Calculates electrical power in watts from voltage and current using P = V × I. Used by electricians, engineers, and DIYers to size fuses, cables, power supplies, and estimate appliance energy use.
Last updated: May 2026
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About this calculator
Electrical power is the rate at which energy is transferred or consumed in a circuit. The fundamental formula is P = V × I, where P is power in watts (W), V is voltage in volts (V), and I is current in amperes (A). Variables: voltage (V), current (A). Combined with Ohm's Law (V = I × R), this expands to two equivalent forms: P = I² × R (Joule's law form, useful when you know current and resistance) and P = V² / R (useful when you know voltage and resistance). One watt equals one joule of energy per second; energy use over time is then E = P × t (in joules, or in watt-hours if t is in hours). In practical terms, power determines how hot a resistor gets, how fast a motor spins, or how quickly a battery drains. Knowing the power consumption of a device is critical for selecting correctly rated wiring (ampacity), fuses, transformers, and power supplies. Edge cases: P = V × I gives apparent power (VA) in AC circuits with reactive loads. Real (true) power in AC is P = V × I × cos(φ), where φ is the phase angle between voltage and current — the cosine factor is called the 'power factor'. For purely resistive AC loads (heaters, incandescent bulbs), power factor is 1 and apparent power equals real power. For inductive or capacitive loads (motors, fluorescent lights), the power factor is less than 1, so apparent power exceeds the true useful power. Three-phase power formulas include a √3 multiplier. For DC and purely resistive AC, the simple V × I formula is exact.
How to use
Example 1: An appliance on a 120 V outlet drawing 5 A. Step 1: P = V × I = 120 × 5 = 600 W. Verify: a 600 W appliance running for 10 hours uses 600 × 10 = 6,000 Wh = 6 kWh of energy. Example 2: A 230 V AC kettle drawing 8.7 A (purely resistive heating element). Step 1: P = 230 × 8.7 = 2,001 W ≈ 2 kW. Verify: 2 kW is typical for an electric kettle. Energy to boil 1 L of water from 20 °C to 100 °C requires roughly 335 kJ = 0.093 kWh; at 2 kW that takes 0.093 / 2 hours = 0.0465 h ≈ 2.79 minutes — matches typical kettle boil times.
Frequently asked questions
How do I calculate the electrical power consumption of a device?
Electrical power consumption is calculated with P = V × I, where V is supply voltage in volts and I is the current draw in amperes. You can find the current on the device's nameplate, in the user manual, or measure it directly with a clamp meter on the live conductor. Multiply the two values to get watts. For example, a device running at 230 V and drawing 3 A consumes 690 W. Over time, energy use in kilowatt-hours equals power (kW) multiplied by hours of operation, which is what utility bills charge for. For AC devices with motors or transformers, also note the power factor — true power is P = V × I × PF, and PF ranges from 0.5 (highly inductive) to 1.0 (resistive).
What is the difference between power in watts and energy in kilowatt-hours?
Power (watts) measures the rate of energy use at any given moment; energy (kilowatt-hours) measures the total amount of energy consumed over time. One kilowatt-hour equals 1,000 watts running for one hour. A 500 W appliance running for 4 hours consumes 2 kWh of energy (0.5 kW × 4 h). Electricity bills charge for energy (kWh), not instantaneous power, so knowing your device's wattage helps you predict and manage costs — multiply wattage by typical run time per month and your tariff to estimate the monthly cost. The analogy: power is like speed (mph), energy is like distance traveled (miles).
Why is it important to know the power rating when choosing a fuse or cable?
Fuses and cables are rated for a maximum current they can carry safely before overheating or failing. Since P = V × I, knowing the power and voltage tells you the current: I = P / V. You then select a fuse rated slightly above that current to protect the circuit without nuisance tripping. Under-rated cables overheat (resistive losses are I²R, scaling with the square of current) and can cause fires; correctly sized cables and fuses ensure long-term reliability and code compliance. Always include a 25% safety margin for continuous loads per NEC 210.20(A) — for example, a 10 A continuous load requires a 12.5 A or larger circuit. Wire ampacity tables (NEC Chapter 9 Table 310) link AWG to maximum continuous current.
What are common mistakes when calculating electrical power?
Confusing apparent power (VA) with real power (W) — they differ by the power factor in AC circuits with inductive or capacitive loads. Using nameplate wattage when actual operating wattage is lower (many appliances spend most time in standby or partial load). Forgetting that the formula uses RMS voltage and current for AC, not peak values (peak is √2 × RMS). Mixing units — milliamps with volts gives milliwatts, not watts; check unit consistency. Ignoring the power factor when sizing a generator or transformer leads to under-capacity selection. For three-phase systems, forgetting the √3 multiplier: P = √3 × Vline × Iline × PF. Treating instantaneous power as the same as RMS average power on devices with cyclic loads (motors during startup draw 5–7× their running current).
When should I NOT use a simple V × I power calculator?
Three-phase AC systems use P = √3 × Vline × Iline × PF (line-to-line voltage) or P = 3 × Vphase × Iphase × PF (phase-to-neutral); the simple V × I formula gives wrong totals. Highly reactive AC loads (motors, transformers, fluorescent lights without electronic ballasts) have power factors below 1.0, so apparent power (VA) overstates real power (W) — use the PF correction. Switching power supplies and LED drivers have time-varying current that requires true-RMS measurement, not peak or average. Pulsed loads (welders, X-ray machines, laser pulses) require considering peak power separately from average. Battery applications use Ah × V to estimate energy, then divide by 1,000 for kWh — straight V × I gives only instantaneous power. Photovoltaic and rectified DC systems have voltage that varies with load and irradiance, requiring MPPT-aware analysis. For any large or safety-critical installation, follow IEEE/IEC standards and consult a qualified electrical engineer.