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Electrical power in watts Enter a valid wattage.
Supply voltage Enter a valid voltage.
Select circuit type Select type.
Current in amps Enter a valid current.
Supply voltage Enter a valid voltage.
Select circuit type Select type.
Power in kilowatts Enter a valid KW value.
Line voltage (phase-to-phase for 3-phase) Enter a valid voltage.
Select circuit type Select type.
Generator or transformer KVA rating Enter a valid KVA value.
Line voltage Enter a valid voltage.
KVA is an AC power measurement Select type.
Used to calculate real KW from KVA Enter PF 0.01–1.0.
Supply or source voltage Enter a valid voltage.
Select second known value Select known value.
Power consumed by the load Enter a valid value.
Current in amps Enter a valid current.
Select second known value Select known value.
Power consumed by the load Enter a valid value.
Result
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⚠️ Disclaimer: Results use standard electrical engineering formulas (P = VI, Ohm's Law). For AC circuits, power factor must be known for accurate results. Always verify with a licensed electrician for safety-critical applications.

📚 Sources & Methodology

All conversion formulas are derived from and verified against the following authoritative references:

Watts, Amps & Volts — Complete Conversion Guide

Watts to Amps: The Core Formula

The relationship between watts, amps, and volts comes directly from Ohm's Law and the power formula. Power (watts) equals voltage times current: P = V × I. To convert watts to amps you divide: I = P / V. This works perfectly for DC circuits and resistive AC loads (heaters, incandescent lights). For inductive AC loads (motors, compressors, transformers), you must include the power factor.

The most common error people make is using the simple formula for motor loads. A 2,000-watt motor at 120V does not draw 16.67 amps — it draws more, because the power factor (typically 0.80–0.90) means the apparent current is higher than the real power suggests. Use the AC with PF mode above for accurate motor calculations.

Watts to Amps Formula — All Circuit Types
DC / Resistive AC: I = P / V Single-phase AC with PF: I = P / (V x PF) 3-phase AC with PF: I = P / (V x 1.732 x PF) Example (DC): 1200W / 120V = 10.0 A Example (1ph): 1200W / (120V x 0.85) = 11.76 A Example (3ph): 1200W / (208V x 1.732 x 0.85) = 3.92 A per phase

Amps to Watts Conversion

To convert amps to watts, multiply amps by volts: P = I × V. For AC circuits with power factor: P = I × V × PF. For 3-phase: P = I × V × 1.732 × PF. A 20-amp 240V circuit with PF=1.0 carries 4,800 watts. The same circuit powering an inductive load at PF=0.85 delivers only 4,080 watts of real power even though 4,800 VA of apparent power is flowing.

KW to Amps — Kilowatt Conversion

KW (kilowatts) is real power — multiply by 1,000 to get watts, then divide by voltage and power factor. For 3-phase systems, also divide by 1.732 (square root of 3). This is the most common calculation for motor starters, panel sizing, and generator load calculations. Always use the nameplate power factor from the motor or equipment data sheet for accuracy.

KW to Amps Formula
Single-phase: I = (KW x 1000) / (V x PF) 3-phase: I = (KW x 1000) / (V x 1.732 x PF) DC: I = (KW x 1000) / V Example: 10 KW 3-phase at 480V, PF=0.90: I = 10000 / (480 x 1.732 x 0.90) = 13.37 A per phase

KVA to Amps — Generator and Transformer Sizing

KVA (kilovolt-amperes) is apparent power — the total power drawn from the source including reactive power. Generators and transformers are rated in KVA because they must supply all apparent power. Converting KVA to amps does not require power factor: I = (KVA × 1000) / V for single-phase, or I = (KVA × 1000) / (V × 1.732) for 3-phase. To find real KW: KW = KVA × PF.

Quick Conversion Reference Tables

Watts to Amps at 120V (PF=1.0)
500W4.17 A
750W6.25 A
1000W8.33 A
1200W10.00 A
1500W12.50 A
1800W15.00 A
2000W16.67 A
2400W20.00 A
Watts to Amps at 240V (PF=1.0)
1000W4.17 A
1500W6.25 A
2000W8.33 A
3000W12.50 A
4000W16.67 A
5000W20.83 A
7200W30.00 A
9600W40.00 A
KW to Amps at 240V 1-Phase (PF=0.85)
1 KW4.90 A
2 KW9.80 A
3 KW14.71 A
5 KW24.51 A
7.5 KW36.76 A
10 KW49.02 A
15 KW73.53 A
20 KW98.04 A
KVA to Amps — Single & 3-Phase
5 KVA @ 120V 1ph41.67 A
5 KVA @ 240V 1ph20.83 A
10 KVA @ 208V 3ph27.76 A
10 KVA @ 480V 3ph12.02 A
25 KVA @ 208V 3ph69.39 A
50 KVA @ 480V 3ph60.11 A
100 KVA @ 480V 3ph120.3 A
500 KVA @ 480V 3ph601.4 A

Why Power Factor Changes Everything

Power factor (PF) is the ratio of real power (watts) to apparent power (VA). A purely resistive load has PF=1.0. Inductive loads like motors, compressors, and fluorescent lighting have PF between 0.7 and 0.95. A low power factor means more current is drawn than necessary to deliver the real power. This increases wire heating (I²R losses), requires larger conductors, and can trigger utility power factor penalties for commercial customers.

💡
Practical tip: When sizing a breaker for a motor, always use the motor nameplate full-load current (FLC) rather than calculating from watts and voltage. NEC Article 430 Table 430.248/250 provides standard FLC values. The nameplate accounts for real operating power factor and efficiency better than any formula.

3-Phase Power: The 1.732 Factor Explained

Three-phase power uses three conductors with voltages 120° apart. The phase relationship means total power is P = V × I × 1.732 × PF, where 1.732 is the square root of 3. A 3-phase 480V 100A circuit delivers P = 480 × 100 × 1.732 × 0.9 = 74,966W (75 KW). The same 100A at single-phase 480V: P = 480 × 100 × 0.9 = 43,200W (43.2 KW). Three-phase delivers 73% more power for the same current — which is why all industrial motors use 3-phase power.

❓ Frequently Asked Questions

Amps = Watts / Volts (I = P/V) for DC and resistive AC. For AC with power factor: I = P / (V x PF). For 3-phase: I = P / (V x 1.732 x PF). Example: 1800W at 120V = 15A. A 1800W motor at 120V with PF=0.85: I = 1800/(120 x 0.85) = 17.65A. Always include PF for motors and other inductive loads.
Watts = Amps x Volts (P = I x V) for DC and resistive AC. With PF: W = I x V x PF. 3-phase: W = I x V x 1.732 x PF. Example: 15A at 120V = 1800W. A motor drawing 15A at 240V, PF=0.85: W = 15 x 240 x 0.85 = 3060W real power, but apparent power VA = 15 x 240 = 3600 VA.
Single-phase: I = (KW x 1000) / (V x PF). 3-phase: I = (KW x 1000) / (V x 1.732 x PF). DC: I = (KW x 1000) / V. Example: 7.5 KW 3-phase motor at 208V, PF=0.85: I = 7500/(208 x 1.732 x 0.85) = 24.5 amps per phase. Use the KW to Amps tab above for instant results at any voltage.
Single-phase: I = (KVA x 1000) / V. 3-phase: I = (KVA x 1000) / (V x 1.732). KVA to amps does not require power factor. Example: 25 KVA 3-phase transformer at 480V: I = 25000/(480 x 1.732) = 30.05 amps. To find real KW: KW = KVA x PF.
1000W at 120V (PF=1.0) = 1000/120 = 8.33 amps. For a motor at PF=0.85: 1000/(120 x 0.85) = 9.80 amps. Use the Watts to Amps tab above for any wattage and voltage combination, with or without power factor.
Watts (W) is real power - the actual useful energy delivered per second. Volt-amps (VA) is apparent power - volts x amps regardless of phase angle. For resistive loads: W = VA (PF=1.0). For inductive loads: W = VA x PF, always less than VA. UPS systems and generators are rated in VA/KVA because they supply all apparent power. Your electricity bill is in KWh (real energy) not KVAh.
20A x 120V = 2,400W total capacity. For continuous loads (NEC 80% rule): maximum 16A x 120V = 1,920W. A dedicated 20A circuit for a refrigerator (120V, 6A, 720W) is well within limits. A 20A circuit for a microwave (1,200-1,500W) plus other loads approaching 1,920W continuous would require its own dedicated circuit.
You cannot convert volts to amps without knowing power (watts) or resistance (ohms). With power: I = P/V. With resistance: I = V/R (Ohm's Law). Example: 12V with 600W load: I = 600/12 = 50A. 24V across 8 ohms: I = 24/8 = 3A. Use the Volts to Amps tab above for both methods.
1.732 is the square root of 3, accounting for the phase relationship in 3-phase power systems. Three conductors carry voltages 120 degrees apart, so total power is not simply 3x single-phase power at the same voltage. P = V x I x sqrt(3) x PF. The 1.732 factor applies to line-to-line voltage calculations. It makes 3-phase systems 73% more efficient than single-phase at the same current and voltage.
At 120V: 1500/120 = 12.5 amps. At 240V: 1500/240 = 6.25 amps. Electric heaters are resistive loads (PF=1.0), so no power factor correction is needed. A 1500W heater at 120V exceeds 80% of a 15A circuit (12A limit for continuous loads), so NEC requires a dedicated 20A circuit. The same heater at 240V needs only a 10A circuit.
3-phase watts: P = I x V x 1.732 x PF. Example: 50A 3-phase at 480V, PF=0.90: P = 50 x 480 x 1.732 x 0.90 = 37,483W = 37.48 KW. Without power factor (apparent power): VA = I x V x 1.732 = 50 x 480 x 1.732 = 41,568 VA = 41.57 KVA. Use the Amps to Watts tab above for any combination.
Lower power factor = higher amps for the same real power. At PF=1.0: 1000W at 120V draws 8.33A. At PF=0.85: same 1000W draws 9.80A (18% more current). At PF=0.7: draws 11.9A (43% more). This extra current heats wires without delivering useful work. Power factor correction capacitors bring PF closer to 1.0 for motors, reducing wasted current and lowering electricity costs.

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