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Volt to Watt (W) Conversion Calculator

Complete guide to converting voltage (V) and current (A) to electrical power (W). Learn formulas for DC, AC single-phase, and three-phase systems with practical examples.

Enginist Engineering Team
Professional electrical engineers with expertise in power systems, circuit design, and electrical code compliance.
Reviewed by PE-Licensed Electrical Engineers
Published: October 17, 2025
Updated: November 9, 2025

Volt to Watt Conversion Guide

Quick AnswerHow do you convert volts to watts?
Convert volts to watts using voltage and current with appropriate formulas for DC, AC single-phase, or three-phase systems. Current must be known—voltage alone is insufficient.
P=V×IP = V \times I (DC) or P=V×I×PFP = V \times I \times PF (AC)
Example

DC circuit: 12V × 5A = 60W. AC circuit at 230V, 10A, PF=0.85: P = 230 × 10 × 0.85 = 1,955W.

Introduction

Every electrical calculation begins with understanding how voltage and current combine to produce power. This fundamental relationship—the product of electrical pressure and current flow—determines how much energy equipment consumes and how electrical systems must be designed.

Why This Conversion Matters

In practical electrical work, you constantly need to translate voltage and current measurements into power values. From verifying that a 230V outlet can support a new appliance to calculating the total load on an industrial distribution panel, power calculations drive every design decision. Understanding volt-to-watt conversion enables you to size transformers correctly, verify equipment compatibility with available supply voltage, calculate energy costs for operational budgets, and ensure circuits aren't overloaded.

The Fundamental Challenge

A critical distinction separates this conversion from simple unit translations: you cannot convert volts to watts using voltage alone. Power requires both voltage (the driving force) and current (the actual flow). A 120V outlet means nothing for power until you know how much current a load draws. Furthermore, AC systems introduce power factor—the phase relationship between voltage and current that determines what fraction of apparent power performs actual work. This guide addresses why the same voltage can result in vastly different power values depending on current and power factor.

What You'll Learn

This guide provides complete volt-to-watt conversion methodology for all electrical system types. You'll master the fundamental power equation P=V×IP = V \times I and its variations for DC, AC single-phase, and AC three-phase systems per IEC 60038 voltage standards. Practical examples demonstrate real-world calculations from residential circuits to industrial motor installations. Reference tables cover standard voltage levels worldwide, typical power factors for common loads, and NEC/IEC compliance requirements.

Quick Answer: How Do You Convert Volts to Watts?

Convert voltage and current to power using P=V×IP = V \times I for DC, P=PF×V×IP = PF \times V \times I for AC single-phase, or P=3×PF×VLL×IP = \sqrt{3} \times PF \times V_{\text{LL}} \times I for AC three-phase. Note: You must know current (amps) to calculate power from voltage.

Core Formula

DC Systems: P(W)=V(V)×I(A)P(\text{W}) = V(\text{V}) \times I(\text{A})

AC Single-Phase: P(W)=PF×V(V)×I(A)P(\text{W}) = PF \times V(\text{V}) \times I(\text{A})

AC Three-Phase (Line-to-Line): P(W)=3×PF×VLL(V)×I(A)P(\text{W}) = \sqrt{3} \times PF \times V_{\text{LL}}(\text{V}) \times I(\text{A})

Where:

  • PP = Power (watts, W)
  • VV = Voltage (volts, V)
  • II = Current (amperes, A)
  • PFPF = Power factor (0 to 1)
  • 31.732\sqrt{3} \approx 1.732 (three-phase factor)

Worked Example

AC Single-Phase: 230V at 10A, PF=0.9

Given:

  • Voltage: V=230V = 230 V
  • Current: I=10I = 10 A
  • Power factor: PF=0.9PF = 0.9

Step 1: Apply AC single-phase formula: P=PF×V×I=0.9×230×10=2,070 WP = PF \times V \times I = 0.9 \times 230 \times 10 = 2,070 \text{ W}

Result: The load consumes 2,070 watts (2.07 kW).

AC Three-Phase: 400V Line-to-Line at 25A, PF=0.85

Given:

  • Line-to-line voltage: VLL=400V_{\text{LL}} = 400 V
  • Current: I=25I = 25 A
  • Power factor: PF=0.85PF = 0.85

Step 1: Apply three-phase formula: P=3×PF×VLL×I=1.732×0.85×400×25=14,740 WP = \sqrt{3} \times PF \times V_{\text{LL}} \times I = 1.732 \times 0.85 \times 400 \times 25 = 14,740 \text{ W}

Result: The three-phase load consumes 14,740 watts (14.74 kW).

Reference Table

ParameterTypical RangeStandard
Residential Voltage (US)120V, 240VNEC 210.6, IEC 60038
Residential Voltage (EU)230VIEC 60038
Commercial Voltage (US)208V, 277V, 480VNEC 210.6
Commercial Voltage (EU)400V (three-phase)IEC 60038
Power Factor (Resistive)1.0Typical
Power Factor (Motors)0.85-0.90Typical
Power Factor (Fluorescent)0.6-0.8Typical

Key Standards

Understanding Power, Voltage, and Current

Key Definitions

Potential (V): The electrical potential difference between two points, measured in volts (V). It represents the "pressure" that pushes electric amp through a conductor.

Electric current (I): The flow of electric charge, measured in amperes (A). It represents the quantity of electrons moving through a conductor.

Load (P): The rate at which electrical energy is consumed or produced, measured in watts (W). It represents the actual work being done by the electrical system.

Capacity Factor (PF): In AC systems, the ratio of real energy to apparent electrical power. It ranges from 0 to 1, where 1 indicates purely resistive load and values < 1 indicate reactive components.

DC Power Calculation

Formula

For DC (Direct I value) systems, wattage solution is straightforward:

P=V×IP = V \times I

Where:

  • P = Load in watts (W)
  • V = Electrical potential in volts (V)
  • I = Amperage in amperes (A)

Example Calculation

Compute the capacity consumed by a 12V DC motor drawing 5A:

P=12V×5A=60WP = 12\,\text{V} \times 5\,\text{A} = 60\,\text{W}

Result: The motor consumes 60 watts of energy.

Applications

DC electrical power calculations are used in:

  • Battery-powered devices
  • Solar panel systems
  • Automotive electrical systems
  • Electronic circuits
  • DC motor applications

AC Single-Phase Power Calculation

Formula

For AC single-phase systems, the formula includes the wattage factor:

P=PF×V×IP = PF \times V \times I

Where:

  • P = Real load in watts (W)
  • PF = Capacity factor (0 to 1)
  • V = RMS V value in volts (V)
  • I = RMS electrical flow in amperes (A)

Example Calculation

Find the energy consumed by a 230V AC appliance drawing 10A with a electrical power factor of 0.9:

P=0.9×230V×10A=2070WP = 0.9 \times 230\,\text{V} \times 10\,\text{A} = 2070\,\text{W}

Result: The appliance consumes 2,070 watts (2.07 kW) of real wattage.

Power Factor Importance

The load factor significantly affects capacity consumption:

  • PF = 1.0 (Unity): Pure resistive load (heaters, incandescent lamps)
  • PF = 0.8-0.95: Typical for motors and inductive loads
  • PF < 0.8: Poor energy factor, increases amp draw and energy costs

AC Three-Phase Power Calculation

Three-phase systems are common in industrial and commercial installations. The formula depends on whether you measure line-to-line or line-to-neutral electric tension.

Line-to-Line Voltage Formula

When measuring volt level between any two phases:

P=3×PF×VLL×IP = \sqrt{3} \times PF \times V_{\text{LL}} \times I

Where:

  • P = Total three-phase load in watts (W)
  • 31.732\sqrt{3} \approx 1.732
  • PF = Capacity factor
  • VLLV_{\text{LL}} = Line-to-line potential (V)
  • I = Line I value (A)

Line-to-Neutral Voltage Formula

When measuring electrical potential from one phase to neutral:

P=3×PF×VLN×IP = 3 \times PF \times V_{\text{LN}} \times I

Where:

  • P = Total three-phase energy in watts (W)
  • VLNV_{\text{LN}} = Line-to-neutral V value (V)
  • I = Line amperage per phase (A)
  • PF = Electrical phase angle

Relationship Between Line Voltages

VLL=3×VLNV_{\text{LL}} = \sqrt{3} \times V_{\text{LN}}

For example, in a 400V three-phase system:

  • Line-to-line electric tension: VLL=400V_{\text{LL}} = 400 V
  • Line-to-neutral volt level: VLN=400V3231VV_{\text{LN}} = \frac{400\,\text{V}}{\sqrt{3}} \approx 231\,\text{V}

Example Calculation

Evaluate the wattage of a three-phase machine with:

  • Line-to-line potential: 400V
  • Line electrical flow: 25A
  • Load factor: 0.85

P=3×0.85×400V×25A=14,740W14.74kWP = \sqrt{3} \times 0.85 \times 400\,\text{V} \times 25\,\text{A} = 14,740\,\text{W} \approx 14.74\,\text{kW}

Result: The drive unit consumes approximately 14.74 kW.

What Are the Practical Applications of?

Industrial Equipment Sizing

When sizing electrical equipment for industrial facilities:

  1. Measure or estimate the operating electrical potential
  2. Determine the amp draw from equipment specifications
  3. Identify the capacity factor (typically 0.8-0.9 for motors)
  4. Measure total energy requirements
  5. Add safety margin (typically 20-25%)

Electrical Panel Design

For electrical panel design:

Energy Cost Estimation

To estimate energy consumption:

  1. Assess wattage consumption in watts
  2. Convert to kilowatts (kW = W1000\frac{W}{1000})
  3. Multiply by operating hours to get kWh

Example: A 2,300W appliance running 8 hours/day consumes:

E=23001000×8=18.4 kWh/dayE = \frac{2300}{1000} \times 8 = 18.4 \text{ kWh/day}

This equals 18.4 kWh per day or about 552 kWh per month.

Standards and References

This calculator follows international standards for load calculations:

  • IEEE Std 1459-2010: Definitions for the Measurement of Electric Capacity Quantities
  • IEC 60364: Low-V value electrical installations
  • NEC Article 220: Branch-Circuit, Feeder, and Service Calculations

Safety Warnings

Common Mistakes to Avoid

  1. Forgetting load factor: AC calculations without capacity factor will overestimate real energy
  2. Mixing volt level types: Don't use line-to-line formulas with line-to-neutral voltages
  3. Using peak instead of RMS: Always use RMS (root mean square) values for AC
  4. Ignoring losses: Real systems have losses (cables, transformers, motors)
  5. Undersizing safety margins: Always include headroom for future expansion

Our calculations follow industry best practices and have been validated against real-world scenarios.

Summary

Converting potential and amperage to energy requires understanding your infrastructure type:

Export as PDF — Generate professional reports for documentation, client presentations, or permit submissions.

  • DC Systems: P=V×IP = V \times I (simplest formula)
  • AC Single-Phase: P=PF×V×IP = PF \times V \times I (include electrical reactive power ratio)
  • AC Three-Phase: P=3×PF×VLL×IP = \sqrt{3} \times PF \times V_{\text{LL}} \times I (industrial systems)

Accurate wattage calculations are essential for:

  • Equipment sizing and selection
  • Electrical setup design
  • Energy cost estimation
  • Safety compliance

Use our free volt to watt calculator for quick, accurate load calculations across DC, AC single-phase, and three-phase systems.

Related tools:

Conclusion

Converting volts to watts requires understanding both voltage and current for your specific system type (DC, AC single-phase, or three-phase). Accurate power calculations are fundamental to electrical system design, equipment sizing, energy cost estimation, and safety compliance. Always account for power factor in AC systems and apply appropriate safety margins for professional installations. The formula varies by system: DC uses P=V×IP = V \times I, AC single-phase uses P=PF×V×IP = PF \times V \times I, and AC three-phase uses P=3×PF×VLL×IP = \sqrt{3} \times PF \times V_{\text{LL}} \times I.

Key Takeaways

  • Convert volts to watts using P=V×IP = V \times I for DC, P=PF×V×IP = PF \times V \times I for AC single-phase, or P=3×PF×VLL×IP = \sqrt{3} \times PF \times V_{\text{LL}} \times I for AC three-phase—you must know current (amps) to calculate power from voltage
  • Power factor is critical for AC systems: resistive loads use PF=1.0, motors use 0.85-0.90, fluorescent lights use 0.6-0.8—forgetting PF gives apparent power (VA) not real power (W)
  • For three-phase systems, use line-to-line voltage (VLLV_{\text{LL}}) with 3\sqrt{3} factor (1.732) for most calculations; mixing line-to-line and line-to-neutral formulas causes 3\sqrt{3} error (73% deviation)
  • Standard voltage levels per IEC 60038: 120V/240V (North America residential), 230V (Europe residential), 208V/480V (North America commercial), 400V (Europe industrial) with ±10% tolerance
  • Lower power factor increases current draw for the same real power output, leading to higher cable losses (I2RI^2R) and potentially higher utility demand charges—power factor correction can reduce current by 20-30%
  • Always use RMS (root mean square) values for AC calculations, not peak voltage—mixing peak and RMS values causes significant errors in power calculations

Further Learning


References & Standards

This guide follows established engineering principles and standards. For detailed requirements, always consult the current adopted edition in your jurisdiction.

Primary Standards

IEC 60038:2009 IEC standard voltages. Defines standard voltage levels for electrical installations worldwide: 120V/240V (North America), 230V (Europe), 208V/480V (commercial), 400V (industrial), with ±10% tolerance.

IEEE Std 1459-2010 IEEE Standard Definitions for the Measurement of Electric Power Quantities Under Sinusoidal, Nonsinusoidal, Balanced, or Unbalanced Conditions. Defines power factor, apparent power, and real power.

IEC 60364 Low-voltage electrical installations. Provides requirements for electrical installations including power calculations and safety.

Supporting Standards & Guidelines

NEC Article 220 Branch-Circuit, Feeder, and Service Calculations. Specifies load calculation methods and demand factors.

IEC 60050 - International Electrotechnical Vocabulary International standards for electrical terminology and definitions.

NEMA Publications National Electrical Manufacturers Association standards for electrical equipment.

Further Reading

Note: Standards and codes are regularly updated. Always verify you're using the current adopted edition applicable to your project's location. Consult with local authorities having jurisdiction (AHJ) for specific requirements.


Disclaimer: This guide provides general technical information based on international electrical standards. Always verify calculations with applicable local electrical codes (NEC, IEC, BS 7671, etc.) and consult licensed electrical engineers or electricians for actual installations. Electrical work should only be performed by qualified professionals. Component ratings and specifications may vary by manufacturer.

Frequently Asked Questions

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