Amp to VA Calculator - Convert Current to Apparent Power

Introduction

Converting current (amps) to apparent power (VA) is essential for properly sizing transformers, selecting UPS systems, rating generators, and planning electrical system capacity. Unlike real power (watts), apparent power does not require power factor—it is simply the product of voltage and current.

Why This Conversion Matters

Understanding the relationship between current and apparent power enables engineers to:

• Size transformers properly — Transformers are rated in kVA, not kW, because they must supply total current regardless of load power factor
• Select UPS systems — Uninterruptible power supplies must handle apparent power for proper load protection
• Rate generator capacity — Generators must supply both real and reactive power components
• Plan electrical capacity — Service entrances and feeders are sized based on current capacity, which relates directly to VA

The Fundamental Challenge

The relationship between current and apparent power differs by system type:

DC / Single-Phase AC: $S = V \times I$

Three-Phase AC (Line-to-Line): $S = \sqrt{3} \times V_{LL} \times I$

Three-Phase AC (Line-to-Neutral): $S = 3 \times V_{LN} \times I$

What You'll Learn

This guide is designed for electrical engineers, technicians, and designers who need to convert current measurements to apparent power for equipment sizing and system design. You will learn:

• Fundamental VA formulas for DC, single-phase AC, and three-phase AC systems
• Difference between VA and watts — Why equipment is rated in VA, not watts
• Transformer sizing applications — Standard kVA ratings and selection criteria per IEC 60076-1
• UPS system selection — Capacity planning and safety margins per IEC 62040-3
• Safety factor guidelines — Proper margin for transients, harmonics, and future growth
• Practical examples — Equipment sizing, capacity planning, and standard ratings

Quick Answer: How to Convert Amps to VA

To convert amps to VA, multiply current by voltage using the appropriate formula for your system type.

Core Formulas

Where:

• $S$ = Apparent power (VA)
• $I$ = Current (A)
• $V$ = Voltage (V)
• $V_{\text{LL}}$ = Line-to-line voltage (V)
• $V_{\text{LN}}$ = Line-to-neutral voltage (V)

Worked Example

Reference Table

Key Standards

Understanding Apparent Power

What is Apparent Power?

Apparent electrical power is the product of RMS electrical potential and RMS amp in an AC circuit, measured in volt-amperes (VA).

Key Differences:

• Real Wattage (W): Actual load consumed
• Reactive Capacity (VAR): Energy oscillating in the circuit
• Apparent Electrical power (VA): Vector sum of real and reactive wattage

Current (A)

Electric current is the flow of electric charge, measured in amperes (A).

Key Points:

• Measured in amperes (A)
• Represents charge flow rate
• Determines conductor sizing
• Affects setup capacity

Voltage (V)

V value is the electrical potential difference, measured in volts (V).

Key Points:

• Measured in volts (V)
• Represents electrical pressure
• Standard values: 120V, 230V, 400V
• Affects load capacity

Apparent Power (VA)

Apparent capacity is the total energy in an AC circuit, measured in volt-amperes (VA).

Key Points:

• Measured in VA or kVA
• Always $\geq$ real electrical power
• Used for equipment sizing
• Includes reactive wattage

The Formulas

DC and AC Single-Phase Apparent Power Formula

For DC and AC single-phase systems:

Where:

• S(VA) = Apparent load in volt-amperes
• I(A) = I value in amperes
• V(V) = Electric tension in volts

AC Three-Phase Apparent Power Formula (Line-to-Line)

For three-phase systems with line-to-line volt level:

Where:

• $V_{\text{LL}}$ = Line-to-line potential
• $\sqrt{3} \approx 1.732$ (three-phase factor)

AC Three-Phase Apparent Power Formula (Line-to-Neutral)

For three-phase systems with line-to-neutral electrical potential:

Where:

• $V_{\text{LN}}$ = Line-to-neutral V value

Step-by-Step Calculations

Example 1: DC Apparent Power

Problem: A DC circuit has 10 A amperage at 100 V. Compute the apparent capacity.

Solution:

1. Identify arrangement type: DC

2. Given values:

   • Electrical flow (I) = 10 A
   • Electric tension (V) = 100 V

3. Apply DC formula:

   $$S(VA) = I(A) \times V(V) S(VA) = 10 \times 100 S(VA) = 1000 \text{ VA}$$

4. Result: The apparent energy is 1000 VA or 1 kVA.

Example 2: AC Single-Phase Apparent Power

Problem: An AC single-phase circuit has 13 A amp at 230 V. Find the apparent electrical power.

Solution:

1. Identify mechanism type: AC Single-Phase

2. Given values:

   • Electric current (I) = 13 A
   • Volt level (V) = 230 V

3. Apply AC single-phase formula:

   $$S(VA) = I(A) \times V(V) S(VA) = 13 \times 230 S(VA) = 2990 \text{ VA}$$

4. Result: The apparent wattage is approximately 2990 VA or 2.99 kVA.

Example 3: AC Three-Phase Apparent Power

Problem: A three-phase circuit has 25 A I value at 400 V line-to-line. Evaluate the apparent load.

Solution:

1. Identify installation type: AC Three-Phase

2. Given values:

   • Amperage ($I$) = 25 A
   • Potential ($V_{\text{LL}}$) = 400 V

3. Apply three-phase formula:

   $$S(VA) = \sqrt{3} \times I(A) \times V_{\text{LL}}(V) S(VA) = 1.732 \times 25 \times 400 S(VA) = 1.732 \times 10000 S(VA) = 17,320 \text{ VA}$$

4. Result: The apparent capacity is approximately 17.32 kVA.

Practical Examples

Example 4: UPS Sizing

Scenario: A server rack draws 8.7 A at 230 V. What UPS capacity is needed?

Given:

• Electrical flow = 8.7 A
• Electrical potential = 230 V
• Equipment = AC Single-Phase

Computation:

Result: A 2 kVA UPS is needed (with 20% safety margin, choose 2.5 kVA).

Example 5: Transformer Sizing

Scenario: A three-phase load draws 50 A at 400 V line-to-line. What transformer rating is needed?

Given:

• Amp = 50 A
• V value (LL) = 400 V
• Infrastructure = AC Three-Phase

Analysis:

Result: A 35 kVA transformer is needed.

Example 6: Generator Capacity

Scenario: A construction site has equipment drawing 72 A at 230 V single-phase. What generator capacity is needed?

Given:

• Electric current = 72 A
• Electric tension = 230 V

Determination:

Result: A 16.6 kVA generator is needed (choose 20 kVA for headroom).

System Types

DC Systems

Characteristics:

• Constant volt level and I value
• Apparent energy = Real electrical power
• Simple evaluation
• Used in batteries, solar systems

Formula: $S(VA) = I(A) \times V(V)$

Applications:

• DC wattage supplies
• Battery systems
• Solar inverters
• Telecommunications

AC Single-Phase Systems

Characteristics:

• Alternating amperage
• Apparent load $\geq$ Real capacity
• Common in residential
• 230 V or 120 V standard

Formula: $S(VA) = I(A) \times V(V)$

Applications:

• Home appliances
• Office equipment
• Small UPS systems
• Lighting circuits

AC Three-Phase Systems

Characteristics:

• Three alternating currents
• Higher capacity
• Industrial applications
• 380 V, 400 V, or 480 V

Formula: $S(VA) = \sqrt{3} \times I(A) \times V_{\text{LL}}(V)$

Applications:

• Large transformers
• Industrial motors
• Data center UPS
• Distribution systems

Apparent Power vs Real Power

Key Differences

Power Triangle

The relationship between powers:

• $S^2 = P^2 + Q^2$
• S = Apparent Load (VA)
• P = Real Capacity (W)
• Q = Reactive Energy (VAR)
• PF = P / S (Electrical power Factor)

Sizing Considerations

For Equipment Sizing:

• Use apparent wattage (VA) for:
  • Transformers
  • UPS systems
  • Generators
  • Circuit breakers
  • Conductors

For Energy Cost:

• Use real load (W) for:
  • Electricity billing
  • Energy consumption
  • Cost calculations

Standards and References

International Standards

• IEEE Std 1459-2010: Definitions for measurement of electric capacity quantities
• IEC 60050-131: International Electrotechnical Vocabulary
• IEC 62040-3: UPS systems performance requirements
• IEC 60076: Energy transformers

Typical Equipment Ratings

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

Conclusion

Converting current (amps) to apparent power (VA) is essential for transformer sizing, UPS capacity selection, generator rating, circuit capacity planning, and electrical system design. The formulas differ for single-phase (S = I × V) and three-phase (S = √3 × I × VLL) systems, but both are straightforward when voltage and current are known. Unlike kW to VA conversion, amp to VA conversion does not require power factor because apparent power is the product of voltage and current. Understanding the difference between apparent power (VA) and real power (W) is crucial for proper equipment selection and system design. Always apply appropriate safety factors (125% for continuous loads) and verify calculations against equipment specifications and electrical codes.

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

Key Takeaways

• Convert amps to VA using $S(\text{VA}) = I(\text{A}) \times V(\text{V})$ for single-phase and $S(\text{VA}) = \sqrt{3} \times I(\text{A}) \times V_{\text{LL}}(\text{V})$ for three-phase—no power factor needed because apparent power is the product of voltage and current
• Power factor is not required for amp to VA conversion—unlike kW to VA conversion, VA represents the product of voltage and current ($S = V \times I$), so apparent power can be calculated directly from current and voltage
• Apply 125% safety factor for continuous loads (>3 hours operation) per NEC 210.19 when sizing conductors and overcurrent protection devices
• Use correct voltage for system type—single-phase uses phase voltage, three-phase uses line-to-line voltage ($V_{\text{LL}}$) with $\sqrt{3}$ factor (1.732)
• Convert kVA to VA first—multiply kVA by 1000 before using in formulas: $S(\text{VA}) = \text{kVA} \times 1000$
• Equipment is rated in kVA/VA—transformers, generators, and UPS systems are sized based on apparent power, so use VA ratings for equipment selection
• Verify against equipment specifications—calculated VA should match or be less than equipment nameplate VA ratings

Further Learning

• VA to Amp Guide - Reverse conversion from apparent power to current
• kVA to kW Guide - Converting kVA to real power
• Transformer Sizing Guide - Sizing transformers based on kVA requirements
• Cable Sizing Guide - Selecting conductors based on current
• Amp to VA Calculator - Interactive calculator for apparent power conversion

Calculator

Use our interactive calculator to convert electric current to apparent energy:

References & Standards

Primary Standards

IEC 60076-1 Power transformers. Specifies transformer kVA ratings and full-load current calculations. Transformer current: I = (kVA × 1000) / (√3 × V) for three-phase systems. Defines transformer sizing based on apparent power (kVA), not real power (kW). Requires 15-25% capacity margin for voltage regulation and future growth.

IEC 62040-3 Uninterruptible power systems (UPS) - Part 3: Method of specifying the performance and test requirements. Defines UPS VA ratings and sizing requirements based on apparent power. Requires 20-30% safety margin for inrush currents and future load growth.

Supporting Standards & Guidelines

IEEE 1459-2010 Standard definitions for the measurement of electric power quantities under sinusoidal, nonsinusoidal, balanced, or unbalanced conditions. Defines apparent power (VA) and specifies that VA represents total current draw, requiring no power factor for current-to-VA conversion.

NEC Article 210.19 Branch circuits. Specifies 125% safety factor for continuous loads (>3 hours operation) when sizing conductors and overcurrent protection devices.

Further Reading

• Electrical Installation Guide - Schneider Electric - Comprehensive guide to electrical installation best practices

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.