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Ampacity Calculator 3 Phase

Three Phase Ampacity Formula:

\[ I = \frac{P}{V \times \sqrt{3} \times PF} \]

watts
volts
(dimensionless)

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1. What is Three Phase Ampacity Calculation?

Three phase ampacity calculation determines the current carrying capacity in a three-phase electrical system. It's essential for proper sizing of conductors, circuit protection, and electrical system design.

2. How Does the Calculator Work?

The calculator uses the three phase ampacity formula:

\[ I = \frac{P}{V \times \sqrt{3} \times PF} \]

Where:

Explanation: This formula calculates the current required in a balanced three-phase system given the power, voltage, and power factor.

3. Importance of Ampacity Calculation

Details: Accurate ampacity calculation is crucial for electrical safety, preventing overheating, ensuring proper circuit protection, and optimizing electrical system efficiency.

4. Using the Calculator

Tips: Enter power in watts, voltage in volts, and power factor (between 0 and 1). All values must be positive numbers with power factor between 0 and 1.

5. Frequently Asked Questions (FAQ)

Q1: What is power factor and why is it important?
A: Power factor represents the ratio of real power to apparent power. It affects the actual current drawn by electrical equipment and is crucial for accurate ampacity calculations.

Q2: When should I use this three-phase calculation?
A: Use this calculation for balanced three-phase AC systems commonly found in industrial and commercial electrical installations.

Q3: What is the significance of √3 in the formula?
A: √3 (approximately 1.732) is the mathematical relationship between line-to-line voltage and line-to-neutral voltage in a balanced three-phase system.

Q4: Are there limitations to this calculation?
A: This calculation assumes a balanced three-phase system. For unbalanced systems or systems with significant harmonics, more complex calculations may be required.

Q5: How does power factor affect the current calculation?
A: Lower power factor results in higher current for the same power output, which means larger conductors and equipment are needed.

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