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Calculate 3 Phase Current

3 Phase Current Formula:

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

Watts
Volts
(0-1)

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1. What is 3 Phase Current?

Three-phase current is the current flowing in a three-phase electrical system, which is commonly used for power distribution and large motors. The calculation considers the balanced load across all three phases.

2. How Does the Calculator Work?

The calculator uses the 3-phase current formula:

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

Where:

Explanation: The formula accounts for the phase relationship in three-phase systems and the power factor which represents the ratio of real power to apparent power.

3. Importance of 3 Phase Current Calculation

Details: Accurate current calculation is essential for proper sizing of circuit breakers, wires, and transformers in three-phase systems to ensure safety and efficiency.

4. Using the Calculator

Tips: Enter power in watts, line-to-line voltage in volts, and power factor (typically 0.8-0.95 for motors). All values must be positive (power > 0, voltage > 0, 0 < PF ≤ 1).

5. Frequently Asked Questions (FAQ)

Q1: What's the difference between line and phase voltage?
A: In a 3-phase system, line voltage is between any two lines, while phase voltage is between any line and neutral. For delta connections, they're equal; for wye, line voltage is √3 times phase voltage.

Q2: Why is power factor important?
A: Power factor affects the actual current drawn for a given real power. Lower PF means higher current for the same real power, increasing losses.

Q3: What's a typical power factor value?
A: Induction motors typically have 0.8-0.9 PF at full load. Resistive loads have PF=1. Capacitors can improve PF.

Q4: Can I use this for single-phase calculations?
A: No, single-phase uses I = P/(V×PF) without the √3 factor. This calculator is specifically for balanced three-phase systems.

Q5: How does voltage affect the current?
A: Current is inversely proportional to voltage for a given power. Higher voltage systems can deliver the same power with lower current, reducing conductor size needed.

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