Calculate 3 phase current in amps from kW, kVA or motor hp at common voltages, or find kW and kVA from amps, using a typical power factor you can change.

Changing basis reinterprets the magnitude. kVA needs no PF for current.

Applies to preset or custom voltage. Balanced wye: line-to-line = line-to-neutral × √3.

0.85 is a typical motor value; use 1 for heaters and other resistive loads, or your nameplate PF.

Optional settings


Related Calculators

3 Phase Current Formula

The following formula is used to calculate the 3‑phase line current from apparent power for a balanced 3‑phase system. 

IL = VA / (√(3) VLL)
  • Where IL is the 3‑phase line current (amps)
  • VA is the apparent power (volt‑amps, VA) 
  • VLL is the line‑to‑line voltage (volts) 

To calculate the 3‑phase line current, divide apparent power (VA) by √3 times line‑to‑line RMS voltage. For balanced wye, VLL = √3 · VLN. Electrical input uses I = kW × 1000 / (√3 × VLL × PF). Mechanical motor hp is converted to input kW using hp × 0.745699871582 / efficiency. Blank motor efficiency uses 90%; replace it with the applicable percentage. PF uses an explicit ratio or percent selector; apparent kVA needs no PF for current. Reverse power uses kW = √3 × VLL × I × PF / 1000. These balanced sinusoidal estimates do not size protection or model imbalance and harmonics.

How to Calculate 3 Phase Current?

The following two example problems outline how to calculate the 3‑phase line current from apparent power and line‑to‑line voltage.

Example Problem #1:

  1. First, determine the apparent power (VA). In this example, the apparent power is measured to be 45 VA.
  2. Next, determine the line‑to‑line voltage (VLL). For this problem, the line‑to‑line voltage is measured to be 5 V.
  3. Finally, calculate the 3‑phase line current using the formula above: 

IL = VA / (√3 · VLL)

Inserting the values from above and solving the equation with the input values gives: 

IL = 45 / (√3 · 5) ≈ 5.1962 (amps). To reproduce this in the calculator, choose apparent power (kVA), enter 0.045 and custom line-to-line voltage 5 V.

Example Problem #2: 

Using the same process as example problem 1, we first define the variables outlined by the formula. In this case, the values are:

apparent power (VA) = 75

line‑to‑line voltage (VLL) = 6

Entering these values into the formula above gives : 

IL = 75 / (√3 · 6) ≈ 7.2169 (amps). Choose apparent power (kVA), enter 0.075 and custom line-to-line voltage 6 V.