Motor Full Load Current Calculator

Calculate full load current (FLA) of a 3-phase induction motor from power, voltage, power factor, and efficiency.

kW
Rated output power on motor nameplate in kW. To convert HP: kW = HP × 0.746
V
Supply voltage (line-to-line). Most Indian dairy plants: 415V
Typical induction motor PF: 0.80–0.90 at full load
Typical efficiency: 0.85–0.94 for modern IE2/IE3 motors
Formula
(motorPower * 1000) / (1.732 * voltage * powerFactor * efficiency)
motorPower Motor Rated Power (kW)
voltage Line Voltage (V)
powerFactor Power Factor (PF)
efficiency Motor Efficiency (η)
Worked Example
1
Given:
efficiency = 0.9
motorPower = 7.5
powerFactor = 0.85
voltage = 415
2
Apply the formula:
(motorPower * 1000) / (1.732 * voltage * powerFactor * efficiency)
3
Result:13.63 A

What is Full Load Current (FLA)?

Full Load Current (FLA or FLC) is the current drawn by an electric motor when running at its rated horsepower/kW output at rated voltage. It is the basis for:

  • Cable sizing: Cables must carry FLA continuously
  • Circuit breaker selection: MCB/MCCB rated at 125–160% of FLA
  • Overload relay setting: Set to 100–105% of FLA
  • Starter sizing: DOL, Star-Delta, VFD selection based on FLA

Formula — 3-Phase Motor

I (A) = P (kW) × 1000 / (√3 × V × PF × η)

Where:

  • P = rated power (kW)
  • V = line voltage (V) — typically 415V in India
  • PF = power factor (typically 0.80–0.90 for induction motors)
  • η = motor efficiency (typically 0.85–0.95)
  • √3 = 1.732

Typical Values for Indian Motors

Motor RatingTypical FLA at 415V
0.37 kW (0.5 HP)1.0–1.2 A
0.75 kW (1 HP)1.8–2.2 A
1.5 kW (2 HP)3.5–4.0 A
3.7 kW (5 HP)8.0–9.5 A
7.5 kW (10 HP)16–18 A
15 kW (20 HP)30–35 A
37 kW (50 HP)72–80 A

Cable Sizing Rule of Thumb

Minimum cable size (mm²) = FLA / 4 for copper cables in conduit at 40°C ambient

Always verify against IS 694 / IS 1554 current ratings and apply applicable derating factors.