Motor Starting Current & Voltage Dip Calculator

How it works

LRC ≈ NEMA code kVA/hp × hp × 1,000 ÷ (√3 V) ≈ 5.5–9 × FLA; starting current = LRC × starter factor; Z_source = V² ÷ kVA × %Z + cable; dip ≈ √3 × I_start × Z × 0.85; torque ∝ (terminal voltage)²

At the instant of starting an induction motor is a stalled transformer with a shorted secondary, drawing six to eight times its running current at a poor power factor until it accelerates. That inrush flowing through the supply transformer and cable impedance drops the voltage for everyone on the bus — the classic light flicker when a compressor starts — and because motor torque falls with the square of voltage, a deep dip also makes the motor itself slower to accelerate. The NEMA code letter on the nameplate gives the locked-rotor kVA per horsepower, so inrush is predictable; the dip then follows from the source impedance. When it exceeds a few percent for frequent starts or about 10% for occasional ones, reduced-voltage starting — soft starter, wye-delta, autotransformer or a drive — trades starting torque for a gentler grid, or the transformer and cables are upsized.

Worked example

A 75 hp, 480 V code G motor (96 A FLA) draws about 595 A locked-rotor. Across a 500 kVA, 5.5% transformer (25 mΩ) plus 80 m of cable (20 mΩ) that pulls the bus down about 4.6% and the motor terminals about 8% — acceptable direct-on-line for occasional starts; a soft starter at 350% would roughly halve the dip.

Frequently asked questions