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Motor

Motor Inrush Current

Calculate motor inrush current, voltage drop during startup, and I²t value for fuse/breaker selection.

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Formula

I_inrush = k × I_FL, ΔV = I_inrush × R_line

k— Inrush multiplier (5–8 typical) (×)
I²t— Fuse energy rating (A²·s)

How It Works

When a motor is first energised, the rotor is stationary and generates no back-EMF. The winding acts like a pure resistor-inductor load, and the initial current — called inrush or locked-rotor current (LRC) — is limited only by the winding impedance at line frequency. For AC induction motors, LRC is typically 5–8× the rated full-load current. For DC motors, inrush is V_supply / R_armature, which can be 10–20× running current. This high current spike must be considered when sizing fuses, circuit breakers, and power supply rails.

Worked Example

A 1.5 kW, 230 V, 50 Hz single-phase induction motor has a rated current of 8.7 A and a locked-rotor current multiplier (LRC ratio) of 6.5. Step 1 — Inrush current: I_inrush = LRC ratio × I_rated = 6.5 × 8.7 = 56.6 A Step 2 — Duration: Inrush decays to rated current in approximately 20–100 ms for small motors (the mechanical time constant). Step 3 — Select a fuse: Use a time-delay (slow-blow) fuse rated at 125–150% of full-load current. Fuse = 1.25 × 8.7 = 10.9 A → select 12 A slow-blow Step 4 — Check wire gauge: Size the supply cable for at least 125% of rated current (per NEC 430.22): I_wire = 1.25 × 8.7 = 10.9 A → 1.5 mm² copper (rated 15 A) Result: A 12 A slow-blow fuse and 1.5 mm² supply cable are appropriate. The 56.6 A inrush spike will clear within 100 ms without blowing the time-delay fuse.

Practical Tips

  • ✓Use a soft-start module or star-delta starter for motors above 3 kW to limit inrush to 2–3× rated current and reduce mechanical stress on couplings
  • ✓Add a bulk capacitor bank (1000–4700 µF) close to the motor driver H-bridge on battery-powered robots to absorb inrush without crashing the main MCU supply
  • ✓For VFD-driven motors, inrush at the motor terminals is negligible because the drive ramps frequency gradually — but the VFD itself still draws a large inrush from the mains on power-up

Common Mistakes

  • ✗Using fast-blow fuses for motor circuits — the inrush spike will blow them on every start; always specify time-delay (Type D or slow-blow) fuses
  • ✗Sizing the power supply for rated current only — an unregulated supply will sag badly during inrush unless it has 3–5× headroom or a soft-start circuit
  • ✗Underestimating inrush in battery-powered systems — the inrush current creates a voltage dip that can reset microcontrollers or crash CAN bus communication

Frequently Asked Questions

The electrical inrush (current spike from zero back-EMF) decays within 1–3 electrical time constants (L/R), typically 10–50 ms. The mechanical load prolongs high current draw until the rotor reaches operating speed — total acceleration time ranges from 100 ms for a small unloaded motor to several seconds for a large loaded machine.
A VFD eliminates motor inrush by ramping up frequency and voltage gradually. However, the VFD's own rectifier and DC bus capacitors draw a large inrush from the mains supply when first powered — this must be managed with pre-charge resistors or AC line reactors.
They are the same current measured under different conditions. Locked-rotor current (LRC) is the steady-state current when the rotor is mechanically held stationary. Inrush current is the transient peak at the moment of switch-on; it may momentarily exceed the LRC value due to transformer-like magnetic flux transients.

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