Motor Starting Voltage Dip
Voltage dip during motor start from locked-rotor current and conductor length.
Example
You enter
- Source voltage (V) 480
- One-way conductor length (ft) 250
- Conductor circular mils (cmils) 250000
- Locked-rotor current (A) 180
- Phase three
- Conductor material (K) 12.9
- Dip limit (%) 15
You get
- Voltage drop during start 4.0 V
- Dip (%) 0.838
- Terminal voltage during start 476.0 V
Details, formula, and sources
With terminal voltage and a pass/fail against the dip limit (contactor dropout risk).
V_drop = (2 for 1-phase, sqrt(3) for 3-phase) x K x LRC x L / cmils; V_terminal = V_source - V_drop; %dip = V_drop / V_source x 100. K is the conductor constant (Cu ~12.9, Al ~21.2 ohm-cmil/ft) and LRC the motor locked-rotor current.
Ohm's-law voltage-drop method (first principles); motor locked-rotor current per NEC Article 430 code-letter tables; contactor pickup/dropout ~85% nominal per NEMA ICS 2, by name.
NFPA 70 free read-only at nfpa.org/freeaccess; LRC user-supplied from the nameplate code letter or estimated as 6x FLA.
Estimate. AHJ and licensed electrician govern. Verify against the NEC edition adopted in your jurisdiction.
Field names used by the API: source_voltage_V, length_ft, cmils, lrc_A, phase, k_const, dip_limit_pct, v_drop_V, dip_pct, v_terminal_V
- Locked-rotor current user-supplied from the NEC 430 code-letter table, or estimated as 6x full-load ampsNEC Article 430
- Conductor constant K 12.9 ohm-cmil/ft copper, 21.2 aluminum (one-way)voltage-drop convention