Inverter AC Output Circuit Conductor and OCPD (NEC 690.8(B))

Continuous output current = AC power / (voltage x [1 single-phase, sqrt(3) three-phase]), then NEC 690.8(B) / 705.60 / 240.4 size the conductor and OCPD at 125% of that current, rounding the OCPD up to the next standard size (240.6). A 9.6 kW inverter at 240 V single-phase puts out 40 A, so the conductor is at least 50 A (a #6 Cu at 75 C) on a 50 A breaker; the same inverter at 208 V three-phase is only 26.6 A. Use the inverter's RATED continuous AC output current from its datasheet if given, and check the 705.12 busbar / point-of-connection limit separately. This is the AC side; the DC side uses the separate 690.8(A) 156% rule. A sizing estimate; the inverter datasheet, the AHJ, and the adopted NEC edition govern.

continuous_current_a = ac_power_w / (ac_voltage_v x [1 single-phase, sqrt(3) three-phase]); min_conductor_ampacity_a = 1.25 x continuous_current_a; ocpd_a = next standard size (NEC 240.6) >= 1.25 x continuous_current_a.

NEC 690.8(B) (continuous current), 705.60 (interconnected-source overcurrent), 240.4 (conductor protection), and 240.6 (standard OCPD sizes), by name; the inverter datasheet and the AHJ govern.

The output-current relation is power / voltage (public); the 125% continuous factor and the standard OCPD sizes are public NEC, and the inverter power, voltage, and phase come from the datasheet.

Estimate. AHJ and licensed professional govern.

Field names used by the API: ac_power_w, ac_voltage_v, phases, continuous_current_a, min_conductor_ampacity_a, ocpd_a

• 125% continuous conductor and OCPD at 1.25 x the continuous output current; OCPD to the next standard size (240.6)NEC 690.8(B) / 240.4 / 240.6
• Busbar limit separate the 705.12 point-of-connection / busbar limit is checked separately (pv-interconnection-busbar)NEC 705.12