AC Motors

Power, efficiency & full-load current

continues from lesson 2 — values defined earlier in the course stay live here

The nameplate gives SHAFT power (what the load receives). The line must supply more, to cover losses (efficiency η) and the magnetizing current that does no work (power factor). Both land in the denominator of the current formula.

P_{shaft} := 7500 W

= 7500 W shaft power — 10 hp

V_{LL} := 480 V

= 480 V line-to-line voltage

PF = 0.86

power factor

\eta = 0.91

efficiency

P_{in} = \frac{P_{shaft}}{\eta}

= 8241.8 W electrical power drawn

P_{loss} = P_{in} - P_{shaft}

= 741.76 W heat — where the missing 9% goes

I_{FL} = \frac{P_{shaft}}{\sqrt{3} \cdot V_{LL} \cdot PF \cdot \eta}

= 11.527 A full-load line current

✓ pass

I_{FL} < 16 A

fits a 16 A motor-rated branch circuit