AC Motors

Power, efficiency & full-load current

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

ELI5: the nameplate lists shaft power (what the load actually gets). The line has to supply more, to cover losses (efficiency η) and the magnetizing current that does no real work (power factor). Both sit in the denominator of the current formula, pushing the line current up.

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

Real-world hook: this full-load current is what an electrician reads off the nameplate to size the wire, breaker, and overload relay for every motor installed.