Voltage is money
What you'll learn
The same 75 HP motor needs 300 kcmil at 208 V and 1 AWG at 480 V. Close the one-line with the disconnect and the ground, and walk the whole recipe.
Same motor, same shaft horsepower, three electric bills of material. Power is voltage times current: raise one, the other falls. At 75 HP the difference stops being academic:
The legacy mill motor runs on a 208 V system: 211 A of FLA, and the 125% rule lands the feed in 300 kcmil — conductor as thick as your thumb, three of them, in 2-1/2″ conduit, for the whole run. Switch the designer to 480 V and the same machine draws 96 A on 1 AWG in 1-1/4″ EMT. That's why industrial plants distribute at 480: amps are copper, and copper is money — in wire, conduit, lugs, and the labor to pull all of it.
Closing the one-line
Two elements we haven't formally met:
- The disconnect (430.110) — rated at least 115% of FLA, and 430.102(B) wants it within sight of the motor, so the mechanic pulling a coupling can see with their own eyes that the circuit is open. On the mill that's a 400 A switch at the machine.
- The equipment ground (250.122) — sized from the breaker protecting the circuit, not the phase conductors. The mill's 600 A breaker calls for a 1 AWG copper EGC riding along in the conduit.
The whole recipe
Five decisions, five rules, one table value driving them all:
| element | rule | keyed to |
|---|---|---|
| conductors | 430.22 — 125%, 75°C, check drop | table FLA |
| breaker | 430.52 — ≤250%, next size up | table FLA |
| overload | 430.32 — 115/125% | nameplate |
| disconnect | 430.110 — ≥115%, in sight | table FLA |
| ground | 250.122 | breaker rating |
That's the course. The designer is the worksheet that never forgets a step — design your own circuit from scratch, tap every element, and check yourself against the code path on each card. When a real install is on the line: verify against your NEC edition, and put a licensed electrician's eyes on it.
Check your understanding
Question 1 of 2
The same 75 HP motor needs 3 × 300 kcmil at 208 V but only 3 × 1 AWG at 480 V. What's the physics behind the savings?