Back to courseLesson 6 of 6

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:

One 75 HP motor, three voltages (Table 430.250)211 A208 V3 × 300 kcmil192 A240 V3 × 250 kcmil96 A480 V3 × 1 AWGAmps = dollars: every extra amp is copper, conduit and lugs down the whole run

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:

elementrulekeyed to
conductors430.22 — 125%, 75°C, check droptable FLA
breaker430.52 — ≤250%, next size uptable FLA
overload430.32 — 115/125%nameplate
disconnect430.110 — ≥115%, in sighttable FLA
ground250.122breaker 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?