NEC 430.24: One Feeder, Several Motors, One Multiplier

The worked example

A shop feeder supplies three motors with table full-load currents of 24 amps, 16 amps, and 10 amps. Minimum conductor ampacity:

• Find the largest: 24 A.
• Give it the margin: 24 x 1.25 = 30 A.
• Add the rest at face value: 30 + 16 + 10 = 56 A.
• Table 310.16 at 75 C: 6 AWG copper (65 A) clears 56 A.

The two wrong answers built into every version of this question: 125% on everything gives (30 + 20 + 12.5) = 62.5 A, and no multiplier at all gives 50 A. Both appear in the choices. Both are one small misreading away from the right method.

Why the code works this way

The 25% adder is thermal headroom for a motor running loaded for hours. Stacking that margin on every machine assumes every motor on the feeder runs flat out at once and overshoots the copper. The code takes the honest middle: full margin for the worst single contributor, face value for the rest. It is the same philosophy as the demand factors in Article 220, applied with motor numbers.

Where candidates lose the point

• Multiplying every motor by 125%. Only the largest gets it.
• Multiplying the sum by 125%. Same error in different clothes.
• Using nameplate currents. Feeder sizing runs on the 430.248/430.250 tables.
• Applying 125% to two motors when they tie for largest. One multiplier, even on a tie.
• Forgetting non-motor loads on the same feeder. They add at their calculated value.
• Sizing the feeder breaker with this formula. Protection is 430.62, a different calculation.

How it shows up on the exam

The question almost always hands you three motors and asks for minimum conductor ampacity, exactly like the worked example. The variations to expect: horsepower given instead of current, which adds a table lookup in 430.250 before the math starts. A tie for largest motor. Or a heater or lighting load riding the same feeder, which simply adds at face value at the end. The skeleton never changes: one motor gets 125%, everything else stacks on top.