9.1 Motor Branch Circuits & Short-Circuit Protection (Article 430 Part IV)

Why the Table-FLC Rule Drives Every Motor Calculation

Virginia's exam is open-book on the 2020 NEC, so the trap is not access to the rule — it is using the wrong current. NEC 430.6(A)(1) is the master rule: for conductors, short-circuit/ground-fault protection, and disconnect ratings, you must use the full-load current (FLC) from Tables 430.247 (DC), 430.248 (single-phase AC), 430.249 (2-phase AC), and 430.250 (3-phase AC) — not the motor nameplate full-load amperes (FLA). The nameplate FLA reflects a specific motor's design; the Code tables are conservative values that account for replacement motors and varying efficiencies. Using nameplate here is one of the most common exam mistakes.

The nameplate FLA does have one critical job: overload protection sizing under 430.6(A)(2) and 430.32. Keep the two currents straight and you will dodge roughly a third of the motor questions on the exam.

Branch-Circuit Conductor Sizing — 430.22

430.22(A) sets the conductor rule for a single continuous-duty motor: the branch-circuit conductor ampacity must be not less than 125% of the motor's Table FLC. The 25% margin covers the fact that motors can run above nameplate under voltage variation and that the overload device is set just above nameplate FLA.

Worked Example — 10 hp, 460V, 3-Phase Motor

1. Find Table FLC. Table 430.250, 460V column, 10 hp row → 14 A.
2. Conductor ampacity = 14 A × 1.25 = 17.5 A.
3. Select conductor. If terminals are 75°C-rated (most modern equipment), use the 75°C column of Table 310.16: #12 AWG copper is rated 20 A, which exceeds 17.5 A. (If only 60°C terminals apply, #12 AWG is 15 A — too small — so you would step up to #10 AWG.)

That 14 A table value stays with you for the rest of the problem — breaker, disconnect, and feeder math all start from it.

Short-Circuit & Ground-Fault Protection — 430.52 and Table 430.52

430.52(C)(1) caps the branch-circuit short-circuit and ground-fault protective device at the percentages in Table 430.52, applied to the Table FLC:

Next-size-up: Exception No. 1 to 430.52(C)(1) lets you round up to the next standard size in 240.6(A) when the calculated value is not a standard rating. Exception No. 2 lets you go higher still if the motor will not start: inverse-time breakers up to 400% (≤100 A) or 300% (>100 A); time-delay fuses up to 225%; non-time-delay fuses up to 400%.

Continuing the 10 hp Example

• Inverse-time breaker: 14 A × 2.50 = 35 A (a standard size, no rounding needed).
• Dual-element time-delay fuse: 14 A × 1.75 = 24.5 A → next standard size 25 A (Exception No. 1).
• Non-time-delay fuse: 14 A × 3.00 = 42 A → next standard size 45 A.

A common trap is applying the 250% to the nameplate FLA or to the conductor ampacity — neither is correct. The percentage is always against the Table FLC.

Inverse-Time vs. Instantaneous-Trip Breakers

Table 430.52 lists two breaker types candidates conflate:

• Inverse-time circuit breakers (250% of FLC) pair a thermal element (bimetallic) with a magnetic trip. The thermal delay lets the breaker ride through the 5–8× FLC inrush a motor draws at start — the default the exam assumes unless it says otherwise.
• Instantaneous-trip circuit breakers (800% of FLC, or 1100% for Design B energy-efficient) have no intentional delay — they trip magnetically the moment current exceeds the pickup. They are permitted only when a separate motor overload relay is in series (430.52(C)(3)) because the breaker alone provides no running-overload protection. They are the standard choice inside a listed combination starter, where the overload block handles running protection and the breaker clears bolted faults only.

Because an instantaneous-trip breaker is set so far above FLC, it clears short circuits, not inrush — that is the overload relay's job. Read the question for the breaker type before picking a Table 430.52 column. The next-size-up allowance (Exception No. 1) lets you round a Table 430.52 result up one standard size in 240.6(A) (27.5 A → 30 A, not 35 A), but it does not apply to conductor ampacity.

Combination Loads — 430.62

When a feeder supplies several motors, 430.62 sizes the feeder short-circuit/ground-fault device as: (largest motor Table FLC × Table 430.52 percentage) + (sum of Table FLC of all other motors). You apply Table 430.52 only to the largest motor; the others are added at 100% of their Table FLC. For a feeder with a 10 hp, 5 hp, and 3 hp all at 460V (14 A + 7.6 A + 4.8 A) using an inverse-time breaker: (14 × 2.5) + 7.6 + 4.8 = 35 + 12.4 = 47.4 A → 50 A standard size.

Worked Example — Feeder to a 25 hp and a 10 hp Motor (460V, 3-Phase)

Table 430.250 gives FLC = 34 A for the 25 hp motor and 14 A for the 10 hp; the 25 hp is the largest motor. Feeder conductor (430.24): (34 × 1.25) + 14 = 42.5 + 14 = 56.5 A → #6 AWG copper (75°C column, 65 A).

Feeder OCPD (430.62) — two device options:

• Inverse-time breaker: (34 × 2.50) + 14 = 85 + 14 = 99 A → next standard size 100 A (Exception No. 1).
• Dual-element (time-delay) fuse: (34 × 1.75) + 14 = 59.5 + 14 = 73.5 A → next standard size 80 A.

The breaker option (100 A) lands higher than the fuse option (80 A) because Table 430.52 allows 250% for a breaker but only 175% for a time-delay fuse on the largest motor. Both are correct for their device — the exam specifies which one to use.

What This Section Prepares You For

Every motor problem on the exam starts with pulling the correct Table FLC. Lock in 430.6(A)(1) for conductor/OCPD/disconnect sizing, 430.6(A)(2) for overload sizing, and the 125% conductor rule of 430.22. The quizzes below test whether you can keep those currents straight under time pressure.