14 AWG on a Long NAC
A new notification appliance circuit leaves a 24 VDC fire alarm control panel whose NAC output is rated 2.0 A. The circuit feeds six wall-mounted horn/strobes rated 0.300 A each, connected in a single Class B string. The first device is 500 ft from the panel and the remaining five are spaced 100 ft apart, with the end-of-line device at the far end. The electrician installed 14 AWG THHN. The panel's published NAC output is a regulated 24.0 V at rated load; the appliances are listed for 16.0 to 33.0 VDC. The panel has one spare, unused NAC. The wire is pulled and the walls are open; the drywall contractor is scheduled for Monday.
1. Will this circuit operate the appliances? Perform the calculation and state your conclusion.
Read the worked answer
Step 1 - total load. 6 × 0.300 A = 1.80 A. Within the 2.0 A NAC rating, but only 0.2 A of margin - also check the panel's total alarm current against its battery calculation.
Step 2 - current per segment. Each device draws 0.300 A, so each successive segment carries 0.300 A less.
| Segment | Length (ft) | Current (A) | I × L (A·ft) |
|---|---|---|---|
| Panel → D1 | 500 | 1.80 | 900 |
| D1 → D2 | 100 | 1.50 | 150 |
| D2 → D3 | 100 | 1.20 | 120 |
| D3 → D4 | 100 | 0.90 | 90 |
| D4 → D5 | 100 | 0.60 | 60 |
| D5 → D6 | 100 | 0.30 | 30 |
| Σ | 1,350 |
Step 3 - resistance. 14 AWG copper ≈ 3.14 Ω per 1,000 ft per conductor at 75 °C. Both conductors are in the loop: 2 × 3.14 = 6.28 Ω/1,000 ft = 0.00628 Ω/ft.
Step 4 - drop. V = 0.00628 × 1,350 = 8.5 V.
Step 5 - voltage at the last appliance. 24.0 − 8.5 = 15.5 V, below the 16.0 V listed minimum.
Conclusion: the circuit fails. The last horn/strobes will be undervoltaged - reduced output, possible failure to operate, and out-of-listing operation. The check must be done at the panel's guaranteed output, not from a field reading taken with a meter on a lightly loaded circuit.
2. What are the practical fixes, and which would you direct given that the walls are still open?
Read the worked answer
Option A - upsize the conductor. Recompute with 12 AWG (≈1.98 Ω/1,000 ft per conductor, 0.00396 Ω/ft loop):
V = 0.00396 × 1,350 = 5.35 V → last device sees 24.0 − 5.35 = 18.65 V. Passes with roughly 2.6 V of margin.
Option B - split the string across the two NACs (three devices per NAC, 0.90 A each). Using the existing 14 AWG on, say, the panel→D1→D2→D3 portion: (500 × 0.90) + (100 × 0.60) + (100 × 0.30) = 450 + 60 + 30 = 540 A·ft; V = 0.00628 × 540 = 3.4 V → 20.6 V at the last device. Passes with the wire already in the wall, but uses the spare NAC and must be reflected in the drawings, the battery calculation, and the panel program.
Decision logic: since the walls are open, either works. Option A costs copper and a re-pull of a 500 ft home run; Option B costs a NAC and panel reprogramming but no wire. Verify the panel's total alarm current and battery calculation under Option B (two NACs energized at once), and confirm the spare NAC is actually spare - not reserved for future equipment.
Do not accept "we'll just measure it after trim-out." By then the correction is a demolition and patching cost, and a field reading taken at nominal voltage does not prove compliance at end-of-discharge battery voltage.
3. Does the 1.80 A load create any other problem the electrician may have missed?
Read the worked answer
Yes - two items beyond voltage drop:
- Panel capacity, not just NAC capacity. 1.80 A on a 2.0 A NAC is legal, but the panel's total alarm current (NACs + SLC + communicator + any auxiliary output) must still fall inside the panel's listed alarm rating and inside what the battery calculation covers. A 0.2 A margin on one circuit evaporates if a second NAC is added later. Record the actual connected load on the panel schedule and the record of completion.
- Inrush versus continuous rating. Strobes draw a much higher current briefly at flash. The appliance and panel listings, and the manufacturer's synchronization requirements, must be satisfied - a circuit that is fine on continuous current can still exceed the panel's instantaneous capability if it is near the rating, and non-synchronized appliances on one circuit can produce a combined flash that is not listed.
Also confirm supervision integrity: the end-of-line device must be at the electrically farthest point of the Class B circuit, and the conductor size change must be shown on the as-built so the next inspector is not recalculating from the wrong wire size. If any device in the string is a silenceable horn paired with a visual device, note that silencing must not disable the visible appliance.