NAC Voltage Drop on a Long Horn/Strobe Run
You are commissioning a new notification appliance circuit in a 260 ft-long warehouse corridor. The panel lists the NAC as 24 VDC regulated, rated 3.0 A maximum, with a 0.6 V maximum line drop allowance. You plan to install 14 wall-mounted horn/strobes rated 0.115 A each at 24 VDC, spaced evenly along the run. The wiring you have on the truck is 16 AWG stranded, which has a resistance of about 4.02 ohms per 1000 ft per conductor. A junior technician suggests simply landing all 14 devices and letting the panel's sync module handle it, arguing that 14 x 0.115 A = 1.61 A is well under the 3.0 A rating, so the circuit is fine.
1. Calculate the total load current and the worst-case two-conductor voltage drop at the end of the run, and determine whether the circuit meets the 0.6 V allowance.
Read the worked answer
Total current = 14 x 0.115 A = 1.61 A. For a two-wire NAC where all devices are beyond the point of calculation, the worst-case drop is computed with the full load flowing through the entire length:
- One-way length = 260 ft; loop resistance uses both conductors: 2 x 260 ft = 520 ft of conductor.
- Resistance = (520 / 1000) x 4.02 ohm = 2.09 ohm.
- Voltage drop = I x R = 1.61 A x 2.09 ohm = 3.37 V.
That is far above the 0.6 V allowance. Even though the current is within the panel's 3.0 A rating, the voltage drop fails badly. The junior technician's reasoning conflates ampacity/current rating with voltage-drop compliance; both must be satisfied. This circuit as designed would likely cause strobes to under-flash and horns to sound weak at the far end.
2. Propose at least two corrective options and evaluate the trade-offs of each.
Read the worked answer
Option A: Upsize the conductors. Going to 12 AWG (about 1.93 ohm/1000 ft) gives R = (520/1000) x 1.93 = 1.00 ohm, drop = 1.61 x 1.00 = 1.61 V - still over 0.6 V. Going to 10 AWG (about 1.21 ohm/1000 ft): R = 0.63 ohm, drop = 1.01 V - still over. Conductor upsizing alone cannot meet 0.6 V at this length with this load, so this option is insufficient by itself.
Option B: Split the circuit into two NACs. Put 7 devices on each of two circuits. Each circuit now carries 0.805 A over the same length: drop = 0.805 x 2.09 = 1.68 V with 16 AWG - still over. Combine with 12 AWG: 0.805 x 1.00 = 0.81 V - still slightly over. Combine split with 10 AWG: 0.805 x 0.63 = 0.51 V - passes.
Option C: Reduce the run or relocate the panel / add a booster power supply nearer the load. A remotely mounted NAC extender or power supply near the corridor midpoint shortens the effective circuit length, which is often the most practical field fix.
The best answer recognizes that voltage drop scales with both current and length, so the fix usually requires reducing one or both - and that splitting without upsizing may still fail.
3. If the panel's listed maximum line drop were instead 2.0 V, would the original 16 AWG single-circuit design pass? Show the reasoning.
Read the worked answer
With a 2.0 V allowance, compare the computed 3.37 V to 2.0 V: 3.37 V > 2.0 V, so it still fails. The margin is not close. To pass at 2.0 V with 16 AWG you would need to reduce current to I = 2.0 / 2.09 = 0.957 A, i.e., no more than 8 devices (8 x 0.115 = 0.92 A, drop = 1.92 V) on that single run. This illustrates why the manufacturer's published line-drop limit - not a generic rule of thumb - governs the design, and why the device count per circuit must be verified against the actual wire gauge and length rather than assumed from the panel's total amp rating.