NAC Voltage Drop on a Waikiki Hotel Retrofit
Your crew is retrofitting a 60-room, four-story hotel in Waikiki. The existing 24 V DC fire alarm control panel has two notification appliance circuits (NACs). NAC 1 serves 12 listed horn/strobes on one Class B path. The nameplate steady current is 0.145 A per device at 24 V, and the manufacturer requires at least 20.4 V at the last appliance. The run from the panel to the last device is 420 ft one way in 14 AWG stranded copper; the first device reads 23.8 V and the last reads 19.1 V during alarm. The owner wants to add three more strobes to the same NAC because the panel has spare capacity. The electrical inspector has asked for voltage-drop calculations and a battery standby check. The ceiling is finished and the owner wants the least invasive fix. You must decide whether to add devices, upsize wire, split the circuit, or add a listed NAC power supply, and document the result.
1. Using the measured values and standard copper conductor data, explain why the last appliance is below its listed minimum. Show the calculation and state whether the measured drop is consistent with 14 AWG copper.
Read the worked answer
The measured drop from the first device to the last is 23.8 V - 19.1 V = 4.7 V. For 14 AWG copper, use about 3.07 ohms per 1,000 ft. The circuit loop is 2 x 420 ft = 840 ft, so R = 3.07 x 0.840 = 2.58 ohms. The steady load is 12 x 0.145 A = 1.74 A. Expected drop = I x R = 1.74 x 2.58 = 4.49 V, so the expected last-device voltage is about 23.8 - 4.49 = 19.3 V. The measured 19.1 V is consistent within measurement and connection tolerance. The appliance is below the listed 20.4 V minimum, so the circuit is already limited by voltage drop, not by panel capacity. Record the first-device voltage and actual current, because strobe inrush and supply regulation can make the worst case slightly lower.
2. The owner proposes adding three devices and keeping 14 AWG. Evaluate this against the manufacturer's minimum and recommend a compliant, least-invasive solution. Include how splitting the NAC or adding a listed power supply changes the calculation.
Read the worked answer
Adding three devices gives 15 x 0.145 = 2.175 A. With the same 2.58 ohm loop, drop = 2.175 x 2.58 = 5.61 V, and the last device would be about 23.8 - 5.61 = 18.2 V, which is worse and non-compliant. Keeping 14 AWG is not acceptable. Splitting the 15 devices into two NACs, for example 8 and 7, drops the current on the loaded circuit to 1.16 A; drop = 1.16 x 2.58 = 2.99 V, leaving about 20.8 V at the last device. That is a compliant least-invasive fix if the panel has a second NAC and battery capacity. A listed NAC power supply near the far end is also acceptable and often better: it shortens the high-current path and lets the panel supervise the extender. Upsizing to 12 AWG would give R = 1.93 x 0.840 = 1.62 ohms; at 2.175 A the drop is 3.52 V, about 20.3 V, which is marginal and harder in a finished ceiling. Recalculate with actual device currents and include inrush and battery standby.
3. What documentation and acceptance tests should be completed before the inspector signs off?
Read the worked answer
Document the final NAC schedule, conductor size, one-way length, device model and tap settings, measured voltage at the first and last appliance under alarm, and actual current. Provide the battery calculation for the code-required standby period plus alarm time, including panel, extender, and NAC loads, with the manufacturer's aging factor. Record that all appliances are listed for the panel and synchronized where required. Before sign-off, test every device on alarm, verify audibility and visual coverage, check NAC supervision for open and short, and confirm the panel reports the extender trouble if used. Give the owner as-built drawings, the record of completion, and the manufacturer's instructions.