Class A Loop Loading and Voltage Drop
A fire alarm contractor is installing a Class A notification appliance circuit (NAC) in a new 12,000 sq ft office building. The panel is a 24 VDC addressable system with a rated NAC output of 3 A per circuit. The design calls for 28 horn/strobes, each rated at 65 mA at 24 VDC, plus 6 remote strobes rated at 45 mA. The wire run is 18 AWG stranded, with a total loop length of 1,200 feet (including the return path). The panel is located in the main electrical room, and the farthest device is approximately 600 feet from the panel along the cable path. The contractor plans to use a single NAC circuit for all devices. The system must comply with NFPA 72 and NEC Article 760. The contractor has a copy of the voltage drop formula: Vdrop = (2 × L × R × I) / 1000, where L is one-way length in feet, R is resistance in ohms per 1000 feet, and I is current in amperes. The resistance of 18 AWG stranded copper is approximately 6.5 ohms per 1000 feet. The panel's low-battery threshold is 21.6 VDC, and the appliances are listed to operate at a minimum of 20 VDC.
1. Calculate the total NAC current and determine whether the single circuit can support all devices. Show your work.
Read the worked answer
First, compute the total current: 28 horn/strobes × 65 mA = 1,820 mA; 6 remote strobes × 45 mA = 270 mA; total = 2,090 mA = 2.09 A. The panel's NAC is rated 3 A, so the load is within the panel's capacity (2.09 A < 3 A). However, we must also check voltage drop. Using the one-way length of 600 ft (since the loop is 1,200 ft total, one-way is 600 ft), R = 6.5 Ω/1000 ft, I = 2.09 A. Vdrop = (2 × 600 × 6.5 × 2.09) / 1000 = (2 × 600 × 6.5 × 2.09) / 1000 = (12,000 × 6.5 × 2.09) / 1000? Wait, compute stepwise: 2 × 600 = 1,200; 1,200 × 6.5 = 7,800; 7,800 × 2.09 = 16,302; divide by 1000 = 16.3 V. That is a 16.3 V drop, which is far too high. The actual voltage at the end would be 24 − 16.3 = 7.7 V, well below the 20 V minimum. Therefore, the single circuit cannot support the load due to excessive voltage drop. The contractor must either use a larger wire gauge, split the devices into multiple circuits, or use a booster power supply. For example, using 14 AWG (R ≈ 3.1 Ω/1000 ft) would yield Vdrop = (2 × 600 × 3.1 × 2.09)/1000 = (1,200 × 3.1 × 2.09)/1000 = (3,720 × 2.09)/1000 = 7,775/1000 = 7.8 V drop, still too high (24 − 7.8 = 16.2 V < 20 V). So even 14 AWG is insufficient; the circuit must be split or a booster used. This illustrates that ampacity alone is not enough; voltage drop must be calculated.
2. If the contractor decides to split the devices into two equally loaded NAC circuits, what is the new voltage drop for each circuit? Assume the same one-way length of 600 ft and 18 AWG wire. Would this meet the 20 V minimum at the end of the circuit?
Read the worked answer
Splitting into two circuits means each circuit carries half the total current: 2.09 A / 2 = 1.045 A. Using the same one-way length (600 ft) and 18 AWG (R = 6.5 Ω/1000 ft), Vdrop = (2 × 600 × 6.5 × 1.045) / 1000 = (1,200 × 6.5 × 1.045) / 1000 = (7,800 × 1.045) / 1000 = 8,151 / 1000 = 8.15 V. The voltage at the end would be 24 − 8.15 = 15.85 V, which is still below the 20 V minimum. Therefore, simply splitting into two circuits is not enough. The contractor would need to further reduce the load per circuit, use a larger wire gauge, or use a booster power supply. For example, using 14 AWG (R = 3.1 Ω/1000 ft) with half the current: Vdrop = (2 × 600 × 3.1 × 1.045)/1000 = (1,200 × 3.1 × 1.045)/1000 = (3,720 × 1.045)/1000 = 3,887/1000 = 3.89 V, giving 24 − 3.89 = 20.11 V, which just meets the 20 V minimum. However, this leaves little margin for battery low condition (21.6 V threshold). At low battery (21.6 V), the end voltage would be 21.6 − 3.89 = 17.71 V, which is below 20 V. Therefore, the design must account for the low battery condition. A better solution is to use a booster power supply closer to the devices or reduce the load further. The key is to calculate voltage drop at the worst-case supply voltage (low battery) and ensure the end voltage is at least 20 V.