Voltage Drop on a 24V Notification Appliance Circuit
A low voltage contractor is calculating voltage drop for a 24V DC notification appliance circuit in a fire alarm system. The circuit has 18 horn/strobes, each rated 0.145A at 24V DC. The one-way cable length is 250 feet, and the contractor plans to use 18 AWG stranded copper. The panel is listed for 24V nominal operation, but its low-battery condition is 20.4V. The appliances are listed to operate at a minimum of 16V. The contractor uses the nominal 24V panel voltage in its calculation and concludes that 18 AWG is acceptable because the drop is less than 10%. The cable will run through a return-air plenum and must be properly listed. The contractor has not checked the manufacturer’s installation instructions for the appliance or the panel, and the first inspection is scheduled for next week.
1. Calculate the total current and the voltage drop for the 18 AWG circuit. Show the formula and state whether the circuit complies.
Read the worked answer
Total current is 18 appliances × 0.145A = 2.61A. The circuit length is the round trip: 2 × 250 ft = 500 ft. Using typical copper DC resistance of 6.51 ohms per 1,000 ft for 18 AWG, the conductor resistance is 500 ÷ 1,000 × 6.51 = 3.255 ohms. Voltage drop = I × R = 2.61A × 3.255 ohms = 8.50V. At the panel’s low-battery condition of 20.4V, the voltage at the end of the circuit is 20.4V − 8.50V = 11.90V. The appliances require a minimum of 16V, so the 18 AWG circuit does not comply. Using nominal 24V and a 10% rule is not the correct test; the listed appliance minimum at the panel’s lowest operating voltage governs.
2. Determine whether 16 AWG or 14 AWG is required, and explain the judgment.
Read the worked answer
Using typical resistance of 4.09 ohms per 1,000 ft for 16 AWG, the resistance is 500 ÷ 1,000 × 4.09 = 2.045 ohms. Drop = 2.61A × 2.045 = 5.34V. At 20.4V, the end voltage is 15.06V, which is still below the 16V appliance minimum. Using 14 AWG at about 2.58 ohms per 1,000 ft, resistance is 1.29 ohms. Drop = 2.61A × 1.29 = 3.37V. End voltage = 20.4V − 3.37V = 17.03V, which exceeds 16V. Therefore 14 AWG is the minimum of these choices. The contractor must use the manufacturer’s actual current draw and resistance data, account for any other devices on the circuit, and verify the panel’s minimum voltage. If the calculation is marginal, a larger conductor or a different circuit arrangement may be needed.
3. What additional cable and installation requirements apply because the cable runs in a return-air plenum?
Read the worked answer
Cable installed in a return-air plenum must be listed for plenum use. For power-limited fire alarm circuits, that generally means an FPLP listing or another listing permitted by the adopted fire alarm code and the AHJ. The contractor must also maintain required separation from power conductors and use the manufacturer’s listed installation instructions. The cable must be supported and protected as required, and any penetration of fire-rated assemblies must be firestopped with a listed system. The contractor should verify that the cable is compatible with the panel and appliances and that the installation follows NFPA 72 and the Florida Fire Prevention Code. Using 14 AWG does not satisfy the listing requirement; the correct cable type and installation method still must be inspected and accepted.