Fire Alarm Systems Practice Questions

1. 0.5 × S
2. 0.7 × S
3. 1.0 × S
4. 1.4 × S

Answer: B. NFPA 72 Chapter 17 allows irregular areas to be covered as long as every point on the ceiling is within 0.7 times the selected spacing of a detector; 0.7S is the radius of the circle that circumscribes an S × S square. Half the spacing (0.5S) is the limit from a detector to a wall, not the coverage radius.

Topic: Initiating devices

1. No, the operable part must be 42 to 48 inches high
2. Yes, any height from 36 to 60 inches is acceptable
3. No, the operable part must be exactly 48 inches high
4. Yes, provided the box is within 5 feet of an exit

Answer: A. NFPA 72 Chapter 17 requires the operable part of a manual fire alarm box to be not less than 42 inches and not more than 48 inches above the floor, so 54 inches is too high. Being within 5 feet (60 inches) of the exit door is a separate location requirement and does not excuse the height.

Topic: Initiating devices

1. 12 inches
2. 24 inches
3. 36 inches
4. 48 inches

Answer: C. NFPA 72 Chapter 17 prohibits locating a smoke detector in a direct airflow or closer than 36 inches (3 feet) to an air supply diffuser or return air opening, because moving air can dilute or divert smoke away from the sensing chamber. The detector in this scenario needs to be relocated.

Topic: Initiating devices

1. 60 to 72 inches above the finished floor
2. 72 to 84 inches above the finished floor
3. 80 to 96 inches above the finished floor
4. 96 to 120 inches above the finished floor

Answer: C. NFPA 72 Chapter 18 requires wall-mounted strobes to have the entire lens not less than 80 inches and not more than 96 inches above the finished floor. Where a low ceiling does not allow that, the lens goes within 6 inches of the ceiling, but a 10-foot ceiling leaves room for the standard range.

Topic: Notification appliances

1. Four 0.1-second pulses, then a 5-second pause
2. A continuous tone interrupted once per second
3. Two 1-second pulses, then a 2-second pause
4. Three 0.5-second pulses, 0.5 s apart, then a 1.5-second pause

Answer: D. The Temporal-3 evacuation signal required by NFPA 72 is 0.5 seconds on, 0.5 seconds off, repeated three times, followed by a 1.5-second pause, for a 4-second cycle. The four-pulse pattern with a 5-second pause is Temporal-4, which is reserved for carbon monoxide alarms.

Topic: Notification appliances

1. 4 hours standby, then 5 minutes in alarm
2. 24 hours standby, then 5 minutes in alarm
3. 24 hours standby, then 15 minutes in alarm
4. 72 hours standby, then 15 minutes in alarm

Answer: B. NFPA 72 Chapter 10 requires 24 hours of standby followed by 5 minutes of alarm at maximum connected load. The 15-minute alarm period applies to emergency voice/alarm communications systems, not to a conventional horn/strobe system.

Topic: Power and batteries

1. 25.2 Ah
2. 18.0 Ah
3. 7.5 Ah
4. 6.25 Ah

Answer: C. Standby is 0.25 A × 24 h = 6.0 Ah, alarm is 3.0 A × (5 ÷ 60) h = 0.25 Ah, and the total of 6.25 Ah × 1.2 = 7.5 Ah, so the next larger standard battery is selected. The 6.25 Ah answer skips the derating, and 25.2 Ah comes from mistakenly using 5 hours of alarm instead of 5 minutes.

Topic: Power and batteries

1. It must be protected by a ground-fault circuit interrupter
2. It may share the circuit with the emergency lighting
3. It must be left unlabeled so it is not switched off
4. It must be marked red and identified as FIRE ALARM CIRCUIT

Answer: D. NFPA 72 Chapter 10 and NEC Article 760 require a dedicated branch circuit whose disconnecting means has a red marking, is identified as FIRE ALARM CIRCUIT and is accessible only to authorized personnel; its location is also recorded at the control unit. The circuit must not supply other loads and must not be fed through a GFCI or AFCI device.

Topic: Power and batteries

1. Class A has a redundant path and survives a single open
2. Class A uses heavier conductors than Class B
3. Class B circuits are not monitored for open conductors
4. Only Class B must be installed in metal raceway

Answer: A. NFPA 72 Chapter 12 defines Class A as a pathway with a redundant path so that devices on both sides of a single open or ground still operate, while Class B has no redundant path and loses devices beyond the open. Both classes are monitored for integrity, so a single open produces a trouble signal in either case.

Topic: Circuits and pathways

1. None, because the resistor value is what matters
2. The circuit now behaves as Class A
3. The panel will report a ground fault on that circuit
4. An open conductor on the circuit would not be detected

Answer: D. The end-of-line device is what lets the panel monitor the whole pathway for integrity (NFPA 72 Chapter 12); with the resistor at the panel, the supervisory current never travels the field wiring, so an open anywhere on the circuit would go unnoticed and devices beyond it would be silently disabled. The resistor must be installed past the last device.

Topic: Circuits and pathways

1. Within 1 year of installation, then every other year
2. Within 6 months of installation, then every year
3. Within 2 years of installation, then every 5 years
4. Only when the detector reports a trouble condition

Answer: A. NFPA 72 Chapter 14 requires detector sensitivity to be checked within 1 year after installation and every alternate year thereafter. Only after the second calibration test, if readings stay within the listed range, may the interval be extended to a maximum of 5 years.

Topic: Inspection and testing

1. For 90 days after the test date
2. For 5 years in every case
3. Until the next test and for 1 year thereafter
4. Only until the AHJ has reviewed them

Answer: C. NFPA 72 Chapter 14 requires records to be kept until the next test of the same kind and for 1 year after that, so a prior result is always available for comparison. A record retained for only 90 days would be gone long before the next annual test.

Topic: Inspection and testing

1. 0.06 Ω
2. 16 Ω
3. 36 Ω
4. 160 Ω

Answer: B. Ohm's law gives R = V ÷ I = 24 ÷ 1.5 = 16 Ω. 36 Ω comes from multiplying instead of dividing, and 0.06 Ω is the current divided by the voltage.

Topic: Basic electricity

1. 1.55 V
2. 3.1 V
3. 6.2 V
4. 12.4 V

Answer: B. The current flows out and back, so the loop is 500 feet: 500 × 3.1 ÷ 1,000 = 1.55 Ω, and 2.0 A × 1.55 Ω = 3.1 V. The 1.55 V answer forgets the return conductor; the voltage reaching the last appliance (about 21 V from a 24 V source) must stay within the appliance's listed operating range.

Topic: Basic electricity

1. The same current flows through every appliance
2. The circuit voltage divides among the appliances
3. Total resistance is the sum of the individual appliance resistances
4. Each appliance sees the same voltage and the branch currents add

Answer: D. In a parallel circuit every branch is connected across the same two conductors, so each appliance sees the full circuit voltage and the panel supplies the sum of the branch currents. Equal current and voltage division describe a series circuit, which is why appliances are never daisy-chained in series.

Topic: Basic electricity