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Try a full lesson: Power supply and loading

This is a complete lesson from the course (module 11 of 12, NICET task 2.3.4), followed by 10 practice questions with full explanations. Nothing to sign up for.

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Why this task matters at Level II

At Level I you install what the drawings show. At Level II you are expected to check that the design actually works: that the batteries will carry the system through an outage, that each notification circuit is not overloaded, and that the last strobe on the circuit still gets enough voltage to flash. NICET puts this under task 2.3.4, inside the submittal and layout domain, which is 20 to 30 percent of the Level II exam.

You will not be asked to recite code text. You will be asked to run the numbers and pick the right result, often with the calculator built into the test software. This lesson walks through the three calculations you need, each with a worked example you can repeat with your own figures.

Keep your own copy of NFPA 72 open while you work. We cite sections by number so you can find them; we never reproduce the code's wording.

1. Secondary power: sizing the standby battery

When normal power fails, the batteries have to run the system in standby for a set time and then still power a full alarm. For a typical fire alarm system that means 24 hours of standby followed by 5 minutes of alarm. An emergency voice/alarm communications system needs 15 minutes of alarm at full load instead of 5. Find these in NFPA 72-2022 Chapter 10, under the secondary power supply requirements (section 10.6.7).

The method is the same every time:

Add up the standby current: the panel's own standby draw plus every device that draws current while the system is quiet (detectors, modules, annunciators).
Add up the alarm current: the panel's alarm draw plus everything that turns on in alarm, which is mostly your notification appliances.
Multiply each by its time in hours. Standby is 24 h. Five minutes of alarm is 5/60 = 0.0833 h; fifteen minutes is 0.25 h.
Add the two amp-hour figures.
Add the capacity margin. NFPA 72 Chapter 10 requires a margin on battery capacity, and most manufacturer worksheets apply 20 percent. Check how your edition and the panel manual state it before you rely on a figure.
Pick the next standard battery size up, never down.

Worked example

A small office building has a conventional panel and one NAC power supply. From the data sheets:

Panel standby 0.120 A, panel alarm 0.250 A
28 smoke detectors at 0.00009 A each in standby (0.0025 A in total)
Notification appliances draw nothing in standby and 1.80 A in alarm

Standby current: 0.120 + 0.0025 = 0.1225 A

Alarm current: 0.250 + 1.80 = 2.05 A

Standby amp-hours: 0.1225 A × 24 h = 2.94 Ah

Alarm amp-hours: 2.05 A × 0.0833 h = 0.171 Ah

Total: 2.94 + 0.171 = 3.11 Ah

With a 20 percent margin: 3.11 × 1.2 = 3.73 Ah

The next standard size up is a pair of 12 V, 4 Ah batteries (wired in series for 24 V). Some designers would choose 7 Ah for future additions, which is fine; choosing 3.5 Ah is not.

What the exam tests here. The common wrong answers come from forgetting to convert 5 minutes to hours (using 5 instead of 0.0833 inflates the result about 60 times), adding the margin to only one of the two terms, or rounding down to a battery that is "close enough".

2. Notification circuit loading

Each notification appliance circuit (NAC) has a current rating, and so does the power supply that feeds it. Before you add appliances, you need to know the circuit can carry them.

Take each appliance's current at the setting you will use. Strobe current changes with the candela setting, so a 75 cd strobe draws more than a 15 cd one. Use the manufacturer's figure for the voltage the circuit will run at; many data sheets list a higher current at the low end of the voltage range, and that is the one to use.
Add them up for the circuit.
Compare the total with the NAC's rating and with whatever is left on the power supply after its other circuits.

Worked example

A NAC is rated 2.5 A. The plan puts these appliances on it (example values, not from a real data sheet):

8 horn-strobes at 15 cd, 0.070 A each = 0.560 A
4 horn-strobes at 75 cd, 0.165 A each = 0.660 A
2 strobes at 110 cd, 0.210 A each = 0.420 A

Circuit total: 0.560 + 0.660 + 0.420 = 1.64 A

That is within the 2.5 A rating with 0.86 A to spare, about 34 percent headroom. If the power supply has two other NACs drawing 1.9 A and 2.1 A and is rated 6.0 A in total, the supply carries 1.64 + 1.9 + 2.1 = 5.64 A, which is within its rating but leaves only 0.36 A. Note that on the submittal: one more circuit of strobes will need another power supply.

3. Voltage drop on a notification circuit

Appliances are listed to work down to a minimum voltage, often 16 V for 24 V regulated appliances. The circuit starts at the lowest voltage the power supply delivers at the end of its battery life (the panel manual gives this; 20.4 V is common for 24 V systems), and the wire eats into that voltage on the way out and back.

Find the wire's resistance per 1,000 ft. For solid uncoated copper, NFPA 70 Chapter 9 Table 8 gives about 3.07 ohms per 1,000 ft for 14 AWG and about 1.93 ohms per 1,000 ft for 12 AWG.
Use the loop length: current goes out and comes back, so double the one-way distance.
Resistance of the run = (ohms per 1,000 ft) × (loop length ÷ 1,000).
Voltage drop = current × resistance (Ohm's law).
Voltage at the last appliance = starting voltage minus drop. It has to be at or above the appliance's listed minimum.

Treating the whole load as if it sat at the far end of the circuit (a "lumped" load) overstates the drop, so if the circuit passes that way it passes. Real designs often split the load into segments for a more accurate figure.

Worked example: does the circuit pass?

Take the 1.64 A circuit above, wired in 14 AWG solid copper, with the last appliance 260 ft from the power supply.

Loop length: 2 × 260 = 520 ft

Resistance: 3.07 × (520 ÷ 1,000) = 1.596 ohms

Voltage drop: 1.64 A × 1.596 ohms = 2.62 V

Voltage at the last appliance: 20.4 − 2.62 = 17.78 V

17.78 V is above the 16 V minimum, so the circuit passes, with 1.78 V to spare.

Worked example: how far can the circuit run?

Turn the question around. With the same 1.64 A load, how long can the 14 AWG circuit be?

Allowed drop: 20.4 − 16.0 = 4.4 V

Maximum resistance: 4.4 V ÷ 1.64 A = 2.68 ohms

Maximum loop length: 2.68 ÷ 3.07 × 1,000 = 873 ft

Maximum one-way length: 873 ÷ 2 = about 437 ft

Moving up to 12 AWG: 2.68 ÷ 1.93 × 1,000 = 1,390 ft of loop, about 695 ft one way. That is the usual fix when a long corridor fails the check, along with splitting the circuit or adding a NAC power supply closer to the load.

What the exam tests here. The classic mistakes are using the one-way length instead of the loop, starting from 24 V instead of the end-of-battery voltage, and using the 15 cd current for a strobe set to a higher candela.

Check yourself

The free practice set below has questions on these three calculations, written in the style of the exam, with a full explanation for every answer.

Where to look in your own books

NFPA 72-2022 Chapter 10: power supplies, secondary power capacity (10.6.7) and battery requirements
NFPA 72-2022 Chapter 18: notification appliances, including how appliances are listed and installed
NFPA 70 (2020) Chapter 9, Table 8: conductor properties
The panel and power supply installation manuals: end-of-battery voltage, NAC ratings and battery calculation worksheets

Practice set

Pick an answer in your head, then open the explanation. Every wrong option is explained too.

1. Submittal Preparation and System Layout

A system draws 0.30 A in standby and 3.2 A in alarm. It needs 24 hours of standby followed by 5 minutes of alarm, and the design adds a 20% margin to the total calculated capacity. What is the minimum battery capacity?

8.91 Ah
8.96 Ah
27.84 Ah
7.47 Ah

Hint: where to look

Secondary power capacity is in NFPA 72 Chapter 10. Work standby and alarm separately in amp-hours, and watch the unit for the alarm time.

Show answer and explanation

Answer: B. 8.96 Ah

Standby is 0.30 A x 24 h = 7.20 Ah. Alarm is 3.2 A x (5/60) h = 0.267 Ah. The total is 7.467 Ah, and 7.467 x 1.2 = 8.96 Ah. 8.91 Ah applies the margin to the standby term only. 7.47 Ah leaves out the margin. 27.84 Ah treats 5 minutes as 5 hours.

Reference: NFPA 72 ch.10 (secondary power supply capacity, 10.6.7); panel manual battery worksheet

2. Submittal Preparation and System Layout

A notification circuit carries 1.2 A, treated as a lumped load at the far end. It is wired in 12 AWG solid copper (1.93 ohms per 1,000 ft) and the last appliance is 400 ft from the power supply. The supply's end-of-battery output is 20.4 V. What voltage reaches the last appliance?

16.00 V
19.47 V
22.15 V
18.55 V

Hint: where to look

Ohm's law plus the conductor resistance table in NFPA 70 Chapter 9. Think about how far the current actually travels in a two-wire circuit.

Show answer and explanation

Answer: D. 18.55 V

Loop length is 2 x 400 = 800 ft, so resistance is 1.93 x 0.8 = 1.544 ohms. Drop is 1.2 A x 1.544 = 1.85 V, and 20.4 - 1.85 = 18.55 V. 19.47 V uses the one-way length. 22.15 V starts from 24 V instead of the end-of-battery voltage. 16.00 V is a typical appliance minimum, not the result.

Reference: NFPA 72 ch.18 (appliance operating voltage); NFPA 70 ch.9 Table 8 (conductor resistance)

3. Submittal Preparation and System Layout

A NAC rated 2.0 A already carries 1.45 A. You want to add strobes that draw 0.095 A each at the chosen setting. How many can you add without exceeding the circuit rating?

Hint: where to look

Start from what the circuit has left, not its full rating, and remember you can't install part of a strobe.

Show answer and explanation

Answer: A. 5

Spare capacity is 2.0 - 1.45 = 0.55 A, and 0.55 / 0.095 = 5.8, so 5 strobes fit. 6 would draw 0.57 A and exceed the rating, which is why you round down. 21 ignores the existing load. 15 divides the existing load by the strobe current.

Reference: NFPA 72 ch.10 (power supplies) and ch.18; manufacturer NAC ratings

4. Installation

Power-limited fire alarm cable will be run exposed, not in raceway, through a ceiling space used for environmental air (a plenum). Which cable type is suitable?

FPL
FPLR
FPLP
CM

Hint: where to look

NFPA 70 Article 760 lists the power-limited fire alarm cable types and the spaces each is permitted in.

Show answer and explanation

Answer: C. FPLP

FPLP is the plenum-rated power-limited fire alarm cable. FPLR is riser-rated and FPL is general-purpose, and neither is permitted exposed in a plenum. CM is communications cable, not a fire alarm cable type.

Reference: NFPA 70 Article 760 (power-limited fire alarm cable types and uses)

5. Installation

A wall-mounted strobe is being installed in a corridor with a 10 ft ceiling. Where should its height be checked?

At the same height as the nearest manual pull station
As close to the ceiling as possible, so it is above furniture
So the entire lens sits within the mounting height range in NFPA 72 ch.18, measured from the finished floor
At any height, as long as the strobe is visible from the corridor entrance

Hint: where to look

NFPA 72 Chapter 18 covers wall-mounted visible appliances. Check what the height range is measured from and which part of the appliance must fall inside it.

Show answer and explanation

Answer: C. So the entire lens sits within the mounting height range in NFPA 72 ch.18, measured from the finished floor

Wall-mounted visible appliances have a required height range measured from the finished floor, and the whole lens has to fall within it (look up the range in NFPA 72 ch.18). Visibility from one point is not the test. Mounting tight to the ceiling can put the lens above the range, and pull stations have their own, different mounting rule.

Reference: NFPA 72 ch.18 (wall-mounted visible appliances, 18.5.5)

6. Installation

During acceptance testing you find several smoke detectors still wearing the dust covers fitted during construction. What should you do?

Remove the covers, test those detectors and note it on the test record
Skip those detectors and mark them as tested
Test the detectors with the covers on, since smoke reaches them eventually
Leave the covers on until the building is occupied

Hint: where to look

Think about what a detector needs to sense smoke, and what NFPA 72 Chapter 14 expects you to record during acceptance testing.

Show answer and explanation

Answer: A. Remove the covers, test those detectors and note it on the test record

Dust covers stop smoke entering the chamber, so a covered detector cannot work. They come off before the detector is put in service, and the detector is then tested and recorded. Testing with the cover on proves nothing, leaving them on leaves the area unprotected, and marking untested devices as tested falsifies the record.

Reference: NFPA 72 ch.14 (acceptance testing); ch.17 (detector protection during construction)

7. Maintenance

Besides an alarm test, what does a periodic test of a duct smoke detector with sampling tubes need to confirm?

That the duct's fan has been shut off before testing
That the detector's sensitivity matches a nearby spot detector
That the detector sounds a local horn
That airflow through the sampling tubes is within the manufacturer's listed range

Hint: where to look

The testing methods table in NFPA 72 Chapter 14 has a row for duct detectors. Ask what makes a duct detector different from a spot detector.

Show answer and explanation

Answer: D. That airflow through the sampling tubes is within the manufacturer's listed range

A duct detector only works if enough duct air passes through its housing, so the pressure differential or airflow is checked against the manufacturer's range. Duct detectors do not need a local horn, the test is normally done with the fan running so airflow can be measured, and its sensitivity is checked against its own listing, not another detector.

Reference: NFPA 72 ch.14 (testing methods table, duct smoke detectors); manufacturer instructions

8. Maintenance

A ground fault trouble appears on one circuit only after heavy rain and clears a day later. Where is the most likely cause?

A failing standby battery
Water getting into a device box, junction box or conductor run on an exterior wall or underground section of that circuit
An incorrect end-of-line resistor value
A detector set to the wrong sensitivity

Hint: where to look

Ask what changes when it rains. Monitoring for integrity is in NFPA 72 Chapter 12.

Show answer and explanation

Answer: B. Water getting into a device box, junction box or conductor run on an exterior wall or underground section of that circuit

A fault that follows the weather points to moisture creating a path to ground, usually where the circuit runs outdoors, underground or through an exterior wall. Battery problems show as a power trouble, a wrong end-of-line value causes a constant open or trouble rather than one tied to rain, and detector sensitivity does not create a ground fault.

Reference: NFPA 72 ch.12 (monitoring for integrity, ground faults); ch.14 (troubleshooting)

9. Maintenance

After an annual inspection and test, how long must the records be kept?

Until the next test, and for one year after that
For the life of the building
For 30 days
Only until the AHJ has seen them

Hint: where to look

Record retention for inspection and testing is in NFPA 72 Chapter 14, section 14.6.

Show answer and explanation

Answer: A. Until the next test, and for one year after that

Inspection, testing and maintenance records are kept until the next test and for one year after it, so there is always a full test cycle on file. Thirty days or until an AHJ visit would leave gaps, and life-of-building retention applies to other documents such as the record of completion, not to every periodic test record.

Reference: NFPA 72 ch.14 (records, 14.6)

10. Management and Supervision

The AHJ acceptance test is booked for Friday, but the elevator contractor cannot attend, so the recall functions cannot be confirmed at the elevators. What is the best action?

Disconnect the elevator recall relays until the elevator contractor is free
Run the test anyway and report recall as passed because the panel showed the relay operated
Tell the general contractor and the AHJ now, and reschedule the elevator part of the test for when the elevator technician can attend
Ask the sprinkler contractor to watch the elevators instead

Hint: where to look

Ask who can prove an elevator actually recalled. Elevator interface testing is covered in NFPA 72 Chapters 14 and 21.

Show answer and explanation

Answer: C. Tell the general contractor and the AHJ now, and reschedule the elevator part of the test for when the elevator technician can attend

Recall is only proven when the elevator responds, which needs the elevator technician, so the right move is to tell the parties early and reschedule that part. Reporting a pass from the panel alone is false, disconnecting the relays creates an impairment, and another trade cannot verify elevator operation.

Reference: NFPA 72 ch.14 and ch.21 (elevator interface testing)

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Questions are original and written to NICET's published content outline. Code is cited by section number so you can look it up in your own copy. Think something is wrong? Report it.