Why this task matters at Level III
At Level II you ran the battery, loading and voltage drop numbers yourself. At Level III you check numbers someone else ran and put your name on the result. Task 3.3.1 covers preparing and approving shop drawings, and the calculations are where a quiet error turns into a dead strobe at the acceptance test or a panel that drops out during an outage.
A reviewer does not redo every line by habit. A reviewer knows where calculations usually go wrong, rebuilds the critical ones from the data sheets, and writes comments a designer can act on without a phone call. This lesson works through three checks with invented numbers, then covers how to accept, comment on or return the package.
Keep your own copy of NFPA 72 open. We cite sections by number and never reproduce the code's wording.
Checking battery and voltage drop calculations
Before checking any arithmetic, check the inputs. Most rejected calculations fail on inputs, not on multiplication.
- Every current figure should trace to a named data sheet and model.
- Strobe currents should match the candela setting on the plans, at the low end of the operating voltage.
- Every device on the plans should appear in the calculation. Count them.
- The durations should match the system type. A typical system needs 24 hours of standby followed by 5 minutes of alarm. An emergency voice/alarm communications system needs 15 minutes of alarm instead. Read NFPA 72-2022 Chapter 10 in your own copy.
- The margin should come from the code edition in force and the manufacturer's worksheet.
Worked calculation 1: checking a standby and alarm battery calculation
Exercise data (invented). A submitted calculation for one panel claims 6.89 Ah and selects a pair of 12 V, 7 Ah batteries. The data sheets give:
- Panel: 0.180 A standby, 0.300 A alarm
- Detectors and modules on the SLC: 0.095 A standby in total
- Remote annunciator: 0.040 A standby, 0.060 A alarm
- Notification appliances: 0 A standby, 3.20 A alarm
- Margin for this exercise: 20 percent. Battery sizes offered: 7, 12 and 18 Ah.
Step 1. Standby current. 0.180 + 0.095 + 0.040 = 0.315 A
Step 2. Alarm current. 0.300 + 3.20 + 0.060 = 3.56 A
Step 3. Standby amp-hours. 0.315 A × 24 h = 7.56 Ah
Step 4. Alarm amp-hours. 5 minutes = 5 ÷ 60 = 0.0833 h. 3.56 A × 5 ÷ 60 = 0.297 Ah
Step 5. Total. 7.56 + 0.297 = 7.857 Ah
Step 6. Margin. 7.857 × 1.20 = 9.43 Ah
Step 7. Select. The next size up is 12 Ah. The submitted 7 Ah battery fails.
Finding the error. Remove the annunciator: standby 0.275 A × 24 = 6.60 Ah, alarm 3.50 A × 5 ÷ 60 = 0.292 Ah, total 6.89 Ah with no margin. That matches the submittal, so the comment is specific: annunciator omitted and no margin applied.
Double check. 0.315 × 24: 0.3 × 24 = 7.2, plus 0.015 × 24 = 0.36, so 7.56. 3.56 ÷ 12 = 0.2967. 7.857 × 1.2 = 7.857 + 1.571 = 9.428. Same result twice.
Worked calculation 2: checking end-of-circuit voltage and maximum run length
Exercise data (invented). One NAC carries 1.85 A of appliances, treated as a lumped load at the far end. Wire is 12 AWG solid copper at 1.93 ohms per 1,000 ft (NFPA 70-2020 conductor properties table). The last appliance is 310 ft from the power supply. The supply's end-of-battery output is 20.4 V and the appliances are listed down to 16.0 V. The submittal claims a maximum run of 1,120 ft.
Step 1. Loop length. 2 × 310 = 620 ft
Step 2. Resistance. 1.93 × 620 ÷ 1,000 = 1.197 ohms
Step 3. Voltage drop. 1.85 A × 1.197 ohms = 2.21 V
Step 4. End voltage. 20.4 - 2.21 = 18.19 V. That is above 16.0 V, so this circuit passes.
Step 5. Maximum run. Allowed drop = 20.4 - 16.0 = 4.4 V. Maximum resistance = 4.4 ÷ 1.85 = 2.378 ohms. Maximum loop = 2.378 ÷ 1.93 × 1,000 = 1,232 ft. Maximum one way = 616 ft.
Finding the error. Starting from 24 V instead: allowed drop 8.0 V, 8.0 ÷ 1.85 = 4.324 ohms, 4.324 ÷ 1.93 × 1,000 = 2,240 ft of loop, 1,120 ft one way. The submittal used nominal voltage. Its claimed maximum is almost double the real one, which matters the day someone extends this circuit.
Double check. 616 × 2 = 1,232 ft. 1.232 × 1.93 = 2.378 ohms. 2.378 × 1.85 = 4.40 V. 20.4 - 4.40 = 16.0 V. Correct.
Worked calculation 3: checking NAC load against a power supply rating
Exercise data (invented). A NAC power supply has four outputs rated 2.5 A each and a total output rating of 6.0 A. The submittal shows each circuit under 2.5 A and marks the supply as adequate. Appliance currents at the low end of the voltage range: 15 cd horn-strobe 0.066 A, 30 cd strobe 0.094 A, 75 cd horn-strobe 0.158 A, 110 cd strobe 0.205 A.
- NAC 1: 10 at 15 cd = 0.660 A, plus 6 at 75 cd = 0.948 A. Total 1.608 A.
- NAC 2: 12 at 30 cd = 1.128 A, plus 4 at 110 cd = 0.820 A. Total 1.948 A.
- NAC 3: 14 at 75 cd = 2.212 A.
- NAC 4: 8 at 15 cd = 0.528 A.
Each circuit is under 2.5 A. Now add them: 1.608 + 1.948 + 2.212 + 0.528 = 6.296 A. That is 0.296 A over the 6.0 A supply rating. The per-circuit check passed; the supply check fails.
Double check. 1.608 + 1.948 = 3.556. 3.556 + 2.212 = 5.768. 5.768 + 0.528 = 6.296. Same total. Comment: move NAC 4 to another supply or a panel NAC, then rerun the battery calculation for whichever supply takes it.
What the exam tests here. Practice finding the input error, not only the arithmetic one: a missing device, nominal instead of end-of-battery voltage, one-way length for loop length, a margin left off, or circuits checked one at a time against a supply that carries them all.
Check yourself
The free practice set below has questions 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: secondary power capacity and duration
- NFPA 72-2022 Chapter 18: notification appliance listing and installation
- NFPA 72-2022 Chapter 7: shop drawing content
- NFPA 70-2020, the conductor properties table: conductor resistance
- IBC 2021, the section on submittal documents: construction documents submitted to the building department
- The panel and power supply manuals: end-of-battery voltage, output ratings, battery worksheet