13.3 Voltage Drop Calculations

13.3 Voltage Drop Calculations

Quick Answer: Voltage drop on a notification or power circuit is I × R_round-trip. Find conductor ohms from AWG resistance per 1000 ft × length/1000 × 2 (for a two-wire out-and-back model), multiply by alarm (or design) load current, subtract from source voltage, and confirm the result is still ≥ listed minimum operating voltage at the last appliance. Blueprint 2.3.4 expects you to work the numbers—not only name the concept.

Battery sizing (13.1–13.2) answers “will the system run long enough on secondary power?” Voltage drop answers “will the farthest appliances still operate at adequate voltage under load?”

Ohm’s Law on Fire Alarm Circuits

V_drop = I_load × R_circuit

V_at_last_device ≈ V_source − V_drop
(for a lumped model that places the entire load at the end—common conservative exam/submittal simplification)

More refined models distribute load along the run; unless the question specifies a method, use the stated method on the calc sheet (Chapter 12 notes) and stay consistent.

Round-Trip Resistance

Current leaves on one conductor and returns on the other. Resistance is approximately:

R_one_way = (Ω per 1000 ft) × (one-way feet / 1000)
R_round_trip = 2 × R_one_way

If someone uses only one-way resistance, drop is about half of reality—false “pass.”

Typical Copper Conductor Resistances (State and Use These)

Approximate DC resistance of copper conductors commonly used in fire-alarm voltage-drop teaching examples (ohms per 1000 ft). Real projects must use the resistance from the cable cut sheet or an accepted copper table for the stranding and temperature; for exam work, state the value you use:

Smaller AWG number → thicker wire → lower resistance → less voltage drop (and larger conduit fill—Section 13.4).

Minimum Voltage at the Last Device

Appliances are listed to operate down to a minimum voltage. Training discussions often mention figures near 16 V on some 24 V notification appliances, or about 85% of nominal (0.85 × 24 V = 20.4 V) as a design rule of thumb used in some specs and older practice. Neither number is a free universal substitute for the listing.

Level II correct approach:

1. Read the appliance cut sheet / listing for minimum operating voltage (and any special NAC compatibility notes).
2. Put that V_min in the calc sheet and in drawing notes (Chapter 12).
3. Show V_end ≥ V_min under the design I_load and length.
4. If the exam gives V_min, use the given value; if it asks conceptually, answer “use the manufacturer/listed minimum,” not a memorized single magic number for every product.

Also confirm the source voltage assumption: some calcs use regulated 24 V; some use a lower battery end-of-discharge voltage. The note and the spreadsheet must match.

Class A vs Class B Path Length

Exam idea: Class A improves survivability against a single open; it does not automatically reduce voltage drop. Extra copper in the loop can increase resistance if the effective path is longer. Always take length from the same plans used for the riser and notes—not by scaling a not-to-scale single-line sketch as if it were a tape measure (Chapter 12 trap).

Complete Numeric Example

Given (state all assumptions):

• NAC source voltage used for calc: 24.0 V
• Total alarm load on the NAC: 1.50 A (sum of appliance currents)
• One-way circuit length to last device: 400 ft
• Conductor: 14 AWG copper at 2.53 Ω/1000 ft (stated)
• Listed minimum at last appliance: 16.0 V (given by product data for this example)
• Model: entire 1.50 A treated at the end (conservative lumped load)

Step 1 — Round-trip resistance

R_one_way = 2.53 × (400 / 1000) = 2.53 × 0.4 = 1.012 Ω
R_round_trip = 2 × 1.012 = 2.024 Ω

Step 2 — Voltage drop

V_drop = 1.50 A × 2.024 Ω = 3.036 V

Step 3 — Voltage at last device

V_end = 24.0 − 3.036 = 20.964 V

Step 4 — Compare to V_min

20.964 V ≥ 16.0 V → PASS for this criterion.

Same Circuit, Smaller Wire (Fail Path)

Change to 18 AWG at 6.39 Ω/1000 ft, same 400 ft, 1.50 A, 24.0 V source:

R_rt = 2 × 6.39 × 0.4 = 5.112 Ω
V_drop = 1.50 × 5.112 = 7.668 V
V_end = 24.0 − 7.668 = 16.332 V

Still above 16.0 V—barely. If V_min were 20.4 V (85% of 24 V design rule on a specification), 16.332 V would FAIL even though 16.0 V listing might pass. This is why notes must lock the pass criterion and wire size cannot be “improved” downward in the field without recalculation (Chapter 12).

Length Sensitivity

Keep 14 AWG, 1.50 A, 24 V, but 800 ft one-way:

R_rt = 2 × 2.53 × 0.8 = 4.048 Ω
V_drop = 1.50 × 4.048 = 6.072 V
V_end = 17.928 V — still ≥ 16.0 V here, but margins shrink fast with length and load.

Mitigations When Drop Fails

• Increase conductor size (lower Ω/kft).
• Shorten the run or split into two NACs (halve current or length per circuit).
• Add a remote booster / NAC power supply closer to the load.
• Reduce candela settings only if still code-compliant and redrawn/recalculated.
• Never “fix” drop by ignoring round-trip or by using standby current instead of alarm current.

Power Circuits vs NAC Circuits

The same I × R method applies to:

• 24 V auxiliary power to door holders, modules, or remote boards (use the current those loads draw).
• SLC voltage drop is often less of a battery-Ah issue and more of a manufacturer distance/branch limit—follow product limits; still understand resistance grows with length and thin wire.

For NAC exam items, assume alarm load current unless told otherwise.

Submittal Documentation Checklist

• Circuit ID matching the drawings.
• AWG, copper type, Ω/kft value used.
• One-way length and how it was measured.
• Load current basis (appliance table).
• Source voltage assumption.
• V_min criterion and product reference.
• Pass/fail and revision if wire or devices change.

Bridge to 13.4

Upsizing from 18 AWG to 12 AWG may fix voltage drop while breaking conduit fill. Always re-check Chapter 9 fill when wire size or count changes.