6.6 Voltage Drop and Wire Resistance

Voltage Drop and Wire Resistance

Voltage drop is the loss of voltage caused by current flowing through conductor resistance. On a NAC, if the drop is too large the farthest horn/strobe may receive less than its listed minimum operating voltage — commonly about 16 V for 24 V-nominal appliances — and may fail to flash, sound at reduced volume, or behave erratically. Voltage drop most often appears with NACs, auxiliary power outputs, and long runs.

The governing relationship is Ohm's law applied to the conductors:

Voltage drop (V) = Circuit current (A) × Total conductor resistance (Ω)

Three details decide the answer:

1. Source voltage. Battery-powered NAC calcs typically start at a worst-case 20.4 V (the battery's discharged/end-of-life floor, ~15% below 24 V), not the nominal 24 V — this is the conservative value AHJs and designers use.
2. Round-trip length. Current flows out and back, so a circuit with a one-way run of L feet has 2L feet of conductor resistance. Forgetting the return path understates the drop by half.
3. Compare remaining voltage, not the drop. The pass/fail test is (source − drop) ≥ appliance minimum.

Standard solid-copper resistance values (per 1000 ft, one conductor):

Read the exhibit's units before calculating: some give Ω/1000 ft, some give Ω/ft, some give total ohms. A unit slip here is a common, costly error.

Fully Worked Voltage-Drop Example

Scenario: A NAC carries 1.62 A of strobe load (from 6.4) to appliances at the end of a 500 ft (one-way) run of 14 AWG copper. Source = worst-case 20.4 V. Appliance minimum = 16 V. Does the end appliance receive enough voltage? (Simplified end-loaded model — all current to the far end.)

Voltage-drop worksheet

Step-by-step:

1. Round-trip conductor length = 2 × 500 = 1000 ft.
2. Total resistance = 1000 ft × 3.14 Ω/1000 ft = 3.14 Ω.
3. Voltage drop = 1.62 A × 3.14 Ω = 5.09 V.
4. Voltage at the appliance = 20.4 − 5.09 = 15.31 V.
5. 15.31 V is below the 16 V minimum → the circuit fails.

Design fixes (each reduces drop): increase wire size (12 AWG cuts resistance from 3.14 to 1.93 Ω/1000 ft → drop falls to about 3.13 V, giving ~17.27 V — a pass); shorten the run; split into two shorter NACs; relocate the panel/booster supply closer; or use lower-current LED appliances. Re-running with 12 AWG: total R = 1000 × 1.93/1000 = 1.93 Ω, drop = 1.62 × 1.93 = 3.13 V, end voltage = 20.4 − 3.13 = 17.27 V ≥ 16 V → PASS.

Trap 1 — answering with the drop: if the drop is 5.09 V, the appliance sees 15.31 V, not 5.09 V; compare the remaining voltage. Trap 2 — one-way length: using 500 ft instead of the 1000 ft round trip halves the drop and produces a false pass. Trap 3 — starting at 24 V: using nominal 24 V instead of the worst-case 20.4 V makes a marginal circuit look acceptable. Trap 4 — mixing units: treating 1.62 A as 1620 (mA) inside an amp equation.

Voltage drop interacts with NAC loading (6.4): a circuit can pass ampacity yet fail voltage drop, or vice versa. The correct exam answer checks the condition the stem actually asks. A plan that shows NAC counts and wire size but no current or voltage-drop basis has a documentation gap a reviewer should flag.

Why 20.4 V, and the Segment-Current Refinement

The 20.4 V starting point deserves a sentence of why: a 24 V-nominal system rides on a battery, and a discharged lead-acid battery near the end of its standby duty can sag to roughly 85% of 24 V, i.e., about 20.4 V. Designing from 20.4 V guarantees the appliance still gets ≥16 V even on the worst day of a power outage. Using the cheerful nominal 24 V hides exactly the condition the calculation is meant to protect against, which is why it is a deliberate trap value in answer choices.

The end-loaded model used in the worked example (all current at the far end) is the conservative case and is what most exam problems use. Real circuits with appliances spread along the run carry less current in the later segments, so a segment-by-segment analysis yields a smaller drop. If an exhibit explicitly gives device positions and per-segment currents, sum the drop segment by segment: each segment's drop = (current in that segment) × (segment round-trip resistance), and the total drop to the last device is the sum.

If the exhibit gives only a total current and total length, the end-loaded model is expected. Choosing the wrong model is a subtle trap; when in doubt, the conservative end-loaded result will never under-state the drop.

A Reusable Voltage-Drop Checklist

1. Identify the critical (farthest) appliance.
2. Determine the current in the conductor path to it.
3. Pull resistance from the table in the correct units (Ω/ft vs Ω/1000 ft).
4. Use the round-trip length (2 × one-way).
5. Compute drop = current × total resistance.
6. Subtract from the worst-case source (20.4 V for battery NACs).
7. Compare the remaining voltage to the appliance minimum (~16 V).
8. If it fails, upsize wire, shorten/split the run, relocate the supply, or use lower-current LED appliances.

Because personal calculators are barred, write each labeled value — amps, ohms, feet, volts — and keep the arithmetic linear. If answer choices are close, rounding matters; if they differ by 2× or 1000×, suspect a forgotten return path or a milliamp/amp slip rather than a rounding issue.