6.7 Principles of Electricity and Notification Circuit Voltage Drop

Why This Section Matters

The SFMO states plainly that the technician test may include "questions based on practical knowledge of general electrical wiring practices or principles of electricity." NFPA 72 assumes this knowledge; it does not teach it. Voltage-drop work is also the calculation most often required on a real Texas plan submittal.

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1. The Three Relationships You Must Know Cold

Where E is volts, I is amperes, R is ohms, and P is watts.

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2. Series and Parallel Behavior

Why it matters. Notification appliances are wired in parallel across the NAC, so their currents add. The conductors are in series with the load, so the conductor resistance eats supply voltage before it reaches the last appliance.

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3. Conductor Resistance

DC resistance of uncoated solid copper, per NEC Chapter 9, Table 8:

Stranded conductors of the same size have slightly higher resistance. Two rules of thumb fall out of the table: going up one wire size cuts resistance by roughly ⅜, and going up two sizes roughly halves it.

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4. Two Ways to Calculate Voltage Drop

Method 1 — Resistance method

   R_total = 2 x (one-way length in ft) x (ohms per 1,000 ft) / 1,000
   V_drop  = I x R_total
   V_end   = V_supply - V_drop

The 2 × is the round trip: current travels out on one conductor and back on the other.

Method 2 — Circular-mil method

   V_drop = (2 x K x I x L) / CM

   K  = 12.9 for copper (ohms per circular-mil-foot, at typical operating temperature)
   I  = total circuit current in amperes
   L  = one-way length in feet
   CM = conductor area in circular mils

Both methods give the same answer within rounding. The circular-mil form is convenient because it lets you solve for the required wire size:

   CM_required = (2 x K x I x L) / V_drop_allowed

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5. Worked Examples

Example 1 — Will the last appliance work?

A NAC leaves a panel that regulates 24.0 VDC at the terminals. The circuit is 18 AWG, the last appliance is 420 feet away, and the total alarm current on the circuit is 0.90 A. Appliances are listed for 16 to 33 VDC.

   R_total = 2 x 420 x 7.77 / 1000 = 6.53 ohms
   V_drop  = 0.90 x 6.53           = 5.87 V
   V_end   = 24.0 - 5.87           = 18.13 V

18.13 V is above the 16 V floor — the circuit passes. Note how little margin remains: adding two more strobes, or the panel regulating at 20.4 V under battery, could push the end of the circuit below listing.

Example 2 — Worst case on battery

Repeat Example 1 assuming the panel delivers only 20.4 VDC at the end of the secondary-power period.

   V_end = 20.4 - 5.87 = 14.53 V   -> BELOW the 16 V minimum. FAILS.

This is why competent designers calculate at the lowest regulated output voltage the panel is listed to deliver, not at 24.0 V. Fire alarm systems are required to perform on secondary power, so the calculation must be run there.

Example 3 — Size the conductor

Same 420-foot run and 0.90 A, but the design must hold the drop to 4.0 V.

   CM_required = (2 x 12.9 x 0.90 x 420) / 4.0
               = 9752 / 4.0
               = 2438 circular mils

2,438 cmil exceeds 18 AWG (1,620 cmil), so step up to 16 AWG (2,580 cmil). Verify:

   V_drop = (2 x 12.9 x 0.90 x 420) / 2580 = 3.78 V   -> under 4.0 V. Passes.

Example 4 — Current draw from candela ratings

Five horn/strobes on a NAC draw 0.075 A each at 15 cd and 0.041 A each for the horn:

   Strobe current = 5 x 0.075 = 0.375 A
   Horn current   = 5 x 0.041 = 0.205 A
   Total          =             0.580 A

Always sum the alarm current — not the standby current — for a NAC voltage-drop calculation, and use the manufacturer's published draw at the candela setting actually installed.

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6. Two Common Errors

1. Forgetting the round trip. Using one-way length halves the calculated drop and produces circuits that fail on the last appliance. The factor of 2 is not optional.
2. Treating the end-of-line resistor as a circuit design element. The EOL resistor exists to prove continuity for supervision. It does not limit load, it does not affect alarm current, and increasing its value does not fix a voltage-drop problem. Fixing voltage drop means shorter runs, larger conductors, fewer appliances per circuit, lower candela settings, or a booster power supply closer to the load.

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7. Related Practical Wiring Points

• Class A and Class X pathways require the outgoing and return conductors to be routed separately so a single physical event cannot damage both; that routing also lengthens the run and increases drop.
• A booster power supply placed near the load is often cheaper and more reliable than upsizing hundreds of feet of conductor — but the booster is itself part of the system, with its own secondary supply and supervision.
• Terminations matter. A loose or oxidized termination adds series resistance exactly where the calculation assumes none.