The Assumed Voltage Formula: Wire Sizing & Drop Calculations

When sizing conductors, relying on the fluctuating voltage measured at the panel is a mistake. Instead, electrical engineering and NEC-style calculations anchor to the assumed voltage (the nominal system voltage, like 120V, 240V, or 48V). The assumed voltage formula calculates percentage voltage drop by deriving load current from this fixed nominal baseline, revealing the inverse-square relationship between system voltage and copper requirements.

Here is the direct answer for single-phase AC (PF ≈ 1.0) and DC circuits: to find the required wire size in Circular Mils (CM), use CM = (2 × K × P × L) / (%VD × Vassumed2) × 100. Below is the complete derivation, symbol table, and step-by-step worked examples to prove the math.

The Core Assumed Voltage Formula and Symbol Table

The unified formula below combines Ohm's Law and the standard resistivity equation into a single expression. It solves for Percentage Voltage Drop (%VD) based on the assumed nominal voltage.

Source reference: Resistivity constants align with Electrical Technology's NEC Chapter 9 Table 8 derivations.

When to Apply It (and When It Breaks)

This formula is highly accurate for steady-state DC systems (like 12V/24V/48V LiFePO4 solar banks) and single-phase AC resistive loads (like baseboard heaters or incandescent lighting).

Core Assumptions

• Power Factor (PF) is 1.0: If calculating for inductive AC loads (motors, compressors), you must divide the Power (P) by the PF before plugging it into the formula, or use the full AC impedance formula incorporating reactance (X).
• Temperature is 75°C: The K constant of 12.9 assumes copper wire operating at 75°C. If your wire is in a freezing environment, resistance drops; if it's bundled tightly in a hot attic, resistance spikes.

Unit Mistakes That Break the Math

• Using AWG instead of CM: You cannot plug "12" (for 12 AWG) into the CM variable. You must look up the exact Circular Mils (12 AWG = 6,530 cmil).
• Mixing Metric and Imperial: The K constant (12.9) is strictly for feet and circular mils. If you use meters and mm², K changes to roughly 0.0172 for copper, and the formula structure shifts entirely.
• Kilowatts vs Watts: Plugging in 1.5 kW instead of 1500 W will undersize your wire by a factor of 1000.

Realistic Answer Magnitudes

A realistic %VD target is ≤ 3% for branch circuits and ≤ 1% for critical DC battery-to-inverter runs. If your calculation yields a 12% drop, your assumed voltage is too low for the distance, or your wire is drastically undersized.

Rearranged Forms for Every Variable

On the bench, you rarely solve for %VD. You usually know your target drop and need to find the wire size or maximum distance. Here are the algebraically rearranged forms:

• Solve for Wire Size (CM):
  CM = (2 × K × P × L) / (%VDtarget × Vassumed2) × 100
• Solve for Max Distance (L):
  L = (%VDtarget × CM × Vassumed2) / (2 × K × P × 100)
• Solve for Max Load Power (P):
  P = (%VDtarget × CM × Vassumed2) / (2 × K × L × 100)

Worked Example 1: 120V Branch Circuit Sizing

Scenario: You are wiring a dedicated 120V outlet for a 1500W space heater in a garage. The one-way wire run from the subpanel is 80 feet. You are using copper wire and want to keep the voltage drop under 3%.

1. Identify Knowns:
   P = 1500 W
L = 80 ft
Vassumed = 120 V
K = 12.9 (Copper)
   %VDtarget = 3
2. Select the Rearranged Formula (Solving for CM):
   CM = (2 × 12.9 × 1500 × 80) / (3 × 1202) × 100
3. Calculate the Numerator:
   2 × 12.9 × 1500 × 80 = 3,096,000
4. Calculate the Denominator:
   3 × 14,400 = 43,200
5. Divide and Multiply by 100:
   (3,096,000 / 43,200) × 100 = 71.66 × 100 = 7,166 cmil
6. Match to AWG (NEC Chapter 9, Table 8):
   14 AWG = 4,110 cmil (Fail)
   12 AWG = 6,530 cmil (Fail - 6530 is less than 7166)
   10 AWG = 10,380 cmil (Pass)

Result: While 14 AWG might handle the 12.5A ampacity on paper, the assumed voltage formula dictates 10 AWG copper to prevent the heater from starving and drawing excess current to compensate for the drop.

Worked Example 2: 48V DC Solar Array String

Scenario: You are wiring a 2400W pure sine wave inverter to a 48V LiFePO4 battery bank. The run is 15 feet. Because DC inverters are highly sensitive to low voltage (which triggers low-voltage disconnects), your target drop is a strict 1%.

1. Identify Knowns:
   P = 2400 W
L = 15 ft
Vassumed = 48 V
K = 12.9 (Copper)
   %VDtarget = 1
2. Apply Formula:
   CM = (2 × 12.9 × 2400 × 15) / (1 × 482) × 100
3. Calculate Numerator:
   2 × 12.9 × 2400 × 15 = 928,800
4. Calculate Denominator:
   1 × 2,304 = 2,304
5. Divide and Multiply by 100:
   (928,800 / 2,304) × 100 = 403.125 × 100 = 40,312 cmil
6. Match to AWG:
   6 AWG = 26,240 cmil (Fail)
   4 AWG = 41,740 cmil (Pass)

Result: You need a minimum of 4 AWG copper. Notice how dropping the assumed voltage from 120V to 48V (a factor of 2.5) caused the required CM to skyrocket due to the inverse-square relationship in the denominator. For a deeper look at DC-specific derating, reference Solar-Electric's wire sizing guidelines.

Decision Tree: Picking Your Wire and Breaker

Use this decision path to finalize your physical components after running the assumed voltage formula.