Calculators › Voltage Drop
Voltage Drop & Ohm's Law
The exam expects basic electrical theory the NEC assumes you know. Ohm's law and voltage drop come up regularly — here are the formulas and a worked example.
Ohm's law & power
V = I × R (volts = amps × ohms)
P = V × I (watts = volts × amps)
P = I² × R (watts = amps² × ohms)
From any two values you can find the rest. Example: a 240 V heater drawing 20 A → P = 240 × 20 = 4,800 W; R = 240 ÷ 20 = 12 Ω.
Voltage drop formula
Using the "K" (resistivity) method, with K ≈ 12.9 for copper and CM = circular mils of the conductor (from NEC Chapter 9, Table 8):
Single-phase: VD = (2 × K × I × L) ÷ CM
Three-phase: VD = (1.732 × K × I × L) ÷ CM
L = one-way length in feet
Example — 120 V branch circuit, long run
A 20 A load, 100 ft one-way, on 12 AWG copper (CM = 6,530). Is the voltage drop acceptable?
- Apply the single-phase formula:
VD = (2 × 12.9 × 20 × 100) ÷ 6,530
= 51,600 ÷ 6,530 = 7.9 V
- As a percentage of 120 V:
7.9 V ÷ 120 V = 6.6%
- Compare to the recommendation: the NEC suggests ≤ 3% on a branch circuit (Informational Note to 210.19). 6.6% is too high.
- Fix it: upsize the conductor. 8 AWG (CM = 16,510) gives:
VD = (2 × 12.9 × 20 × 100) ÷ 16,510 = 3.1 V (2.6%) → OK
12 AWG = 6.6% (too high); 8 AWG = 2.6% (acceptable)
Exam note: voltage-drop limits (3% branch, 5% total) are recommendations in NEC Informational Notes, not mandatory rules — a common true/false trap. But you're still expected to be able to calculate it.