CSLB C-10 Electrical ContractorWiring Methods and MaterialsHard
A 208V three-phase feeder supplies a continuous 50A load located 200 feet from the panel, wired with 8 AWG copper conductors (16,510 circular mils). Using the formula VD = (√3 × K × I × D) / CM, where K = 12.9 ohms·cmil/ft for copper, what is the approximate voltage drop, and does it comply with the recommended 3% (6.24V) limit?
- A13.5V, exceeds the limit
- B6.2V, complies
- C9.8V, exceeds the limit
- D3.1V, complies
Show answer & explanationAnswer & explanation
Correct answer: A. 13.5V, exceeds the limit
VD = (1.732 × 12.9 × 50 × 200) / 16,510 ≈ (1.732 × 12.9 = 22.34; 22.34 × 50 = 1,117; 1,117 × 200 = 223,400; 223,400 / 16,510 ≈ 13.5V). Since 13.5V far exceeds the recommended 3% limit of 6.24V (208V × 0.03), the 8 AWG conductors are undersized for this run and a larger conductor is needed.
Why the other options are wrong
- B. 6.2V would comply, but this is not the calculated result for 8 AWG.
- C. 9.8V understates the actual voltage drop for this run.
- D. 3.1V is far too low for this length and load with 8 AWG conductors.
Three-Phase Voltage Drop Calculation
Three-phase voltage drop uses the formula VD = (√3 × K × I × D) / CM, incorporating the square root of 3 factor unique to three-phase circuits.
- √3 (1.732) factor distinguishes 3-phase from single-phase VD formulas
- K = 12.9 for copper, 21.2 for aluminum (approximate constants)
- Recommended max voltage drop is typically 3% for branch circuits/feeders combined 5%
Memory trick: Three phases need the root-three factor before you find the drop.