A client requests a dedicated 20A, 120V circuit for a new medical imaging device. The device manufacturer specifies a maximum voltage drop of 3% at the farthest receptacle, which is 150 feet from the panel. Assuming the circuit uses copper conductors, which AWG conductor size is the minimum required to meet this voltage drop requirement? (Assume a conductor temperature of 75°C and ignore any voltage drop in the service entrance conductors.)
- A12 AWG
- B8 AWG
- C10 AWG
- D6 AWG
Show answer & explanationAnswer & explanation
Correct answer: B. 8 AWG
Maximum allowable voltage drop = 3% of 120V = 3.6V. Using the approximate voltage drop formula for single-phase circuits: VD = (2 × K × I × L) / CM, where K for copper is 12.9 (at 75°C), I is 20A, L is 150ft. Rearranging for Circular Mils (CM): CM = (2 × K × I × L) / VD. So, CM = (2 × 12.9 × 20A × 150ft) / 3.6V = 25,800 CM. Comparing this to AWG conductor tables: 10 AWG ≈ 10,380 CM, 8 AWG ≈ 16,510 CM, 6 AWG ≈ 26,240 CM. Therefore, 6 AWG is the minimum size to meet the voltage drop, as 8 AWG is too small.
Why the other options are wrong
- A. 12 AWG (6,530 CM) is too small and would result in a much higher voltage drop.
- C. 10 AWG (10,380 CM) is too small for this distance and current, resulting in a voltage drop exceeding 3%.
- D. 6 AWG (26,240 CM) provides sufficient circular mils to keep the voltage drop below 3% (CM calculated as 25,800).
Voltage Drop Calculation
Voltage drop is the reduction in electrical potential along the length of a conductor due to its resistance, calculated to ensure proper appliance operation.
- VD = (2 × K × I × L) / CM for single-phase circuits (K is conductor constant).
- Maximum allowable voltage drop is typically 3% for branch circuits and 5% for feeder + branch circuits combined.
- Larger conductor sizes (smaller AWG number) reduce voltage drop.
Memory trick: VD: Very Distant wires need Very Durable (Large) conductors.