A client plans to install a new central air conditioning unit requiring a 240V, 30A dedicated circuit. The unit is located 60 feet from the service panel. The contractor estimates the total voltage drop should not exceed 3%. What is the minimum AWG conductor size required, assuming copper conductors (K=12.9) and a 75°C termination rating?

Correct answer: A. 10 AWG

1. Maximum allowable voltage drop: 3% of 240V = 7.2V. 2. Formula for CM: CM = (2 * K * I * L) / VD. 3. K = 12.9 (copper, 75°C), I = 30A, L = 60 ft, VD = 7.2V. 4. CM = (2 * 12.9 * 30 * 60) / 7.2 = 46440 / 7.2 = 6450 CM. 5. From CEC Chapter 9, Table 8 (Conductor Properties): - 10 AWG has 10,380 CM. - 8 AWG has 16,510 CM. - 6 AWG has 26,240 CM. Since 10 AWG (10,380 CM) is greater than the required 6450 CM, it satisfies the voltage drop requirement. 6. For a 30A circuit, 10 AWG copper conductor is rated for 30A at 75°C (CEC 310.16). While AC units have motor loads, the question implies a straight 30A circuit. If the AC unit is a motor load, CEC 430.22(A) would require sizing the branch-circuit conductors at 125% of the motor's full-load current. If the 30A is already sized for the 125% factor, then 10 AWG is sufficient. Assuming the 30A is the final calculated current after all adjustments (or the question is simplified to focus on voltage drop for a 30A load), 10 AWG is the correct answer. If 30A is the motor FLC, then 30A * 1.25 = 37.5A, requiring a 40A OCPD and 8 AWG conductor. However, given the options and common exam simplifications, 30A is often taken as the *circuit* current for the voltage drop calculation. For a 30A circuit, 10 AWG is the standard wire size and satisfies the voltage drop calculation.