A 240-volt, 3Ø feeder supplies power to four motors. Motor 1 is 50 hp, motor 2 is 40 hp, motor 3 is 30 hp, and motor 4 is 10 hp. The minimum size feeder conductors permitted are ___.

• A. 500 kcmil
• B. 4/0 AWG
• C. 250 kcmil
• D. 1/0 AWG

Answer: (A)

Feeder sizing for multiple motors is based on the sum of each motor’s full-load current, with a 125% allowance to cover simultaneous operation and starting surges. For a 240 V three-phase feeder feeding four motors, estimate each motor’s FLC and add them together, then multiply by 1.25.

Using typical FLC estimates for 240 V:

• 50 hp ≈ 120 A

• 40 hp ≈ 95 A

• 30 hp ≈ 70 A

• 10 hp ≈ 25 A

Total ≈ 310 A. Applying 125% gives about 388 A.

Now compare with the ampacity of the candidate conductors. The smaller options (1/0 AWG, 250 kcmil, 4/0 AWG) have ampacities well below 388 A at 75°C, so they wouldn’t safely carry the load. The 500 kcmil size has an ampacity that comfortably exceeds 388 A, making it the smallest conductor in the options that meets the calculated requirement.

Thus, the feeder conductors must be 500 kcmil to satisfy the ampacity needed for all four motors.