For a feeder supplying one 5 hp and three 2 hp motors, what is the minimum size conductor allowed?

For a feeder supplying one 5 hp and three 2 hp motors, what is the minimum size conductor allowed?

• A. 4 AWG
• B. 6 AWG
• C. 8 AWG
• D. 2 AWG

Answer: (C)

To determine the minimum size conductor needed for a feeder supplying one 5 hp and three 2 hp motors, it's essential to calculate the total load in amperes and then refer to the appropriate NEC (National Electrical Code) tables to select the correct conductor size. First, convert the horsepower to kilowatts using the formula: - 1 hp ≈ 0.746 kW. Thus: - The total load for the motors can be calculated: - One 5 hp motor: 5 hp × 0.746 kW/hp = 3.73 kW. - Three 2 hp motors: 3 × (2 hp × 0.746 kW/hp) = 3 × 1.492 kW = 4.476 kW. Now, add the power of all motors: - 3.73 kW + 4.476 kW = 8.206 kW. Next, to convert the total power (in kW) to amperes, use the following formula assuming a 480 V system (for example): - Amperes = Watts / Volts. - Total watts = 8.206 kW × 1000 = 8206

To determine the minimum size conductor needed for a feeder supplying one 5 hp and three 2 hp motors, it's essential to calculate the total load in amperes and then refer to the appropriate NEC (National Electrical Code) tables to select the correct conductor size.

First, convert the horsepower to kilowatts using the formula:

• 1 hp ≈ 0.746 kW.

Thus:

• The total load for the motors can be calculated:

• One 5 hp motor: 5 hp × 0.746 kW/hp = 3.73 kW.

• Three 2 hp motors: 3 × (2 hp × 0.746 kW/hp) = 3 × 1.492 kW = 4.476 kW.

Now, add the power of all motors:

• 3.73 kW + 4.476 kW = 8.206 kW.

Next, to convert the total power (in kW) to amperes, use the following formula assuming a 480 V system (for example):

• Amperes = Watts / Volts.

• Total watts = 8.206 kW × 1000 = 8206