What is the maximum current for a single-phase, 230-volt branch circuit supplying a load at a distance of 140 feet with 4 AWG copper conductors and holding the voltage drop to 4.6 volts?

What is the maximum current for a single-phase, 230-volt branch circuit supplying a load at a distance of 140 feet with 4 AWG copper conductors and holding the voltage drop to 4.6 volts?

• A. 50.00 A
• B. 53.16 A
• C. 56.75 A
• D. 60.00 A

Answer: (B)

To determine the maximum current for a single-phase, 230-volt branch circuit with specific parameters, it's essential to use the voltage drop formula while taking into account the length of the circuit and the specifications of the conductors. The relevant voltage drop formula for a single-phase circuit is: \[ \text{Voltage Drop (VD)} = \frac{2 \times I \times L \times R}{1000} \] Where: - \( I \) is the current in amperes - \( L \) is the length of the conductor in feet (total length for the round trip) - \( R \) is the resistance of the conductor per 1000 feet For 4 AWG copper conductors, the resistance \( R \) is approximately 0.2485 ohms per 1000 feet. Given the load is at a distance of 140 feet, we need to consider the total distance, which is 280 feet (140 feet out and 140 feet back). The acceptable voltage drop is specified as 4.6 volts. Now, rearranging the voltage drop formula to solve for current \( I \): \[ I = \frac{VD \times 1000}{2 \times L \

To determine the maximum current for a single-phase, 230-volt branch circuit with specific parameters, it's essential to use the voltage drop formula while taking into account the length of the circuit and the specifications of the conductors.

The relevant voltage drop formula for a single-phase circuit is:

[ \text{Voltage Drop (VD)} = \frac{2 \times I \times L \times R}{1000} ]

Where:

• ( I ) is the current in amperes

• ( L ) is the length of the conductor in feet (total length for the round trip)

• ( R ) is the resistance of the conductor per 1000 feet

For 4 AWG copper conductors, the resistance ( R ) is approximately 0.2485 ohms per 1000 feet.

Given the load is at a distance of 140 feet, we need to consider the total distance, which is 280 feet (140 feet out and 140 feet back). The acceptable voltage drop is specified as 4.6 volts.

Now, rearranging the voltage drop formula to solve for current ( I ):

[ I = \frac{VD \times 1000}{2 \times L \