For a 19.8-amp, 3Ø, 208-volt load 152 feet from the panel, what is the smallest circuit conductor you can use to limit voltage drop to 1.8%?

To determine the smallest circuit conductor that can be used to limit the voltage drop to 1.8% for a load of 19.8 amps at 208 volts over a distance of 152 feet, it is essential to calculate the allowable voltage drop.

1. Calculate the Maximum Allowable Voltage Drop:

The percentage voltage drop is calculated based on the total voltage:

[

\text{Maximum Voltage Drop} = 208 \text{ volts} \times 0.018 = 3.744 \text{ volts}

]

2. Calculate the Voltage Drop per Conductor:

Since the load is 3-phase, we need to account for the circuit configuration when calculating the voltage drop. The formula for voltage drop in three-phase systems is:

[

V_d = \frac{1.732 \times I \times L \times R}{1000}

]

Where ( V_d ) is the voltage drop in volts, ( I ) is the current in amps, ( L ) is the one-way length in feet, and ( R ) is the resistance of the conductor in ohms per thousand feet.