What is the maximum length of a single-phase, 110-volt branch circuit supplying a 40-ampere load with 6 AWG aluminum conductors permitted without exceeding a voltage drop of 3.3 volts?

To determine the maximum length of a single-phase, 110-volt branch circuit supplying a 40-ampere load with 6 AWG aluminum conductors, we need to consider the allowable voltage drop. The permissible voltage drop for this circuit is 3.3 volts, which is 3% of the given voltage (110 volts).

Voltage drop in a circuit occurs due to the resistance of the conductors, and it can be calculated with the formula:

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

Where:

• ( L ) is the one-way length of the circuit in feet,

• ( I ) is the current in amperes,

• ( R ) is the resistance of the wire per 1000 feet.

For 6 AWG aluminum conductors, the resistance is approximately 0.393 ohms per 1000 feet. Plugging the values into the formula, we can rearrange it to solve for ( L ):

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

Substituting the voltage drop (3.3 volts),