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?

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?

• A. 40.25 ft
• B. 51.06 ft
• C. 65.75 ft
• D. 75.50 ft

Answer: (B)

To determine the maximum length of a single-phase, 110-volt branch circuit supplying a 40-ampere load with 6 AWG aluminum conductors without exceeding a voltage drop of 3.3 volts, we can apply the voltage drop formula: Voltage Drop (VD) = (2 * K * I * L) / 1000 In this formula: - K is the resistance factor for aluminum conductors, which is approximately 12.9 ohms per thousand feet. - I is the load current in amperes (in this case, 40 A). - L is the one-way length of the conductor in feet. We need to rearrange this formula to find L: L = (VD * 1000) / (2 * K * I) Substituting the values: - VD = 3.3 volts - K = 12.9 ohms/1000ft - I = 40 A L = (3.3 * 1000) / (2 * 12.9 * 40) Calculating this gives: L = 3300 / (2 * 12.9 * 40) L = 3300 / 1032

To determine the maximum length of a single-phase, 110-volt branch circuit supplying a 40-ampere load with 6 AWG aluminum conductors without exceeding a voltage drop of 3.3 volts, we can apply the voltage drop formula:

Voltage Drop (VD) = (2 * K * I * L) / 1000

In this formula:

• K is the resistance factor for aluminum conductors, which is approximately 12.9 ohms per thousand feet.

• I is the load current in amperes (in this case, 40 A).

• L is the one-way length of the conductor in feet.

We need to rearrange this formula to find L:

L = (VD * 1000) / (2 * K * I)

Substituting the values:

• VD = 3.3 volts

• K = 12.9 ohms/1000ft

• I = 40 A

L = (3.3 * 1000) / (2 * 12.9 * 40)

Calculating this gives:

L = 3300 / (2 * 12.9 * 40)

L = 3300 / 1032