Calculating Voltage Drop in 2 Wire AC Circuit

What is the voltage drop in a 2 wire AC circuit using #6 uncoated wire with a load of 5,050 VA at 230 volts and a one way length of 450 ft in IMC?

1. 5.5 volts

2. 10.25 volts

3. 11.45 volts

4. 16.60 volts

Final Answer:

The calculated voltage drop for a 2 wire AC circuit using #6 uncoated wire, assuming a one way length of 450 ft, is approximately 7.81 V.

Explanation:

Finding the Voltage Drop in an AC Circuit

To calculate the voltage drop in a 2 wire AC circuit with #6 uncoated wire over a one way length of 450 ft, you need to determine the current, the resistance of the wire, and then use Ohm's law. The power (VA) and voltage information is provided which allows us to find the current using the formula P = VI, where P is the power in volt-amperes (VA), V is the voltage, and I is the current in amperes (A).

Calculating the current (I):

I = P / V = 5,050 VA / 230 V = 21.96 A (approximately)

Determining the resistance of the #6 uncoated wire over 450 ft:

The American Wire Gauge (AWG) standard states that a #6 wire has a resistance of approximately 0.3951 ohms per 1000 feet. So for 450 ft, the resistance (R) would be 0.3951 ohms x (450/1000) = 0.177795 ohms.

Applying Ohm's law (V = IR) to find the voltage drop (V_drop):

V_drop = I x R = 21.96 A x 0.177795 ohms = 3.905 V approximately for one way. Since it is a 2 wire circuit, the total voltage drop must be multiplied by 2.

Total voltage drop: 3.905 V x 2 = 7.81 V approximately.

Therefore, out of the given options, none exactly match our calculated voltage drop. However, option 1 stating a 5.5 volts drop is the closest to our calculated value, suggesting that there might be additional factors or approximations in the given problem that we are not aware of, or a possible error in the provided options.