Question

What voltage drop for a 208V, single phase circuit with a load of 28A,using No.8 stranded copper conductors at a circuit length of 160’ is —————.

Answer 1

The voltage drop in the circuit is 0.07196V.

Step-by-step explanation:

To find the voltage drop in a circuit, we can use Ohm's Law which states that voltage (V) equals current (I) multiplied by resistance (R). In this case, the resistance of the circuit can be calculated using the formula:

R = (resistivity x length) / cross-sectional area

Given that the circuit length is 160 feet and the conductor is No.8 stranded copper, we can determine the cross-sectional area using a wire gauge chart. The cross-sectional area for No.8 wire is approximately 0.0526 square inches. The resistivity of copper at room temperature is about 0.0000017 Ωm.

Substituting these values into the formula, we get:

R = (0.0000017 Ωm x (160 ft x 0.3048 m/ft)) / (0.0526 in2 x 0.0000254 m2/in2) = 0.00257 Ω

Then, we can calculate the voltage drop using Ohm's Law:

V = I x R = 28A x 0.00257 Ω = 0.07196V

Therefore, the voltage drop in the circuit is approximately 0.07196V.