S2E6: Modeling

Joe has a barn that is feet from his house. He needs to supply 1000 Watts at 240V to a resistive load at his barn from the 60Hz power line at his house. Note that the circuit from the house to the barn requires two lengths of the interconnecting wire. He proposes to use number AWG wire to connect his house to his barn. Number AWG copper wire has a resistance of per 1000 feet.

What is the total resistance (in Ohms) of the transmission line?

(190.0 / 1000) * 1.588 = 0.30 Ohms

What is the resistance (in Ohms) of Joe's load at his barn?

V=IR

240 = I * R

P=IV

1000=I*240

(1000/240) = I

I = (1000/240) = 4.17

240 = (1000/240) * R

240 / (1000/240) = R

R = 240 / (1000/240) = 240 / 4.17 = 57.6 Ohms

What is the voltage drop (in Volts) from the house to the load at the barn due to the resistance in Joe's transmission line?

V=IR

V = (4.17)*(0.30) = 1.25 Volts

1.25 Volts * 2 lengths of wire = 2.5 Volts