Question

A 4160-V, three-phase, Y-connected source supplies power to two inductive loads connected in parallel via a three-wire transmission line. Load A is Y-connected and has an impedance of 30 + j40 R/phase. Load B is Aconnected and has an impedance of 150 + j360 R/phase. For each load, determine (a) the current, power, and power factor, (b) the total power supplied by the source, and (c) the power factor of the combined load. A Y connected synchronous motor that takes 30 kW at 0.2 pf leading is connected in parallel with the source. What is the current intake of the motor? What is the overall power factor?

Answer 1

For load A, the current is 76.70 + j102.43 A, the power factor is 0.8, and the power is 5888 + j7851 W. For load B, the current is 1.4053 + j3.3812 A, the power factor is 0.248, and the power is 522.31 + j1260.33 W. The total power supplied by the source is 6410.31 + j9111.33 W, and the power factor of the combined load is 0.8006. The current intake of the motor is 0.036 A, and the overall power factor is 15.06 + j21.42.

Step-by-step explanation:

For load A:

1. Calculate the current by using Ohm's Law: I = V/Z = 4160 / (30 + j40) = (4160 / sqrt(30^2 + 40^2)) * cos(arctan(40/30)) + j(4160 / sqrt(30^2 + 40^2)) * sin(arctan(40/30)) = 108.41 / sqrt(2) + j144.55 / sqrt(2) = 76.70 + j102.43 A
2. Calculate the power by using the formula P = I^2 * R = (76.70 + j102.43)^2 * 30 = 5888 + j7851 W
3. Calculate the power factor by using the formula PF = cos(arctan(40/30)) = 0.8

For load B:

1. Calculate the current by using Ohm's Law: I = V/Z = 4160 / (150 + j360) = (4160 / sqrt(150^2 + 360^2)) * cos(arctan(360/150)) + j(4160 / sqrt(150^2 + 360^2)) * sin(arctan(360/150)) = 8.229 / sqrt(34) + j19.832 / sqrt(34) = 1.4053 + j3.3812 A
2. Calculate the power by using the formula P = I^2 * R = (1.4053 + j3.3812)^2 * 150 = 522.31 + j1260.33 W
3. Calculate the power factor by using the formula PF = cos(arctan(360/150)) = 0.248

(a) To find the total power supplied by the source, add the powers of load A and load B: P_total = 5888 + j7851 + 522.31 + j1260.33 = 6410.31 + j9111.33 W

(b) To find the power factor of the combined load, calculate the apparent power S and divide it by the total power P: S = V * I = 4160 * (76.70 + j102.43 + 1.4053 + j3.3812) = 425.5 kW, P = V * I * PF = 425.5 * 0.8 = 340.4 kW, power factor = P / S = 340.4 / 425.5 = 0.8006

(c) To find the current intake of the motor, we need to calculate the apparent power S and divide it by the voltage V and power factor PF: S_motor = P_motor / PF_motor = 30 kW / 0.2 = 150 kVA, I_motor = S_motor / V = 150 / 4160 = 0.036 A

To find the overall power factor, we need to calculate the total power P_total and divide it by the apparent power S: power factor = P_total / S = 6410.31 + j9111.33 / 425.5 = 15.06 + j21.42