Final answer:
To find the total complex power, power factor, and supply current of the three loads connected in parallel, calculate the individual values for each load and add them up. For an inductive load, use the formula S = √(P² + Q²) to find the apparent power. For a capacitive load, use S = P / power factor. For a resistive load, the apparent power is equal to the real power. The total complex power is the sum of the individual apparent powers and the power factor is Ptotal / Stotal. The supply current is Stotal / the supply voltage. Plugging in the values, we get = 82.7 A.
Step-by-step explanation:
To find the total complex power, power factor, and supply current of the three loads connected in parallel, we can calculate the individual values for each load and then add them up. For Load 1, since it is an inductive load, we can calculate the apparent power using the formula S = √(P² + Q²) where P = 60 kW and Q = 660 kvar. Plugging in the values, we get S1 = √((60,000)² + (660,000)²) = 670,677 VA.
For Load 2, the apparent power can be calculated using the formula S = P / power factor where P = 240 kW and the power factor is given as 0.8. Plugging in the values, we get S2 = 240,000 / 0.8 = 300,000 VA.
For Load 3, the apparent power is equal to the real power since it is a resistive load. So S3 = 60 kW = 60,000 VA.
The total apparent power is the sum of the individual apparent powers, so Stotal = S1 + S2 + S3 = 670,677 + 300,000 + 60,000 = 1,030,677 VA.
The total power factor can be calculated using the formula power factor = Ptotal / Stotal where Ptotal is the sum of the real powers of the individual loads. In this case, since Load 3 is the only resistive load, Ptotal = 60 kW. Plugging in the values, we get power factor = 60,000 / 1,030,677 ≈ 0.058.
The supply current can be calculated using the formula I = Stotal / V where V is the supply voltage. In this case, the supply voltage is 12.47 kV = 12,470 V. Plugging in the values, we get I = 1,030,677 / 12,470 ≈ 82.7 A.