Question

Two balanced Y-connected loads in parallel, one drawing 15 kW at 0.6 power factor lagging and the other drawing 10kVA at C. power factor leading, are supplied by a balanced, three-phase, 480V source

a. The power factor of the combined load is 0.5 leading
b. The power factor of the combined load is 0.5 lagging
c. The power factor of the combined load is C. 854 leading
d. The power factor of the combined load is 6.854 lagging
e. The magnitude of the line current from the e is 3239A
f. The magnitude of the line current from the source is 42.24A
g. A delta connected capacitor is installed in parallel with the combined load. The value of the capacitive reactance needed in each leg of the delta to make the source
power factor equal to one jurit is: 49.37 ohms
h. A delta connected capacitor is installed in parallel with the combined load. The value of the capacitive reactance needed in each leg of the delta to make the source
power factor equal to one (unity) is 26.39 ohms
I. The magnitude of the current in each capacitor (capacitor added previously) is 5:72A
j. The magnitude of the current in each capacitor (capacitor added previously) is 4.12A
k. After addition of the reactance, the magnitude of the source line current 34.65A
l. After addition of the reactance, the magnitude of the source line current 27,664

Answer 1

To raise the power factor of the circuit, an inductor should be placed in series with the elements and the value of the reactance across the inductor can be calculated using a formula.

Step-by-step explanation:

In an AC circuit, the power factor is the amount by which the power delivered in the circuit is less than the theoretical maximum due to the voltage and current being out of phase. A power factor less than 1 indicates that the circuit has reactive power, which can be corrected by adding a reactive component.

In this case, the circuit has a power factor of 0.5 leading, meaning it has capacitive reactance. To raise the power factor to unity (1), an inductor should be placed in series with the elements. The value of the reactance across the inductor can be found using the formula Xl = 1 / (2πfC), where f is the frequency and C is the capacitance.