Final answer:
To raise the power factor of a load connected to a 120Vrms, 60 Hz power line from 0.8 to 0.95 lagging, a capacitor with a value of 312 microfarads should be placed in parallel with the load.
Step-by-step explanation:
When a load is connected to a power source and absorbs power at a certain power factor, the value of the capacitor needed to adjust the power factor can be calculated using the concepts of reactive power and power factor correction. In the given scenario, a load is connected to a 120Vrms, 60 Hz power line, absorbing 4 kW at a power factor of 0.8 lagging. To raise the power factor to 0.95 lagging, we need to determine the required capacitive reactance that compensates for the excess inductive reactive power.
First, calculate the current initial reactive power (Qi):
Qi = P(tan(cos-1(power factori)))
Qi = 4000 W * tan(cos-1(0.8)) = 3000 VAR (since tan(cos-1(0.8)) = 0.75)
Next, calculate the final reactive power (Qf) needed for a power factor of 0.95:
Qf = P(tan(cos-1(power factorf)))
Qf = 4000 W * tan(cos-1(0.95)) = 1310.17 VAR
The change in reactive power (ΔQ) needed from the capacitor is:
ΔQ = Qi - Qf
ΔQ = 3000 VAR - 1310.17 VAR = 1689.83 VAR
The capacitive reactance (XC) is given by:
XC = V2 / ΔQ
XC = (1202) V2 / 1689.83 VAR = 8.511 Ω
Finally, calculate the capacitance (C):
C = 1 / (2πfXC)
C = 1 / (2π*60*8.511) = 312 μF
The value of the capacitor required to raise the power factor to 0.95 lagging is 312 microfarads.